Chemistry of Solutions Study Huide for Test 1
Chemistry of Solutions Study Huide for Test 1 202-NYB-05
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CHEMISTRY OF SOLUTIONS STUDY GUIDE FOR TEST 1 LECTURES 1 THROUGH 6 LECTURE 1 REVIEW Topics Covered: Solution composition: a short review of significant figures (students are graded for their use of significant figures, further abbreviated as “sig figs”); review of mass percent, the mole fraction, molarity, molality and parts per million (ppm). How to Count Sig Figs Do your best to memorize these as marks will be deducted if you use the incorrect number of sig figs in your calculations! Scenario no. 1: when a number doesn’t contain a decimal or any zeros, all of the digits count as sig figs. For example, 1234 has 4 sig figs. Scenario no. 2: when zeroes precede nonzero numbers, only the nonzero digits count as sig figs. For example, 0.01 has 1 sig fig. Scenario no. 3: zeroes that are sandwiched between nonzero digits always count as sig figs. For example, 20.04 has 4 sig figs. Scenario no. 4: zeroes that follow nonzero digits are only significant if there is a decimal in that number. For example, 1.200 has 4 sig figs. Scenario no. 5: exact numbers (e.g.: data obtained from a laboratory experiment) are 2 counted, not measured. These numbers have an infinite number of sig figs. For example, there are exactly 2.54 cm in one inch. Using Sig Figs in Calculations There are certain rules to follow when counting sig figs while doing calculations. • If you’re multiplying or dividing, your final answer should have the same amount of digits as the number with the least number of sig figs. For example, 1.202 (4 sig figs) ÷ 2.1 (2 sig figs) = 0.572381 (7 sig figs), final answer: 0.6 (2 sig figs). • If you're adding or subtracting, your final answer should have the same amount of decimal spaces as the value with the least amount of decimal spaces. For example, 0.12 (2 decimal spaces) + 1.123 (3 decimal spaces) = 1.243 (3 decimal spaces), final answer: 1.24 (2 decimal spaces) When we discuss solutions,consider solutes and solvents. This course mainly deals with liquid solutions. o A solution is comprised of solutes and solvents; o The solute in a solution is the substance that is being dissolved; o The solvent in a solution is what is doing the dissolving. Water is the universal solvent. 3 Units of Concentration 1) Fractional Composition (sum of all the individual parts of a solution equals the whole solution). a) Calculating Mass Percent: To calculate the mass percent of a solution, divide the given mass of the solute by the mass of the whole solution, then multiply this value by 100. Mass % = ( Mass of Solute ÷ Mass of Solution) × 100 Note: the sum of all the individual solutes divided by the mass of the solution should equal to 1. For instance, (Mass of Solute A/Mass of Solution) + (Mass of Solute B/Mass of Solution) + (Mass of Solute C/Mass of Solution) = 1. The mass percent gives us the mass of a solute per total mass of solution. b) Calculating Mole Fraction: To calculate the mole fraction, find the number of moles of the solute, and divide this number by the total number of moles in the solution. Calculate the total number of moles of solution by adding up all the moles of the individual solutes in the solution. Mole Fraction = ( Moles of Solute A ) ÷ ( Moles of Solute A + Moles of Solute B + Moles of Solute C) Note: the sum of all the moles of solutes in the solution should equal to 1, 4 which means Moles of Solute A + Moles of Solute B + Moles of Solute C = 1. 2) Other Units of Concentration c) Molarity: Molarity is defined by moles of solute per litre of solution. This unit of concentration is temperature-dependent because it is based on the volume of a solution. This value represents the volume of solute + the volume of solvent. Molarity = M = Mole/Litre d) Molality: Molality is defined by moles of solute per kilogram of solvent. This unit of concentration is not temperature-dependent because mass is unaffected by changes in temperature. Molality is denoted by a lowercase cursive “m”. Molality = m = Moles of solute/Kilograms of solution. e) Parts per million: Parts per million (ppm) is defined by the mass of a chemical per litre. Ppm and mg/L are interchangeable units of concentration; they mean the same thing. 5 Exercises From Lecture Slides & Their Solutions 1) A solution of sodium chloride was made by dissolving 7.00 g of NaCl in 50.0 mL of water. Calculate the mole fraction of sodium chloride, assuming that water has a density of 1.00 g/mL. Step 1: calculate the molar mass of NaCl: Molar Mass of NaCl = (22.99 g) + (35.45 g) = 58.44 g Step 2: calculate the number of moles of NaCl in 7.00 g: 1 mol. NaCl = 58.44 g X mol. NaCl = 7.00 g 7.00 g NaCl x 1 mol. NaCl ÷ 58.44 g NaCl = 0.1198 mol. NaCl. Step 3: calculate the molar mass of water. Molar Mass of H O = (2 x 1.008 g) + (15.99 g) = 18.006 g. 2 Step 4: calculate the number of moles of water in 50.0 mL using its density. (50.0 mL H 2 ) x (1.00 g H 2)/(mL) x (mol. H O2)/(18.006 g) = 2.777 mol. H O. 2 6 Step 5: calculate the mole fraction of sodium chloride: X = mol. of solute/ mol. of solution X = (0.1198 mol. NaCl)/ (2.777 mol. H2O + 0.1198 mol. NaCl) = 0.04136 mol. Final answer with correct number of sig figs: 0.0414 mol. (4 sig figs). 2) A solution of sodium chloride was made by dissolving 7.00 g (3 sig figs) of NaCl with 50.0 mL (3 sig figs) of water. Calculate the molality of sodium chloride, assuming that water has a density of 1.00 (3 sig figs) g/mL. Step 1: convert 50.0 mL of water to 50.0 g of water, then convert grams to kilograms. 50.0 mL H2O x (1.00 g/ mL) = 50.0 g of H2O 50.0 g/ 1000 mL = 0.05 kg. Step 2: Use the number of moles of NaCl in 7.00 g (0.11978 mol. NaCl in 7.00 g) and the mass of water in kilograms to calculate m. m = mol. of solute/ kg of solution m = (0.11978 mol.)/ (0.05 kg) = 2.3956 mol./ kg. 7 Final answer with the right number of sig figs: 2.40 mol./kg (3 sig figs). LECTURE 2 REVIEW Topics Covered: solution composition, review of intermolecular forces, energies of solution formation, effects of temperature and pressure on solubility. Solution Composition -- Continued • It’s important to note that mass percent, molarity, molality, and the mole fraction are values that are not dependent on temperature; a change in temperature will not alter the mass of a solution. • A change in temperature will change a solution’s volume. Therefore, molarity is temperature dependent. • Remember: Molarity = M = moles of solute / liters of solution Mass percent = ( mass of solute / mass of solution ) × 100 Mole Fraction = X = Aoles of A / total moles in solution Molality = m = moles of solute / kilograms of solvent • Normality: the normality of a solution is the number of equivalents in a given solution. For example, consider H SO : this compound contains 2 hydrogen atoms, so it can 2 4 + produce 2 protons (2H ). 