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EKU - Chem 111 - Study Guide

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Chemistry 111 Study Guide(Dr. Patton) HIGHLIGHT: Important Try solving the example problems on your own and then check to see if you were correct. This is a great study technique that will help you on the upcoming exam. 4.4- Oxidation-Reduction Reactions ❖ Oxidation reduction (or redoDon't forget about the age old question of serialism in art

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x) reactions: involve transfer of electrons between reactants (Rules on page 112-113 in textbook) ⮚ Oxidation: Occurs when an atom or ion loses electrons ▪ Oxidation Is Loss of Electrons (OIL) ⮚ Reduction: Occurs when an atom or ion gains electrons ▪ Reduction Is Gain of Electrons (RIG) ⮚ Oxidizing agent ▪ Cause another reactant to be reduced ⮚ Reducing agent ▪ Cause another reactant to become oxidized ❖ Rules for assigning oxidation numbers 1. Elements have an oxidation number of 0 ▪ EXAMPLE: Ca, H2, S8 2. The sum of the oxidation numbers of each atom in a neutral compound will add up to zero 3. Oxygen is -2. In peroxides oxygen is -1 4. Hydrogen is +1. In halides hydrogen is -1 5. Sum of oxidation numbers for polyatomic ion equals the charge on ion ▪ EXAMPLE: Find the oxidation number of N in the nitrate (NO3) and nitrite(NO2) ion We can solve this like an algebra problem by solving for x.… NO3-: N=x, the oxidation number of O here is -2 and the subscript of oxygen is 3 so we multiply 3 by -2. We set the equation equal to -1 because the charge of NO3- is - 1… x+3(-2)=-1 x-6=-1 x=+5 NO2-: N=x, the oxidation number of oxygen here is still -2 but the subscript is 2 so we multiply 2 by -2. We set the equation equal to -1 because the charge of NO2- is - 1… x+2(-2)=-1 x-4=-1 x=+3 6. Fluorine always has an oxidation number of -1. Other halogens have multiple possible oxidation numbers but are usually -1 7. The oxidation number of a monatomic ion is the same as its charge▪ EXAMPLE: Ca2+ (oxidation number is +2) S2- (oxidation number is -2) 8. Metallic elements have positive oxidation numbers while non-metals may be negative or positive but cannot be higher than the group number it is in. 9. Oxidation numbers can be fractions. EXAMPLE: Using the information from above, which substance is being reduced, oxidized and which substance is the reducing agent, oxidizing agent… Let’s say we are working with the formation of water 2H2(g) + O2(g) 2H2O(l)… 1. First we assign oxidation numbers… PRODUCT SIDE H2: 0 (rule 1) O2: 0 (rule 1) REACTANT SIDE H2O: H2: +1 (rule 4) O: -2 (rule 3) 2. Now let’s determine the change in oxidation numbers… H2: Hydrogen is now +1, Hydrogen gained an electron O2: Oxygen is now -2, Oxygen loss electrons 3. Now from the info above we can determine which substance was oxidized, reduced…. Refer back to the definitions of oxidation. Oxygen is oxidized because it loses electrons Refer back to the definition of Reduction. Hydrogen is reduced because it gains electrons 4. Now we can determine which is the reducing agent, oxidizing agent… Refer back to the definition of Oxidizing agent. Oxygen is being oxidized which lets us know that hydrogen is the oxidizing agent. Refer back to the definition of Reducing agent. Hydrogen is being reduced which lets us know that oxygen is the reducing agent ______________________________________________________________________________ ❖ Combination reaction ⮚ Definition: two or more substances combine to give a single product ▪ EXAMPLE: 6Li(s) + N2(g) Li3N(s) REACTANTS Li: 0 N2: 0PRODUCTS Li:+1 N: -3 ***Remember electrons are negatively charged*** ▪ Li is being oxidized because it loses an electron. Remember electrons are negatively charged. ▪ N2 is being reduced because it is gaining electrons (3). ❖ Decomposition reaction ⮚ Definition: formation of two or more compounds from the breakdown of one compound ▪ EXAMPLE: 2KClO3(s) 2KCl(s) + 3O2(g) REACTANTS K:+1 Cl:+5 (we found the oxidation number of Cl by solving for x like what we did in the previous notes) O3:-2 PRODUCTS K:+1 