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Chemistry of solutions, lecture 5 notes

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Chemistry of solutions, lecture 5 notes 202-NYB-05

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These notes explain the law of mass action, the equilibrium constant, and how to use the quadratic equation to solve for x in equilibrium problems.
Chemistry of Solutions
Nadia Schoonhoven
Class Notes
#EquilibriumConstant, quadratic formula
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This 16 page Class Notes was uploaded by CatLover44 on Sunday September 25, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 5 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.


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Date Created: 09/25/16
Chemistry of Solutions Course Number : 202-NYB-05 Lecture no. 5 Date : Thursday, September 8, 2016 Professor : Nadia Schoonhoven 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 =ceqm constant for concentrations, K = eqmpconstant 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)+ 3Cl2(g) 2NCl 3(g)is written as: K = [NCl 3 /[N 2[Cl2] where the concentrations of N 2(g)and 3Cl 2(g)are multiplied. Example: given the following K, what is its corresponding chemical rxn? All the species are in the gaseous phase. K = [SO ]3([SO ][O2] ) 2 1/2 Solution: SO + 2/2O = SO2 3 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. [NH3] = 3.1x10 M -2 -1 [N2] = 8.5x10 M -3 [H2] = 3.1x10 M -2 2 -1 -3 K = (3.1x10 ) /((8.5x10 )(3.1x10 )) K = 3.84x10 4 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. -1 -3 -2 K = (8.5x10 )(3.1x10 )/(3.1x10 ) K = 2.6x10 -5 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. 4 -5 K’ = 1/K = 1/(3.84x10 ) = 2.6x10 Example: calculate the value of K for the following rxn. All the species are in the gaseous phase. 1/2N 2 3/2H = N2 3 Solution: K = (NH )/3(N ) (2 ) ) 2 3/2 2 1/2 3 1/2 K = ((NH ) 3 /((N )(H )2) 2 2 = 196.082x10 = 1.96x10 2 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. Equilibrium Constant Summary • The eqm expression for a rxn (K ) is 1he reciprocal (1/K ) of th1t for the reverse rxn. n • If the equation for a rxn is multiplied by some number, K new = (Kinitial where n is the number that the rxn is multiplied by. • 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)= K1 x K2 -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. 3 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 K = (0.1766)x10 7 6 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. Equilibrium Expressions for Gases • Equilibrium constants can be calculated in terms of partial pressures. 4 • When calculating the eqm constant for gaseous solutions, K , only phe particles that are gaseous count, so solids and liquids are ignored when calculating K . p We know that K = 0.p33, and P(N O ) = 2.712atm4 2 Kp = [NO 2 ÷ [N2O4] 2 (1.33) = [NO 2 ÷ (2.71) 1/2 ((1.33)(2.71)) = 1.898 atm = 1.90 atm (3 sig figs) Consider the eqm expression for the rxn (reaction): 2 K p= [NO ] 2 ÷ [N 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) 2 (0.133) (2.71) = [NO ] 2 2 √(0.36043) = [NO ] , s2 the pressure of NO = 0.6003582atm = 0.600 atm (3 sig figs). 5 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, andpK values to K valuep with the following equations: -Δn Δn K c= K (pT) and K = K (Rc) p Where Δnrepresents 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 )3(RT)) ÷ (P(N )/R2)(P(H )/RT)2 2 -2 -1 3 -3 K c= (P(NH )3)(RT) ÷ (P(N ))(2T) (P(H )) (RT2 K = ((P(NH )) ÷ (P(N )) (P(H ) )(RT) 2 p 3 2 2 • note: the power that RT is raised to in the previous step equals the value obtained by using -Δn (-(2- 4)) = 2. 6 • 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 for the rxn: 2 2 K p= [H 2 [S 2 ÷ [H 2] The eqm partial pressures are: P(H 2 = 0.25 atm 7 P(S 2 = 0.25 atm P(H S) = 1.19 atm 2 So the eqm expression is: 2 2 K p= (0.25) (0.25) ÷ (1.19) = (0.015625) ÷ (1.4161) = 0.0110338 = 0.011 (2 sig figs). Use the equation K =pK(RT) -Δ: -1 K p= (0.011) (0.08206 L atm/ mol K) (1373.15 K) = (0.011) / (112.68 L atm/ mol) = 0.000097621 = 9.8x10 (2 sig figs) 8 Write the eqm expression for the rxn: K = (0.24)(1.1) ÷ (0.15) = 0.2904 ÷ 0.15 = 1.936 -Δn Use the equation K = p(RT) : K = (1.936) ((0.08206 L atm / mol K)(600.15 K)) 2 p 3 K c= 4.4x10 Note: in chemical reactions (at eqm) where more than one phase is present, as long as some of each phase is present, adding more solid to the system will not disturb its state of eqm. 9 When calculating K , solidp and liquids are ignored because: - the concentrations of pure liquids depend on their densities, which remains constant - When pure solvents are present in excess in a gaseous solution, Kp isn't altered - In solids, the concentrations of molecules is determined by the solid’s density. Experiments show that the amount of solid in these rxns are insignificant. 