1 mole of H SO th2n 4as 2 equivalents. This means that a solution of H SO with 3.75 M 2 4 will contain 2 × 3.75 = 7.50 equivalents per litre of solution, so the normality of this solution is 7.50 N. 8 • Note: equivalents mean different things in different reactions. For example: an equivalent for an acid-base reaction is defined by the amount of an acid or base that's required to give or lose one mole of protons, whereas in an oxidation-reduction reaction (or redox reaction) an equivalent is defined by the amount of reducing agent or oxidizing agent required to give or lose one mole of electrons. Review of Intermolecular Forces Types of Intermolecular Forces: 1. Dipole-Dipole Interactions: these types of interactions occur between molecules that have dipole moments, or polar molecules. A polar molecule is a molecule that has a partial positive charge and a partial negative charge. Molecules with these types of forces between them want to maximize attraction and minimize repulsion , meaning the partial positive charges will align themselves with the partial negatives, and vice versa. They are 1% as strong as ionic or covalent bonds, and become weaker as the dipoles between molecules get further apart from each other. These forces are mainly important when we analyze aqueous solutions. • Hydrogen bonds are subtype of these intermolecular forces. Strong dipole-dipole interactions occur between molecules when hydrogen is bound to fluorine, oxygen, or nitrogen. A good way to remember which atoms hydrogen forms strong bonds with is thinking “hydrogen bonds are fun,” where fun is spelled F-O-N for fluorine (F), oxygen (O), and nitrogen (N). Hydrogen bonds are very strong and require a lot of energy to break. 9 • Note: hydrogen bonds do not occur between hydrogen atoms! It may seem obvious, but, as an example, consider two Lewis structures for formaldehyde (CH O2 and water. The hydrogen atoms for both structures are on the outside (surrounding carbon in formaldehyde, and on either side of oxygen in water), but they don't bond with each other. A hydrogen bond is formed between the oxygen atom in formaldehyde and one of the hydrogen atoms in water. 2. London Dispersion Forces: these forces form between non-polar (NP) molecules, like noble gases. As electrons move around a nucleus, their distribution becomes asymmetrical which can cause a momentary dipole moment between two NP molecules. This dipole moment is very weak and short-lived, but can change the electron distribution of the atom next to it. These forces can also be called induced dipole interactions. These forces are most useful when examining large molecules; as the surface area of a molecules increases, it has more points of contact with other molecules, and this strengthens the bonds between them. Note: ionic compounds easily dissolve other ionic compounds. Remember, like dissolves like. Solution Formation • There are three steps in the process of solution formation: 1. The intermolecular forces in the solute are overcome, which causes the solute to expand. This step entails separating the solute into its individual components. 10 2. The intermolecular forces in the solvent are overcome, which causes the solvent to expand. This step entails separating the solvent into its individual components. 3. The solute and solvent mix. The first and second steps are usually endothermic (require energy) while the third step is usually exothermic (releases energy), but this mostly depends on the entropy of the solution you're dealing with. Energies of Solution Formation (Enthalpy) • Stronger intermolecular forces require more energy to break because the molecules are more difficult to separate. • When a solute and solvent are mixed, new intermolecular forces can occur between the solute and solvent. • Mixing can yield an exothermic or endothermic reaction, depending on how compatible the solute and solvent are . • The heat of a chemical reaction is denoted by ΔH, which is also called enthalpy change. A negative ΔH tells us that a reaction is exothermic (releases energy), while a positive ΔH tells us that a reaction is endothermic ( requires energy). • Enthalpy is the sum of energy that was absorbed or released during the three steps of solution formation. Below are two examples of enthalpy diagrams. The diagram on the left represents an exothermic reaction, while the one on the right demonstrates an endothermic reaction. 11 • Consider a solution made up of oil and water. Oil is NP, while water is polar. Since oil is NP, it is immiscible in water meaning that it can't form a solution with water. Oil molecules are held together by London dispersion forces which, despite being the weakest types of intermolecular forces, are very strong between oil molecules because of their size. Therefore, they require a lot of energy to break. Breaking these bonds then results in a strong endothermic reaction (ΔH > 0)1 The erosion of London dispersion forces between oil molecules demonstrates the first step of solution formation: expanding a solute. • The second step of this process of solution formation, expanding the solvent, requires a lot of energy because the hydrogen bonds in water are very strong (ΔH > 0). 2 • The enthalpy of the resulting solution is large and positive because the first and second reactions required a lot of energy. Therefore, ΔH + 1H = ΔH2 solution0. In this 12 case, the sum of a large and positive ΔH and Δ1 results2 in a large and positive ΔH solution Note: no solution actually forms when oil and water interact because of the great amount of energy that is required to produce such a solution. This combination isn't favourable. whenever any NP and polar substances interact, they aren't expected to form a solution because of the great amount of energy that would be required to mix them. • When sodium chloride, which is polar, is dissolved in water, which is also polar, the ΔH and 1H are bot2 large and positive, but ΔH is large and negative. ΔH is solution solution exothermic despite the first two steps being endothermic because of the strong reaction that occurs between the Na ions and the water molecules in the solution when the solute and solvent are mixed. Consider the chemical equations that coincide with this reaction: NaCl = Na + + Cl - , ΔH = 786 kJ/mol (s) (g) (g) 1 + - + - H2O (l) Na (g)+ Cl (g)= Na (aq) Cl (aq) ΔH hyd= -783 kJ/mol ΔH solution 786 kJ/mol – 783 kJ/mol = 3 kJ/mol o ΔH hydrepresents the change in enthalpy when gases occur in a solute with water in it. It combines the values of ΔH2 and ΔH3, so we can calculate ΔH solution with ΔH hyd • Even though the dissolution of NaCl (s)requires energy, a solution still forms because it is favourable for water to form a solution with salt (solvation increases entropy: solvation is another word for dissolution; molecules of solvent are assigned to molecules of solute during solvation). 