Cl:-1 O2:0 ▪ Cl is reduced ▪ O is oxidized ▪ K remains the same ❖ Combustion ⮚ Definition: substance that reacts with oxygen which usually releases heat (formation of H2O and CO2) ▪ EXAMPLE:C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) REACTANTS C3:-2.7 (solved for x) H8:+1 O2:0 PRODUCTS C:+4 (solved for x) H2:+1 O2:-2 O:-2 ▪ C3 is oxidized ▪ H8 remains the same ▪ Oxygen is ❖ Disproportionation reaction ⮚ Definition: Oxygen is reduced AND oxidized ▪ EXAMPLE: 2H2O2(aq) 2H2O(l)+ O2 (g)REACTANTS H2:+1 O2:-1 PRODUCTS H2:+1 O:-2 O2:0 ▪ H2 remains same ▪ O2 is reduced AND oxidized ❖ Displacement reaction ⮚ Definition: redox reaction in which atom or ion is displaced by another atom or ion (3 types; hydrogen displacement, metal displacement, halogen displacement) ▪ Hydrogen displacement • Metal +H2O Metal(OH) +H2O(g) • Metal +2HCl MCl2 +H2O 4.5- Concentration ❖ Molarity (M in mol/L) ⮚ Equation: M=n/V in Liters. You can rearrange this equation to solve for a variety of applications. ▪ V=n/M ▪ n=M x V ⮚ EXAMPLE: How many moles of MgCl2 are present in 60.0 ml of .100M MgCl2 solution? Convert ml to Liters- 60.0ml (1L/1000ml)= .060 L Find moles using molarity… n=(.100M mol/LMgCl2)(.060L)= .006mol ⮚ EXAMPLE: How many grams of KCl are present in 250.0 ml of a .125 M solution of KCl Convert ml to Liters- 250.0ml (1L/1000ml)= .250 L Find mass of KCL- 74.55g Find moles of KCl using molarity of solution… n= (.125m/L)(.250L)= .03125 m KCl Find grams of KCl using moles of KCl… (.03125 mol KCl)74.55 g KCl/ 1 mol KCl=2.32 grams of KCl ⮚ EXAMPLE: A chemist needs .0725 moles of acetic acid for a reaction. What volume of a 1.50 M acetic acid solution is needed to provide this amount? Given- .725 mol (n) 1.50 mol/ L Need- V=? V=.725mol/1.50mol/L= .48333 L, Convert to milliliters=483 ml ❖ Dilution⮚ Definition: Often conducted in lab for preparing a less concentrated solution from a more concentrated stock solution (concentrated solution that will be diluted) ▪ Equation: MiVi=MfVf • M=Molarity • V=Volume • i=initial • f= final ▪ EXAMPLE: How many milliliters of a 5.070 M KOH stock solution are required to prepare 200.00ml of .8660 M KOH lab sample? Let’s rearange the dilution equation to solve for the unknown… MiVi/Mf=Vf Now lets plug in numbers…. (5.070M)(200.00ml)/.8660M=Vf Vf=1.17ml 4.6- solution stoichiometry ❖ Acid-base titration ⮚ Definition: Process in which an acid/base is purposefully and carefully added to a base/acid to bring about neutralization ⮚ Phenolphthalein is a common indicator dye which helps to identify the equivalence point. ⮚ EXAMPLE: 10.00ml sample of HCL of unknown concentration is titrated with a .100M NaOH solution requiring 12.54ml to reach the equivalence point. Find the molar concentration of HCL. First lets find the mols of NaOH… (M)(V)=n (.100mol/liter)(.01254L)=.001254mol NaOH Now lets find the molar concentration of HCl… [HCl]: .001254 molHCl/.0100L=.125M 5.1- Substances that exist as gases ❖ Noble gases many molecular compounds and diatomic gases exist as gases under normal conditions of temperature and pressure ⮚ Noble gasses (monatomic): He, Ne, Ar, Kr, Xe, Rn ⮚ Diatomic gasses: H2, N2, O2, F2, Cl2 ⮚ Molecular compounds: H2S, CO, NH3 5.2- Pressure of a gas ❖ Pressure ⮚ Definition: Amount of force applied to an area (P=F/A) ▪ Applied by atmosphere. The following are units used to report the pressure of a gas. • 1 atm(atmospheric pressure)= 760 mmHg = 760 torr = 101.325 kPa = 14.7 pounds/square inch(PSI) ▪ EXAMPLE: Which of the following represent the highest pressure? a. 735mmHg b. 1.06 x 105 Pa c. 678 torr d. .926 atm First, let’s convert each to the same unit. We can convert each to atm… a. 735 mmHg x 1 atm/760mmHg=.967 atm For b we have to convert Pa to kPa and then convert kPa to atm. There is one 1000 Pa in 1kPa… b. 1.06 x 105 Pa x 1kPa/1000Pa x 1 atm/101.325 kPa=1.05 atm c. 678 torr x 1 atm/760 torr d. .926 atm The correct order is… b, a ,d, c ⮚ Partial pressure ▪ Remember the equation for partial pressure • Partial pressurei=Xi total pressures 5.3- The gas laws ❖ Boyles law ⮚ Definition: Pressure and volume are