10 Solids and liquids are excluded when calculating K : p K p= (SO 2(g) / (O 2(g) The answer is B. Note: K only tells us two things: 1) whether a reversible rxn shifts to the left or right. A small K value tells us the system’s eqm shifts left (production of reactants is favoured), while a large K value tells us that the system’s eqm shifts to the right (production of products is favoured). 2) how far away a system is from reaching eqm. A small K indicates that the system is reactant-favoured, a K between 10 and 10 means that a lot of both (reactants and products) are produced, and a large K indicates that the system is product favoured. Note: K doesn't tell us the speed of a chemical reaction. 11 Only if we know the value of K, and the initial concentrations of the chemical species in a rxn, we can calculate a reaction’s eqm concentrations. 0.0087 M 0.298 M 0 M 0.0087 - x 0.298 + x + 0.002 M 0.0067 M 0.301 M 0.002 M Since we know the eqm concentration of C2 , we can use this value as the value of x. Then, once we have the eqm concentrations for all the chemical species, we can calculate K: K = (0.002)×(0.301) ÷ (0.0067) 12 -2 = 0.0898507463 = 8.99×10 K = 6.0×10 -2 The system shifts right because no Cl 2(g) is present initially. K = (2.0x10 )(3.0x10 )÷(6.7x10 ) = 8.96x10 -2 -2 -4 Example II: the eqm constant at 500°C is 6.0x10 . The initial concentrations of NH , N , and H are 1.0x10 3 2 2 -2 M, 5.0 M, 1.0x10 M respectively. Is the system at equilibrium? The system is at eqm if Q = K. -4 2 -2 -7 Q = (1.0x10 ) ÷ (5.0)(1.0x10 ) = 2.0x10 Q doesn't equal to K, so the system isn't at eqm. Q is like K, except it's calculated using the initial concentrations of chemical species instead of the eqm concentrations. 13 Example II Continued: N 2(g)+ 3H 2(g) = 2NH 3(g) -2 K = 6.0x10 at 500°. Which way will the system shift if the following are present? [NH ]3 o 1.0x10 M -4 [N 2 o= 5.0 M [H 2 o= 1.0x10 M 2 Q = (0.0001) ÷ (5.0)(0.01) = 2.0x10 -3 Q < K, so the system will shift to the right. Example III: at 700 K, K = 5.10. Calculate the equilibrium concentrations of all the species if initially 1.00 mol of each is mixed in a 1.00 L flask. CO (g) O 2 C(g) 2(g)+ H 2(g) Q = (1.00 M)(1.00 M) ÷ (1.00 M)(1.00 M) = 1 Q < K, so the system favours the formation of products. (Shifts right). CO H 2 CO 2 H 2 Initial M 1.00 1.00 1.00 1.00 Change 1.00-x 1.00-x 1.00+x 1.00+x Equilibrium M 0.61382 0.61382 1.38618 1.38618 Calculate the value of x: (5.10) = (1.00+x)(1.00+x)/(1.00-x)(1.00-x) 2 2 (5.10) = (1.00+x) /(1.00-x) √(5.10) = √(1.00+x)2/√(1.00-x) 2.2583 = (1.00+x)/(1.00-x) 2.2583(1.00-x) = (1.00+x) 2.2583 – 2.2583x = 1.00 – x 1.2583 = 3.2583x 14 x = 0.38618 The Eqm Concentrations are: CO + H O = CO + H (g) 2 (g) 2(g) 2(g) [CO] = [H O2 = 1.00 – 0.38618 = 0.61382 M [CO ]2= [H2] = 1.00 + 0.38618 = 1.38618 M Check The Value of x: 2 2 K = (1.38618) /(0.61382) K = 5.0998, so the value of x is correct. Exercise IV: The production of gaseous HF from F (g) and H (g) has K = 1.15x10 = 115. If 3.00 mol of each 2 2 species is initially in a 1.50-L flask, what will their eqm concentrations be? Initial Concentrations: [H 2 o= [F 2 o= [HF] o 3 mol/1.5 L = 2.00 M Calculate Q: Q = (2.00) /(2.00)(2.00) Q = (4.00)/(4.00) Q = 1, Q<K, so the system shifts right. H 2 F2 H F Initial M 2.00 2.00 2.00 Change 2.00-x 2.00-x 2.00+2x Equilibrium M 0.4715 0.4715 5.057 Calculate the Value of x: (115) = (2+2x) /(2-x) 2 √(115) = √(2+2x) /√(2-x) 2 10.724 = (2+2x)/(2-x) 15 10.724(2-x) = 2+2x 21.448 – 2 = 10.724x + 2x 19.448 = 12.724x x = 1.5285 Calculate the Eqm Concentrations: [H 2 = [F ]2= 2 – 1.5285 = 0.4715 M [HF] = 2 + 2(1.5285) = 5.057 M Check the Value of x: K = (5.057) /(0.4715) 2 K = 115.03, so our x value is correct. 16


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