13 Factors Affecting Solubility • There are three factors that determine whether or not a substance is soluble (miscible) or insoluble (immiscible): 1. The structure of a molecule determines its polarity, which in turn determines whether it is miscible or immiscible. 2. Pressure also affects whether or not a substance is soluble. For example, CO d2ssolves in water at high pressures. 3. For aqueous solutions, temperature impacts solubility. • We can tell whether a molecule is polar or not by the way it looks. The structural form of vitamin A tells us that is it non-polar, and is therefore immiscible in water (vitamin A is fat-soluble). The structural form of vitamin C tells us that it is polar, and is therefore water soluble (miscible). • Polar solutes are hydrophilic (water-loving). This explains why water molecules interact with themselves as easily as they do with other polar molecules. Polar solvents dissolve polar solutes. • NP molecules are hydrophobic (water-fearing), which means that interactions between two water molecules are stronger than interactions between a water molecule and a NP molecule. NP solvents dissolve NP solutes. • When choosing a good solvent, it's helpful to draw the Lewis structure of the substance you want to dissolve. Doing so will give you a better idea of its structure, which determines its polarity. • Soft drinks go flat if their cans (or bottles) are left open in a room-temperature environment because the increase 14 in temperature and decrease in pressure cause the CO 2n our drinks to dissolve faster. Without CO , t2e fizziness of a soft drink diminishes. • Temperature is an important factor to consider when solids, liquids, and gases are being dissolved , while pressure is important to consider specifically when gases are being dissolved. o The partial pressure of a gas determines the frequency of the collisions between gas molecules and a solution’s surface. o Pressure also determines the equilibrium concentration of a gas. The Effect of Pressure on Solubility of Gaseous Solutions 15 o The figure above tells us that when pressure was increased, the gas molecules collide more with the surface of the solution. This causes the rate at which the gas is being dissolved to increase, which in turn causes the concentration of the dissolved gas to be greater than it was initially. The system compensates by increasing the rate of escape of the gas molecules from the solution. • Henry’s law states that the concentration of a dissolved gas is directly proportional to the partial pressure of the gas above the solution. o Pressure is related to the number of gas molecules in a solution per litre of volume. o An increase in pressure causes molecules to enter a gaseous solution more quickly, while the rate at which gas molecules leave a solution remains unaffected. • The equation for Henry’s law is C = k P, Hhere C is the concentration (mol/L) of the dissolved gas, k is H the Henry’s law constant for a given gas (mol/L atm) and P is the partial pressure of the gas above the solution (atm). The table below provides the constants for a few gasses. 16 • Using Henry’s Law: A soft drink is bottled so that the bottle at 25°C contains CO g2s at a pressure of 5.0 atm over the liquid. Assuming the partial pressure of CO in2the -4 atmosphere is 4.0 x 10 atm, calculate the equilibrium concentration of CO in the2bottle before and after the bottle is opened. The Henry's law constant for CO in aqueous solution is -2 2 3.1 x 10 mol/L atm at 25°C. Step 1: list the information you're given. - Soft drink is bottled at 25°C - When the soft drink is closed, the pressure over the liquid is 5.0 atm - The pressure of the air around the soft drink is 4.0 -4 x 10 atm - The Henry's law constant for CO2 at 25°C is 3.1 x 10 mol/L atm Step 2: calculate the concentration of CO2 before the bottle is opened. - Henry’s law: C = k P = H0.031 mol/L atm) x (5.0 atm) = 0.16 M Step 3: calculate the concentration of CO2 after the bottle is opened. - Henry’s law: C = k P = H0.031 mol/L atm) x⋅ (0.0004 atm) -5 = 1.2 x 10 M. 17 • Another Henry’s Law Exercise: The solubility of gaseous CO in water at 10°C is 0.240 g 2 per 100.0 mL, under a pressure of 1.0 atm of carbon dioxide. During manufacturing, a soft drink at 10°C is saturated with CO u2der a pressure of 4.0 atm, and then sealed. (a) What mass of CO is2dissolved in a 355 mL can of this drink? (b) What volume of CO wi2l be released from this drink? Step 1: list the given information. - Solubility of CO 2n water at 10°C is 0.240 g per 100.0 mL. - Initial pressure of CO 2s 4.0 atm. - Pressure of CO a2ter the drink is opened is 1.0 atm. - Volume of the can is 355 mL. Step 2: use dimensional analysis to find the mass of dissolved CO i2 a 355 mL can. - Convert 100.0 mL to g: (0.240 g)/ (100.0 mL⋅1.0 atm) x (4.0 atm) x 355 mL = 3.408 g = 3.4 g (2 sig figs) (a)The mass of dissolved CO in 2 355 mL can is 3.4 g. Step 3: use Dalton’s law of partial pressures to calculate the mass of CO re2aining after the can has been opened. - Dalton’s law: P CO2 = XCO2⋅P totalwhere X CO2is the mole fraction of CO , P is the partial pressure of 2 CO2 18 CO 2 and P totals the total pressure of the solution. Isolating for PCO2, we get: PCO2 =(0.00031 mol CO )/21 mol of air) x 1 atm = -4 3.1 x 10 atm (2 sig figs). -4 - 3.1 x 10 atm represents the pressure of CO once 2 it escapes the can. - For the mass of CO rem2ining after the can is opened: (0.240 g)/(100.0 mL⋅1.0 atm) x (3.1 x 10 -4 atm) x 355 mL -4 = 2.64 x 10 g -4 = 2.7 x 10 g (2 sig figs). Step 4: use the amount of CO rel2ased and the amount of CO2 remaining to calculate the amount in litres of CO re2eased. - Total mass of CO rele2sed = 3.4 g – 2.7 x 10- 4 g = 3.3997 g = approximately 3.4 g (2 decimal spaces). - Convert this total mass of CO2 to moles of CO 2 (3.4 g CO 2 x (1 mol CO )/244.01 g) = 0.077256 mol CO .2 - Use the ideal gas law to calculate how many litres of CO were released. 2 o Ideal gas law: PV = nRT § P = 1 atm § n = 0.077256 mol 19 § R = 0.08206 atm⋅L/ K⋅mol § T = 283.1 K - PV = nRT becomes V = nRT/P o V = [(0.077256 mol)⋅(0.08206 atm⋅L/ K⋅mol)⋅(283.1 K)]\ (1.0 atm) § V = 1.77 L (b)1.77 L of CO w2re released from the 355 mL can. • Another Henry's Law exercise: A minimum concentration of 1.3 x 10 M O must b2 maintained in fresh water in order to sustain aquatic life. In the mountains of Montana, the partial pressure of O may 2 drop to as low as 0.15 atm. (a) What is the concentration of O in2water under these conditions? K for O = 1.3 x 10 -3 H 2 mol/L⋅atm. Using Henry's law, C = (1.3 x 10 -3mol/L⋅atm)(0.15 atm) = 0.000195 M -4 = 2.0 x 10 M (2 sig figs) (b) At lower elevations, would more or less O be2 dissolved in water? More O w2uld be dissolved. The lower below sea level water you travel, more pressure is exerted on your lungs. Since pressure increases with depth, and increasing pressure improves the solubility of gases, the solubility of O 2 increases. 