inversely related ⮚ Equation: P1V1=P2V2 ⮚ EXAMPLE: a fixed volume of gas at room temperature occupies a volume of 1.8 L at pressure of 775 mmHg. What would be the pressure in atm if gas is compressed to a volume of 0.85 L at the same temperature? Let’s rearrange the Boyles law equation… P1V1/V2=P2 Now we can plug in numbers… (775mmHg)(1.8L)/.85 L=P2 P2=1641.18 Now we need to convert our answer to atm since that is what our question asks for… 1641.18 mmHg x 1 atm/760=2.2 atm ❖ Charles law ⮚ Definition: Volume and temperature are directly proportional at a constant pressure. Remember temperature is supposed to be in K(273.15). ⮚ Equation: V1/T1=V2T2 ⮚ EXAMPLE: Under constant pressure a fixed amount of gas occupies 9.6 L at 88 degrees Celsius. What volume(L) will the same gas sample occupy at a temperature of -143 degrees Celsius?We can rearrange the equation to help us solve for the unknown… (T2)V1/T1=V2 Let’s convert 88 to K and -143 to K… 88+273.15=361.15 K -143+273.15=130.15 K Now we can plug in the numbers… (130.15)(9.6 L/361.15)=V2 V2= 3.5L ⮚ An extension to Charles law… ▪ Pressure and temperature are constant at a constant volume. Temperature is still supposed to be in Kelvin. ▪ Equation: P1/T1=P2T2 ❖ Avogadro’s law ⮚ Definition: Volume and moles are directly proportional at constant temperature and pressure ⮚ Equation: n1/V1=n2/V2 ⮚ EXAMPLE: a participant in a kinesiology study has a lung volume of 6.15L and 0.254 moles of air during deep inhalation. During exhalation the participant’s lung volume decreases to 2.55 L. How many moles of gas does the participant exhale? We can rearrange the equation to solve for the unknown… (V2)n1/V1=n2 Now we can plug in numbers… (2.55L).254 mol/6.16 L=n2 n2=.1053 mol We have solved for n2 but we are not finished with the problem. The question asks us how many moles of gas the participant exhales. So we need to subtract the number of moles of air in the participant’s lungs during exhalation from the number of moles of air in the participant’s lungs during exhalation… .254-.1053=.149 moles ❖ STP ⮚ Definition: (Standard Temperature and Pressure) the volume occupied by one mole of any gas is 22.4L which is known as the standard molar volume ▪ EXAMPLE: The volume of .280L of gas has a mass of .400g @ STP. What is the molar mass of the gas? Write the equation of molar mass then plug in given?... Molar mass= mass/moles =.400g/? moles We can find the amount of moles by using a conversion factor of liters to moles. Remember @ STP there is 22.4 liters in one mole… .280L x 1 mol/22.4L=.012494 moles Now we can plug in .012494 into our molar mass equation above… Molar mass=.400g/.012494 =32.01 g/mol❖ Ideal Gas ⮚ Definition: model that can be used to present most real gases fairly accurately. Used when conditions do not change ⮚ Equation: PV=nRT ▪ P: Pressure (atm) ▪ V: Volume ▪ n: moles ▪ R: Constant; .0821 (L-atm/ mol-K ▪ T: Temperature (Kelvin) ⮚ EXAMPLE: Dry ice is solid CO2. A 0.050-g sample of dry ice is placed in an evacuated 4.6 L vessel at 30.0 °C. What is the pressure inside the vessel when all the dry ice is converted to CO2 gas? We can rearrange our equation to solve to pressure… P=nRT/V Now let’s complete our conversions… 30 degrees Celsius to Kelvin= 30 + 273.15 =303.15 K .050g of CO2 to moles of CO2= .050g CO2 x 1 mol/44.01g CO2 =1.136 x 10-3 moles Now we can plug in numbers. Our units will cancel… P= (1.136 x 10-3 mol) (.0821 L-atm/mol-K) (303.15 K)/4.6 L =6.1 x 10-3 atm ⮚ The relation of molar mass and density can be derived from PV=nRT ▪ Equation: PM=dRT ▪ EXAMPLE: A compound has the empirical formula of SF4. At 20°C a 0.100gram gaseous sample of this compound occupies a volume of 22.1 ml exerting a pressure of 775 torr. What is the compounds molecular formula? We should know how to convert our units to the correct units so we can just go ahead and list what each known variable means in its converted form from now on (where needed) … T: 293.15 Mass: .0100 g V:.0221L P:1.02 atm ❖ Combined Gas law ⮚ Used when conditions change ⮚ Equation: P1V1/n1T1=P2V2/n2T2 ▪ EXAMPLE: Assuming the gas is fixed, a