20 The Effect of Temperature on Solubility - Generally, solubility of solids in water increases when temperature increases, but this isn't always true. The solubility of solids depends on enthalpy. - The solubility of gaseous solutions decreases if temperature is increased. Vapour Pressure of Solutions - By the end of this section, we will know why people throw salt on icy roads, why people add salt to water they're trying to boil, and why people add antifreeze to their car’s radiator fluid. - Properties of liquid solutions are very different from those of pure solvents. • Pure volatile liquids: the molecules in these liquids are more likely to leave the liquid solution to become gaseous molecules (vaporize). In order for molecules to vaporize, they have to reach the surface of the liquid, and then overcome the intermolecular interactions. • A solution consists of a solute and a solvent . If the solute is nonvolatile, not all the molecules in the liquid can vaporize because their access to the surface is blocked. o Nothing is stopping vapour from returning to the liquid solution! • Note: the presence of a nonvolatile solute lowers the vapour pressure of a solvent. • To predict the vapour pressure of a solution, ask yourself what fraction of the molecules in that solution are volati le. 21 • Consider a closed environment containing one beaker of water, and another beaker of some aqueous solution. Water molecules (in the form of vapour) leave the beaker and enter the aqueous solution, which lowers its vapour pressure. Therefore the vapour pressure of a pure solvent is greater than that of a solution. • In the same closed system, the aqueous solution absorbs water vapour to lower the vapour pressure of water in the other beaker. The system wants to have the same vapour pressure in each beaker. • This system is only at equilibrium once all the water is transferred to the aqueous solution. • At any temperature, the vapour pressure of a solution (P ) is less than the vapour pressure of a soln pure liquid (P°). • If a liquid-vapour solution is initially in equilibrium, the presence of a nonvolatile solute will cause the rate of vaporization to be less than the rate of condensation. • Eventually, the rate of condensation will decrease until the rates of the two reactions are equal again (new equilibrium). The vapour pressure at this new equilibrium will be lower than it was initially. • Raoult’s Law states that the presence of a nonvolatile solute lowers the vapour pressure of a solvent. The equation for Raoult’s law is: P soln x solvent° solvent Where P solnis the solution's vapour pressure, xsolvents mole fraction of the solvent, and P°solvents the vapour pressure of the pure solvent. 22 • The equation for Raoult's law is in the form y = mx + b where y = P soln x = xsolventand m = P° solvent so if you were to draw a corresponding graph for this equation it would be a straight line. o If Raoult's law is obeyed, the solution is considered ideal. Example of Raoult's Law: in a warm room, the compound beta-pinene may have a vapour pressure of approximately 97 torr. The vapour pressure of this compound can be detected from solutions used in products where a pine scent is desired. If you had a solution containing 0.10 moles of beta-pinene and 0.050 moles of a nonvolatile compound, what would be the new vapour pressure of beta-pinene? Step 1: list the valuable information you're given. - Vapour pressure of beta-pinene = 97 torr. - 0.10 mol. beta-pinene - 0.050 mol. nonvolatile compound - Find the new vapour pressure of beta-pinene Step 2: use the equation for Raoult's law: what are you missing to solve for P°solvent - P soln x solvent° solvent - We are missing the mole fraction of beta -pinene o Calculate xsolvent: § x solvent =oles of Beta-pinene) / (Moles of Beta-pinene + Moles of Nonvolatile Compound) x = (0.10)/(0.15) = 0.66667 § solvent 23 - P soln= (0.66667) (97 torr.) = 64.667 torr. = 65 torr. (2 sig figs) The new pressure of the beta-pinene is 65 torr. LECTURE 3 REVIEW Topics Covered: determining molar mass from Raoult's law, solution composition, deviations from Raoult's law, vapour pressure, boiling-point elevation, and freezing-point depression. Exercise: Calculate the expected vapour pressure above a solution containing 158.0 g of sucrose (table sugar, molar mass = 342.3 g) in 643.5 mL of water at 25°C. At this temperature, the density of water is 0.9971 g/mL and the vapour pressure is 23.76 torr. Given: 158.0 g of sucrose, molar mass = 342.3 g, 643.5 mL water, 25°C, density of water at 25°C is 0.9971 g/mL. Calculate the expected vapour pressure above the solution. Raoult's law equation: P solution x solvent° solvent We want to calculate the value of P solution We are missing the mole fraction of water (x solvent 1. Calculate the moles of sucrose in 158.0 g. 158.0g × (1 mol. / 342.3 g) = 0.46158 mol. sucrose. 2. Calculate the moles of water in 634.5 mL. 24 0.9971 g/mL × 643.5 mL = 641.6339 g water 641.6993 g × 1 mol. / 18.015 g = 35.6203 mol. water. 3. Calculate the mole fraction of water. 0.46158 mol. sucrose + 35.6203 mol. water = 36.082 mol.. Xwater= 35.6203 mol. water/ 36.082 mol. solution = 0.98720 4. Calculate the pressure above the solution: 0 Psolution xsolvent solvent(0.9872) × (23.76 torr) = 23.46 torr. Note: keep the following plot in mind when you're asked to solve problems involving Raoult's law: 25 THE IMPORTANCE OF RAOULT'S LAW Ø Raoult's law can be used to determine molar mass of a compound. If you're given the change in vapour pressure of a solution, and you're told that a certain amount (in grams) of a certain substance is added to some amount of a solvent, solve for solvent the equation for Raoult's law. From here you should be able to solve for the number of moles of the compound you're interested in, so all you need to do at this point is convert moles to grams using the given information. 26 Ø You can use Raoult's law to characterize compounds since variations in vapour pressures depend on the total number of solute particles in a solution. Vapour pressure tells you about the solute that was dissolved in a solution. Take note of whether a compound exhibits molecular or ionic dissolution, because counting particles is essential in problems like these. For example, NaCl is an ionic + - compound, and it produces Na ions and Cl ions (2 particles), so two moles of ions are produces for 1 mole of NaCl. On the other hand, sugar doesn't dissociate in water, so one mole of sugar produces only one mole of solute (1 particle). Exercise: Predict the vapour pressure of a solution prepared by mixing 35.0 g solid Na S2 (m4lar mass = 142 g) with 175 g of water at 25°C. The vapour pressure of pure water is 23.76 torr at 25°C. P solution x solvent° solvent 1. Calculate the moles of water in 175 g. 175 g × 1 mol./ 18.015 g = 9.7141 mol. water. 2. Calculate the moles of Na SO in235.04g. 35.0 g × 1 mol./ 142 g = 0.24648 mol. Na SO . 2 4 2- Use the equation: Na SO2( )4 s2Na( )+ aqSO ( ) 4 aq Three moles of solute per mole of Na SO , so 3(0.24648)= 2 4 0.73944 mol. Na SO2. 4 3. Calculate the mole fraction of water. 27 For total moles in solution, 9.7141 mol. water + 0.73944 mol. Na 2O 4 10.4495 mol. Xwater= 9.7141 mol./ 10.4495 mol. = 0.92962. 4. Calculate the vapour pressure using Raoult's law. Psolution x solvent0solvent Psolution =.92962) × (23.76 torr) = 22.078 torr, round to 3 sig figs for the final answer: 22.1 torr. LIQUID-LIQUID SOLUTIONS We’ve applied Raoult's law to solutions with non -volatile solutes, like solutes with no vapour pressure, but Raoult's law can also be applied to solutions made of solvents with different vapour pressures. When you're faced with a problem that has two vol atile components, both components add to the pressure’s solution, or P solutionso use the following equation when you're solving problems involving 2 volatile components: Psolution (x A° A + (x PB )B Where P solution still the observed pressure of the solutionA x is the mole fraction of the first component, and P°Ais the partial pressure of A. The same logic extends to the second part of the equation, except you'd calculate the mole fraction of B and the partial pressure of B. A two volatile-component solution is only ideal if the components react as strongly as they do with each other as 28 they would on their own. Consider the intermolecular forces and polarity of the components. For example, hexane (C H ) an6 14 octane (C 8 )18re both non-polar, so they would share London dispersion interactions. The molecules of these compounds don't prefer to interact with themselves or each other, so the intermolecular interactions between octane and hexane are as strong as they would be if they were bonding with themselves. This constitutes an ideal two volatile-component solution (no deviation from Raoult’s law). It's very helpful to visualize the plots of reactions when you're dealing with Raoult's law because the graphs can tell you important information about the vapour pressures of the individual components of a solution and the solution itself. From this information, you can draw conclusions about boiling points and the solution’s composition, even if the problem you're faced with doesn't provide you with a lot of information. Keep the graphs in this lecture in mind when solving these problems. Exercise: the vapour pressures or pure benzene and toluene at 25 degrees Celsius are 95.1 torr and 28.4 torr respectively. A solution is prepared in which the mole fractions of benzene and toluene are both 0.500. What are the partial pressures of the benzene and toluene above the solution? And what's the total vapour pressure of the solution? Benzene and toluene are both volatile. Use the equation: P solution (x A ) A (x P )BtoBcalculate their partial pressures, and the pressure of the solution. 29 The fact that there are 0.500 moles of benzene and toluene is very important. Any time you're doing a problem like this one, and you find the moles of one of the volatile components, just subtract the value you have from 1 to get the number of moles for the other component of the solution. 0 (xAP )A= (0.500) × (95.1 torr) = 47.6 torr 0 (xBP )B= (0.500) × (28.4 torr) = 14.2 torr 0 0 Psolution (xAP )A+ (x PB) B 47.6 torr + 14.2 torr = 61.8 torr How to know when to use Raoult's law? Ø Think about what's being asked of you in the problem; o What kind of information is given? Vapour pressure? o What are they asking for? Partial pressure? Total pressure of he solution? o These hints are meant to make it more simpler to distinguish between scenarios where you should use the equation for Raoult's law, or its variation. POSITIVE AND NEGATIV E DEVIATIONS FROM RAOULT'S LAW • A positive deviation from Raoult's law is characterized by a solution pressure that yields a pressure that's higher than anticipated. In these cases, the mixing of these components isn't favoured, and they bond more effectively with themselves than with each other. • Negative deviations from Raoult's law are characterized by a solution pressure that is lower than expected. Mixing is highly favourable in these cases, and the components of 30 these solutions interact more strongly with each other than with themselves. • Negative and positive deviations are demonstrated by certain plots: • Always look at the structure and intermolecular interactions between a solution’s components. • In non-ideal solutions, molecules are being kicked out of the solution, which partly explains why un-mixing is favourable in these cases. • Solutions composed of components of similar strength are ideal. Two non-polar components can either produce a negative deviation from Raoult's law, or an ideal solution. • Below is a table that summarizes the deviations from Raoult's law. You can use it to predict whether or not a solution is ideal. 31 • Note: In problems where you are asked if a solution is ideal, and you're given a value for the solution's vapour pressure, calculate the vapour pressure of that solution, and compare it to the vapour pressure the problem provides you with. If the value you calculated is less than the one you were given, the lower vapour pressure is ideal. This problem then shows a positive deviation from Raoult's law Problem: mixing some solutes and solvents can lead to non -ideal solutions where Raoult's law does not predict the solution vapour pressure accurately. Which of the following scenarios best depicts this type of situation? 1. Two polar liquids when mixed have a good possibility of producing negative deviations (actual vapour pressure is less than expected) from Raoult’s Law. (This statement is true. If you examine the two graphs that are just above the table explaining deviations from Raoult's law, you’ll see that when two polar liquids are combined, mixing is highly favoured and a negative deviation from Raoult's law occurs. 2. When two polar liquids are mixed, they have a good possibility of producing positive deviations (actual vapour pressure greater than 32 expected) from Raoult’s Law. (False because a mixture of two polar liquids yields a negative deviation from Raoult's law; the components of this kind of mixture react more strongly with each other than they do with themselves, so mixing is highly favoured). 