gas initially at 4.0 L, 1.2 atm, and 66 °C undergoes a change so that it’s final volume and temperature are 1.7 L and 42 °C. What is the final pressure? We can rearrange the equation to solve for final pressure (P2)… P2=(P1V1/T1)(T2/V2) Let’s determine what each known variable equals…P1: 1.2 atm V1: 4.0L T1: 66 degrees Celsius T2: 42 degrees Celsius V2: 1.7L Now let’s plug in numbers… P2=( (1.2atm)(4L)/339.15K)(315.15K/ 1.7L) =2.6 atm ❖ Relation of molar mass and density ⮚ This relationship can be derived from PV=nRT ⮚ Equation: PM=dRT ⮚ EXAMPLE: A compound has the empirical formula of SF4. At 20 degrees Celsius a .100 gram gaseous sample of this compound occupies a volume of 22.1 ml with a pressure of 775 torr. What is the molecular formula of this compound? Let’s list our givens and do our conversions where needed… T: 293.15 K Mass: .100 g V: .0221L P: 1.02 atm Let’s use the equation used to find the integer… n= molar mass of molecular formula/ molar mass of empirical formula n=molar mass/108 The next two steps will help us find the molar mass. In this step we can find density which will later help us find molar mass… d=mass/volume d=.100g/.0221 L =4.53 g/L Now we can use density to find molar mass which will help us find molar mass… M=dRT/P M=(.453g/L)(.0821 L-atm/K-mol)(293.15K)/1.02 atm (units will cancel out) =107g/mol Now we can refer back to our previous equation used to find the integer… n=molar mass/108 n=107/108 =1 Now we can multiply the integers of our empirical formula by the integer we got… SF4 S1 x 1=1 F4 x 1= 4 =SF4 Our empirical formula is the same as our molecular formula ❖ Ideal gas equation (one reactant or product is a gas)⮚ EXAMPLE: What is the volume of CO2 produced at 37 degrees Celsius and 1.00 atm when 5.60 g of glucose react completely according to the following reaction? Let’s find the number of moles of CO2 using stoichiometry… (5.60 g C6H12O6) (1 mol C6H12O6/180.16 g C6H12O6) (6 mol CO2 / 1 mol C6H12O6) =..187 mol CO2 Now that we know the number of moles of CO2 we can use the ideal gas law to find the volume of CO2. Lets rearrange the ideal gas law equation to find volume… V=nRT/P V=(.250 mol)(.0821)(310 K)/ 1.00 atm = 6.36 L CO2 Ch. 6- Thermodynamics 6.6 Standard enthalpy of rotation and reaction ❖ Standard Enthalpy of Formation ⮚ We can use the heat of formation of substances to determine the change in enthalpy for a chemical reaction (we use the standard enthalpy formation chart) ⮚ ΔHfo ▪ The degree symbol indicates standard condition of a pressure at 1 atm, and a temperature at 25 degrees Celsius. ▪ f signifies the change in enthalpy for the formation of 1 mol of a substance ▪ standard enthalpy of formation of substances in their most stable elemental form is zero. (ΔHfo) • EXAMPLE: H2(g), Br2(l), Cu(s), S(rhombic) ⮚ Two ways to calculate enthalpies of a reaction ▪ Direct method: uses charted data of ΔHfo • Equation: ΔHrxno=∑n ΔHfoproducts-∑m ΔHforeactants (n and m are stoichiometric coefficients in the reaction),( ∑ means the sum of) • EXAMPLE: Using the reference table from textbook, calculate the heat of combustion for the following reaction C2H4(g)+3O2(g) 2CO2(g)+2H2O(l) PRODUCTS: H2O(l): 2(-286 kJ mol-1)= CO2(g): 2(-395.5 kJ mol-1)= REACTANTS: C2H4(g): 52.30 kJ mol-1=52.30 kJ mol-1 O2(g): 3 (0 kJ mol-1 )=0 ▪ Indirect method (Hess’s Law): process where substance is formed through an indirect route❖ Remember that the change in enthalpy (delta H) is delta H of products minus change in delta H of reactants ❖ Endothermic: system absorbs the heat from surroundings (+; energy of reactants<energy of products) ❖ Exothermic: system loses heat to surroundings (-; energy of products>energy of reactants) ❖ Electronic structure of atoms ⮚ Energy level is listed as s,p,d,f ⮚ Remember to study the periodic trends. ▪ Ionization energy ▪ Atomic radius ▪ Ionic radius ▪ Electron affinity ❖ Practice writing Lewis dot structures ⮚ Find sum of valence electrons of all atoms ⮚ Connect outer shells to central atom ⮚ Fill octets of outer atoms ⮚ Fill octet of central atom

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