3. When a non-polar liquid and polar liquid are mixed they have a good possibility of producing a negative deviation from Raoult’s law. (False because a mixture of a polar and non-polar liquid yields a positive deviation from Raoult's law; the components of this type of solution react stronger with themselves than they do with each other. Un-mixing is favoured). 4. None of these seem logical. (False because we know, from analyzing the plots for nonideal mixtures of two volatile liquids that the first statement is true). Exercise: a solution contains 2.43 g acetone (C H O,3mo6ar mass = 58.08 g) and 3.95 g of carbon disulfide (CS , mol2r mass = 76.15 g). At 35°C, this solution has a total vapour pressure of 645 torr. Is this an ideal solution? The vapour pressure of pure acetone and pure carbon disulfide at 35°C are 332 and 515 torr, respectively. Both components of the solution are molecular and volatile . Acetone is polar, carbon disulfide is non-polar. 1. Calculate the moles of C H O 3n6 CS . 2 2.43 g × 1 mol. / 58.08 g = 0.04184 mol. C H O3 6 3.95 g × 1 mol. / 76.15 g = 0.05187 mol. CS . 2 2. Calculate the total number of moles in the solution . 33 ntotal 0.04184 mol. + 0.05187 mol. = 0.0937 mol. 3. Calculate the mole fractions of acetone and carbon disulfide. xA= 0.04183 mol./ 0.0937 mol.= 0.4465 x = 0.05187 mol./ 0.0937 mol. = 0.5535 B 4. P total (0.4465) × (332 torr) + (0.5535) × (515 torr) = 433 torr (3 sig figs) 433 torr < 645 torr, which means this solution is ideal, and the vapour pressure of 645 torr exhibits a positive deviation from Raoult's law. The components have opposite types of intermolecular interactions (CS2is polar, and C 3 6 is non-polar), so un-mixing is favoured. Exercise: If a litre of wine contains 120 mL ethanol & 880 mL water, what is the wine’s vapour pressure at 20°C ? We want to find the pressure of the solution (wine and water) at 20°C. Use Raoult's law, Psoln = (X PA ) A (X P° B. B 34 Psolution P ethanol P water= X ethanol P° ethanol X water P° water We have 12% alcohol by volume, so: 12/ 100 = 0.12 x 1 L = 0.12 L of ethanol in the bottle of wine. What percentage of the wine is water? 1 L – 0.12 L = 0.88 L of water, so 88% of the wine is water. Calculate the moles of ethanol in the wine: The density of ethanol is 0.785 g/ mL, and the density of water is 1.000 g/ mL. For the moles of ethanol: 120 mL × 0.785 g/ mL × 1 mol. / 46.07 g = 2.048 mol. ethanol. For the moles of water: 880 mL × 1.000 g/ mL × 1 mol. / 18.05 g = 48.83 mol. water. Mole fraction of ethanol: XA= 2.048 mol. / 50.878 mol. = 0.04025. Mole fraction of water: XB= 48.83 mol. / 50.878 mol. = 0.9597. For the pressure of the solution: 35 Psolution (0.04025)(48.39) + (0.9597)(17.535) = 18.597 mm Hg = 19 mm Hg (2 sig figs). v If the wine is forgotten overnight in your trunk in the winter, and the temperature is -15°C, will it freeze? o The wine will freeze. Since it's made up of mostly water, and water expands when it freezes (solids take up more space than liquids), the bottle will break. LECTURE 4 REVIEW TOPICS COVERED: OSMOTIC PRESSURE, COLLIGATIVE PROPERTIES OF ELECTROLYTE SOLUTIONS, EQUILIBRIUM. OSMOTIC PRESSURE • Osmosis is the net flow of solvent that passes through a semipermeable membrane. o All molecules in a solution bump into a semipermeable membrane, but only the solvent can go through it. The side of the solution containing more water molecules has a higher rate of escaping water molecules. The solvent wants to move to the side of the membrane that has more solute (less solvent). • So far, the colligative properties we know are boiling-point elevation, and freezing-point depression. A third one is osmotic pressure. o Osmotic pressure is the pressure that must be applied to stop the solvent from passing a semipermeable membrane (to make the solution reach equilibrium). o We calculate osmotic pressure with the following formula: § ∏ = MRT, where ∏ is osmotic pressure, M is molarity (moles of solute per litre of solution), R is 36 the ideal gas constant (0.08206 L atm / mol K) and T is temperature in Kelvins. § The units for R must match with the units of the other values. For example, if you're using R = 0.08206 L atm / mol K, temperature must be in Kelvins and osmotic pressure must be in atmospheres. § This equation is very similar to the ideal gas law equation, PV = NRT • When an osmotic system reaches equilibrium, no osmosis is actually occurring (no net flow of solvent across semipermeable membrane). • The concentration gradient (low solvent concentration to solvent high concentration) is countered when pressure is exerted by the higher collusion of liquid on one side. This is the osmotic pressure, ∏. • Osmometry is useful because it lets us precisely determine solute concentration. -3 Exercise: to determine the molar mass of a protein, 1.00 x 10 g of it are dissolved in 1.00 mL of solution. The osmotic pressure of this solution was 1.12 torr at 25°C. Calculate the molar mass of this protein. ∏ = MRT ∏ = 1.12 torr, V = 0.001 L, T = 298.15 K. ∏ = MRT becomes M = ∏ / RT 1.12 torr x 1 atm / 760 torr = 0.0014737 atm, because 1 atm = 760 torr. 37 M = (0.0014737 atm) / (0.08206 L atm/ mol K) (298.15 K) = -5 0.0000602341 mol/L, or 6.02 x 10 mol/L. -8 M x V = n, so (0.0000602341 mol/L) x (0.001 L) = 6.02 x 10 mol. Mass/ molar mass = moles, and mass / moles = molar mass, so: MM = (1.00 x 10 g) / (6.02 x 10 -8mol) = 16.6019 g/mol = 1.66 x 104 g/mol (3 sig figs). • Dialysis is an application of osmotic pressure. In dialysis, the membrane lets solvent and other small molecules pass through it. This occurs naturally in plant and animal cells. Blood is cleaned by using a solution that has the same osmotic pressure (isotonic). o If a solution with an osmotic pressure that is greater than that of blood (hypertonic) this can cause blood cells to shrivel up. § The concentration of solute is greater outside the blood cells than it is inside the cell, so the blood inside the cell seeks to leave it so it can lower the concentration of solution outside of it. o If a solution with an osmotic pressure less than blood is injected into the bloodstream (hypotonic), this can cause blood cells to explode. § The concentration of solute is lower inside the cell than it is on the outside of the cell, so the solvent around the cell wants to enter to cell to increase its concentration. § Remember: concentration is directly related to osmotic pressure! 38 Exercise: what concentration of sodium chloride in water is needed to produce an aqueous solution that's isotonic with blood, whose osmotic pressure is 7.70 atm at 25°C. ∏ = MRT becomes M = ∏ / RT ∏ = 7.70 atm, T = 298.15 K, R = 0.08206 L atm / mol K M = (7.70 atm) / (0.08206 L atm/ mol K) (298.15 K) = 0.31472 mol/L + - NaCl dissolves into two ions, Na and Cl , so the concentration of sodium chloride is: 0.31472 M / 2 = 0.157 M. Note: each particle affects the concentration, osmotic pressure, boiling point and freezing point of a solution. • Reverse osmosis occurs if the external pressure of a solution is larger than the osmotic pressure, causing solvent to leave a solution instead of enter it. o Reverse osmosis is useful because it is used in the process of desalination of sea water; applying a pressure that's greater than the sea water’s osmotic pressure produces fresh water. Exercise: in an ideal solution, what happens when a non-volatile solute is added to a solvent? Solution: vapour pressure decreases, freezing point decreases, boiling point increases, and osmotic pressure increases. 39 • Electrolyte solutions demonstrate colligative properties, which only involve the numbers of particles in an ideal solution. o Consider a solution of 0.100 mol/kg glucose (molecular dissolution) with a freezing-point depression of 0.186°C, and another solution of NaCl (ionic dissolution) of the same molality. o The freezing-point of the NaCl solution is not double that of the glucose solution, even though NaCl dissolves into two ions in aqueous solutions, and glucose doesn't. This occurs because of ion pairing: some ions stay paired up during the process of dissolution. • The van’t Hoff factor, i, relates the number of particles released when a substance dissolves. Study the table below: Exercise: which solution should have the highest boiling point: 0.20 m ethylene glycol (HOCH CH2OH,2non-electrolyte); 0.12 m NaI; 0.10 m CaCl 2 or 0.12 m Na SO2? 4 Solution: 0.12 m Na SO ,2usi4g the van't Hoff factor. 0.20 m ethylene glycol: 1 x 0.20 = 0.2 (molecular dissolution) 40 0.12 m NaI: 2 x 0.12 = 0.24 (ionic dissolution, releases 2 ions) 0.10 m CaCl : 3 x 0.10 = 0.30 (ionic dissolution, releases 3 ions) 2 0.12 m Na S2 : 4nd 3 x 0.12 = 0.36 (ionic dissolution, releases 3 + 2- ions: 2Na and SO 4 ). Exercise: complete the table, and rank the solutions from lowest to highest vapour pressure, boiling point, freezing point, and osmotic pressure. Solute Solution Species van't Total concentration in Hoff concentration Solution Factor of particles KCl 0.25 M K , Cl- 2.00 0.500 M (0.25 M x 2) C6H 12 6 0.40 M C , H , 1.00 0.40 M 2- O (0.40 M x 1) + (NH 4 2O 4 0.20 M 2NH 2- 3.00 0.60 M SO 4 (0.20 M x 3) Recall: osmotic pressure is directly related to concentration. o The solution with the lowest vapour pressure is (NH ) S4 2 4 Concentration is directly related to vapour pressure, and (NH 4 2O h4s the greatest concentration, while C H O6h12 6 41 the lowest. Therefore, from lowest to highest vapour pressure, (NH )4 2 < 4Cl < C H O 6 12 6 o The presence of a nonvolatile solute increases the boiling point of a solution. C H6O12as6the least amount of particles in solution so it's boiling point is the lowest, while (NH )4 2 4 releases three particles when dissolved so it has the highest boiling point. In order of lowest to highest boiling point, C 6 12< 6Cl < (NH ) SO 4 2 4 o From lowest to highest, the freezing points of the solutions are (NH ) SO < KCl < C H O ., the opposite of the ascending 4 2 4 6 12 6 order of boiling points. o Osmotic pressure is dependent on concentration, so from lowest to highest, the osmotic pressures of the solutions are C 6 12< 6Cl < (NH ) SO 4 2 4 Exercise: calculate the freezing point and the boiling point of a 0.050 m FeCL aqueous solution using the observed and expected 3 van't Hoff factors. The molal freezing and boiling point constants for water are 1.86°C kg/mol and 0.51°C kg/mol. Expected i = 4, observed i = 3.4 ΔT = i Km 1. FP, expected: ΔT = (0°C- ((4) (1.86°C kg/mol) (0.050 mol/kg)) = 0°C – 0.37°C = – 0.37°C 2. FP, observed: ΔT = (0°C – ((3.4) (1.86°C kg/mol) (0.050 mol/kg)) = 0°C – 0.32°C = -0.32°C 3. BP, expected: ΔT = (100°C + ((4) (0.51°C kg/mol) (0.050 mol/kg)) = 100°C + 0.102°C = 101.10°C 4. BP, observed: 42 ΔT = (100°C + ((3.4) (0.51°C kg/mol) (0.050 mol/kg)) = 100°C + 0.0867°C = 100.87°C • Study the following diagram because understanding it can help you solve problems involving the van't Hoff factor: EXERCISES 1. A 1.48 M solution of 3 C6H5O7(citric acid) in water has density 1.10 g/mL. Calculate the mass percent, molality, mole fraction, and normality of citric acid. 43 Citric acid has three acidic protons and its molar mass is 192.124 g/mol. (i = 4). Mass percent = mass of solute / mass of solution x 100% Molality = moles of solute/ kg of solvent Mole fraction of citric acid = moles of citric acid / (moles of citric acid + moles of water) Normality = number of equivalents. M = 1.48 mol. solute per litre of solution, so moles of citric acid = 1.48 mol. 192.124 g/mol x 1.48 mol. = 284.3 g of citric acid. Assuming 1 L, 1000 mL x 1.10 g/mL = 1100 g of solution. 1100 g solution – 284.3 g citric acid = 815.7 g water = 0.8157 kg solvent. 815.7 g x 1 mol. / 18.02 g = 45.27 mol. water. Solution: - Mass percent = (284.3)/(1100) x 100% = 25.9% - Molality = 1.48 mol./ 0.8157 kg = 1.81 m - Mole fraction of citric acid = (1.48)/(1.48 + 45.27) = 0.0317 - Normality = 1.48 M x 3 acidic ions = 4.44 N 2. A 25.00 mL solution containing 4.562 mg of an ionic compound with formula MCL3 exhibits an osmotic 44 pressure of 83.1 +\- 0.6 mm Hg at 22°C. What's the most likely identity of M: Al, Cr, Mn, Fe, or Co? Use osmotic pressure to find the moles of solute and the total number of moles of dissolved ions. ∏ = MRT => ∏ = (n/v)RT => n = (∏V)/(RT) = ((83.1 mm Hg/760 mm Hg/atm) x (0.02500 L/ (0.08206 L atm/ mol K)(295.15 K)) -4 = 1.1286 x 10 mol. ions. Calculate the molar mass based on the 4 ions that MCl 3 releases. 1.1286 x 10 mol. / 4 = 2.823 x 10 -5mol. MCl 3 Mass / Molar Mass = moles => mass / moles = molar mass, so: 0.004562 g / 2.823 x 10-5mol. = 161.6 g/mol = Molar Mass of Compound. For the Molar Mass of M: 161.6 g/mol – 3(35.45g/mol) = 55.3 g/mol. % error in osmotic pressure: 100% x (0.6/88.1) = 0.72% 45 55.3 g/mol +/- (0.0072 x 55.3 g/mol) = 55.3 g/mol +/- 0.4 g/mol, so M can either be Fe or Mn due to the percent error. 3. The electrolyte in car batteries (12 V lead storage batteries) is a 3.75 M sulfuric acid solution with a density of 1.230 g/mL. Calculate the mass percent and molality of the sulfuric acid, and find its freezing point. A problem with the same values was solved in an earlier lecture, so they will be used in this solution. m = 3.75 mol. / 0.8623 kg = 4.35 mol/kg H S2 .4 Using the van't Hoff Factor with the estimated value of 3, because H2SO 4roduces 3 ions in solution: ΔT = 0°C – (3 x 4.35 mol/kg x 1.86°C kg/mol) = 0°C - 24.3°C = - 24.3°C. 3. Cigars are best stored in a humidor at 18°C and 55% relative humidity. This means the pressure of water vapour should be 55% of the vapour pressure of pure water at the same temperature. The proper humidity can be maintained by placing an aqueous solution of glycerol, C3H 5OH) (3M = 92.09 g/mol) inside the humidor. Calculate the concentration of glycerol (in mass%) required to lower the vapour pr sure of water to be desired value. Assume ideal gas behaviour, with glycerol as a nonvolatile component. Determine the desired vapour pressure of water in the humidor. 46 P = 55%, so P = X solventwater => X solvent (0.55 atm)/(1 atm) = 0.55. Convert the mole fraction of glycerol to mass %: Assume the total number of moles in solution is 1 (ideal), so 1 mole – 0.55 mole = 0.45 moles of solvent. Mass of solute = 0.55 mol x 92.09 g/mol = 50.6495 g Mass of solvent = 0.45 mol x 18.02 g/mol = 8.109 g Mass % = (8.11 g) / (50.65 g + 8.11 g) x 100% = 14% (2 sig figs) CHEMICAL EQUILIBRIUM (SHORT REVIEW) • Any reversible reaction can reach equilibrium. • We know a system reached equilibrium if the concentration of its reactants and products stopped changing, and if the rate of the forward reaction equals to the rate of the reverse reaction. • There are some instances where chemical reactions go so slowly that their reactants and products don't seem to be changing, but they really are. 47 LECTURE 5 REVIEW Topics Covered: Law of mass action, the equilibrium constant, using the quadratic equation to solve for x, solving equilibrium problems. LAW OF MASS ACTION In a reversible rxn (reaction) at eqm (equilibrium) and a constant temperature, a certain ratio of products and reactants has a constant value, K (K = eqm c constant for concentrations, p = eqm constant for pressures) Eqm is a dynamic state where he rate of formation of reactants is equal to the rate of formation of products. The ratio K is calculated as follows: n m K = [products] / [reactants] Where the square brackets indicate we are calculating with concentrations, and the exponents m and n are the coefficients of the reactants and products, respectively, in a chemical rxn. The K for the chemical rxn N 2(g)+ 3Cl 2(g)= 2NCl 3(g)is written as: K = [NCl ]3/[N ]2Cl ]2 3 where the concentrations of N 2(g)and 3Cl 2(g)re multiplied. Example: given the following K, what is its corresponding chemical rxn? All the species are in the gaseous phase. K = [SO ]3([SO ][2 ] 2 1/) Solution: SO +21/2O = SO2 3 48 Example: calculate the value of K for the following rxn given the eqm concentrations. The system is at eqm at 127°C and all species are gaseous. -2 [NH3] = 3.1x10 M [N2] = 8.5x10 M-1 -3 [H2] = 3.1x10 M -2 2 -1 -3 K = (3.1x10 )4/((8.5x10 )(3.1x10 )) K = 3.84x10 Example: calculate K for the following rxn: 2NH 3(g)= N 2(g)+ H 2(g)using the same concentrations as in the previous example. K = (8.5x10 )(3.1x10 )/(3.1x10 ) -2 -5 K = 2.6x10 You could have also taken the reciprocal of the previous K value to get this answer because this rxn is the reverse of the previous one. K’ = 1/K = 1/(3.84x10 ) = 2.6x10 -5 Example: calculate the value of K for the following rxn. All the species are in the gaseous phase. 1/2N + 3/2H = NH 2 2 3 Solution: K = (NH )3((N )2(H ) )2 3/2 2 1/2 3 1/2 K = ((NH )3) /((N )(2 ) )2 2 = 196.082x10 = 1.96x10 2 Note: You can factor out ½ in the exponents to make your calculations easier. Calculate your value for K as usual and then take the square root for the final answer. You could have also just taken the square root of K = 3.84x10 to get the same answer as the two methods are essentially the same. 49 EQUILIBRIUM CONSTANT SUMMARY • The eqm expression for a rxn (K ) i1 the reciprocal (1/K ) 1f that for the reverse rxn. • If the equation for a rxn is multiplied by some number, K new = (K ) , where n is the number that the rxn is multiplied by. initial • If two rxns occur in sequence, the overall eqm constant for the sum of the processes is he product of the K values. To add these chemical equations, the second one must be reversed, so its constant becomes 1/K . 2 The overall constant is: K (1+2) K 1 K 2 -9 -11 = (6.6x10 ) ÷ (4.8x10 ) = 137.5 137.5 isn't one of the answers, but 140 is close to this value. In this case, the answer is 140. Note: Reviewing Hess’ law could be very useful, because you use it when calculating the equilibrium constant for the sum of two reversible reactions. To calculate the equilibrium constant of the sum of two reversible reactions, multiply the constants of each individual reaction to get the overall equilibrium constant. 50 Use the equilibrium concentrations of reactants and products to calculate K. K = (3.99x10-2)2/(3.0x10-5)2 K = 1.59x10-3/9.00x10-10 7 K = (0.1766)x60 K = 1.77x10 One temperature can only have one corresponding eqm constant, but there is an infinite number of eqm positions. The same reversible reaction can have its eqm shift left or right, but it's constant will always be the same. You can only calculate the eqm constant using the eqm concentrations of the given species. 51 EQUILIBRIUM EXPRESSIONS FOR GASES • Equilibrium constants can be calculated in terms of partial pressures. • When calculating the eqm constant for gaseous solutions, K , p only the particles that are gaseous count, so solids and liquids are ignored when calculating K . p We know that K = 0p133, and P(N O ) = 2274 atm. K = [NO ] ÷ [N2O4] p 2 2 (1.33) = [NO ] 2 ÷ (2.71) 1/2 ((1.33)(2.71)) = 1.898 atm Consider the eqm expression for the rxn (reaction): = 1.90 atm (3 sig figs) 2 Kp= [NO ] 2 [N O ] 2 4 Since we know the pressure of N O , and the value for K , 2 4 p 2 (0.133) = [NO ] 2 (2.71) 52 2 (0.133) (2.71) = [NO ] 2 2 √(0.36043) = [NO ] , s2 the pressure of NO = 0.600352 atm = 0.600 atm (3 sig figs). Remember, when you're trying to find the concentration of a gas, use the ideal gas equation PV = nRT. To calculate concentration, use n/V = P/RT. You can convert K values to K values, and K values to K p p values with the following equations: -Δn Δn K c K (Rp) and K =cK (RTp Where Δn represents the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants. Pay attention to the signs of Δn! The preceding equations were derived as follows: 2 3 K c (P(NH )/(3T)) ÷ (P(N )/RT)(P2H )/RT) 2 K = (P(NH ) )(RT) ÷ (P(N ))(RT) (P(H )) (RT) 3 -3 c 3 2 2 K p ((P(NH )) 3 (P(N )) (P(2 ) )(RT)2 3 2 • note: the power that RT is raised to in the previous step equals the value obtained by using -Δn (-(2-4)) = 2. 53 • If all the products in a given reaction are gaseous, to calculate Δn, add the coefficients of the products and the coefficients of the reactants, and then subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products. Consider the eqm expression fo
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