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Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K= 16a³(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº=-RT.In(K): (16a² (2-a)²) 27(1-a)* you can solve for a given that AG°= -32.9 kJ/mol. Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following equation: (1-1.497) a² +2 1.497a1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: Solutions: x = 0.550263 -4,Gº RT x = 5,47388 There are two answers since this is a quadratic equation. Now do the same for: -A-Gº In (16a²(2-a)²) 27(1-a)¹ to solve for a. You will also get two answers; make sure you pick the RT right one. = If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K 16a²(2-a)² p=2 27(1-a) p² and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a)-1 if a-0 and P(a)-0.5 if a-1?

Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K= 16a³(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº=-RT.In(K): (16a² (2-a)²) 27(1-a)* you can solve for a given that AG°= -32.9 kJ/mol. Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following equation: (1-1.497) a² +2 1.497a1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: Solutions: x = 0.550263 -4,Gº RT x = 5,47388 There are two answers since this is a quadratic equation. Now do the same for: -A-Gº In (16a²(2-a)²) 27(1-a)¹ to solve for a. You will also get two answers; make sure you pick the RT right one. = If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K 16a²(2-a)² p=2 27(1-a) p² and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a)-1 if a-0 and P(a)-0.5 if a-1?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K = 16a²(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº = -RT.In(K): (16a² (2-a)²) 27(1-a)* Solutions: x = 0.550263 In you can solve for a given that AG = -32.9 kJ/mol. equation: (1-1.497) a² +2 Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following 1.497a- 1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: -A,Gº RT right one. x = 5.47388 There are two answers since this is a quadratic equation. Now do the same for: -4,Gº In (16a²(2-a)²) 27(1-a)4 - to solve for a. You will also get two answers; make sure you pick the RT If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K = 16a²(2-a)² poz 27(1-a) p2 and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a) = 1 if a=0 and P(a)=0.5 if a = 1?
Transcribed Image Text:Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K = 16a²(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº = -RT.In(K): (16a² (2-a)²) 27(1-a)* Solutions: x = 0.550263 In you can solve for a given that AG = -32.9 kJ/mol. equation: (1-1.497) a² +2 Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following 1.497a- 1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: -A,Gº RT right one. x = 5.47388 There are two answers since this is a quadratic equation. Now do the same for: -4,Gº In (16a²(2-a)²) 27(1-a)4 - to solve for a. You will also get two answers; make sure you pick the RT If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K = 16a²(2-a)² poz 27(1-a) p2 and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a) = 1 if a=0 and P(a)=0.5 if a = 1?
N2(g) + 3H₂(g) → 2NH3(g) How much does this reaction proceeded if you start with n moles of nitrogen and 3n moles of hydrogen? Also, let's assume that the reactor maintains a pressure of 1 bar like the reaction is in a balloon, and that the temperature is maintained at 25 °C. Hint: We use the table below to calculate mole fractions and partial pressures: N₂ n(1-a) n(1-a) 4n-2na (1-a) 4- 2a Amount at equilibrium Mole fractions Partial pressures P K H₂ 3n(1-a) 3n(1-a) 4n-2na 3(1-a) 4-2a P NH3 2na 2na 4n-2na 2α 4- 2a "₁₂ where a represents the extent of the reaction. K=II, where P is the partial pressure of species i. As usual, Pº is just 1 bar. Now we can show: -P 2a (420P)² 4- 2a Po. ((1-2). P) (³(1-za). P.) ³ 4 2a 4 2a Simplification gives: 2α 4-2α P 4-2α 3 (+²2 F-)² (1-²8 - F) (+-²a - F-)³ = 16a²(2-a)³² pm² 4-2α 3(1-a) 27(1-a)* P2¹
Transcribed Image Text:N2(g) + 3H₂(g) → 2NH3(g) How much does this reaction proceeded if you start with n moles of nitrogen and 3n moles of hydrogen? Also, let's assume that the reactor maintains a pressure of 1 bar like the reaction is in a balloon, and that the temperature is maintained at 25 °C. Hint: We use the table below to calculate mole fractions and partial pressures: N₂ n(1-a) n(1-a) 4n-2na (1-a) 4- 2a Amount at equilibrium Mole fractions Partial pressures P K H₂ 3n(1-a) 3n(1-a) 4n-2na 3(1-a) 4-2a P NH3 2na 2na 4n-2na 2α 4- 2a "₁₂ where a represents the extent of the reaction. K=II, where P is the partial pressure of species i. As usual, Pº is just 1 bar. Now we can show: -P 2a (420P)² 4- 2a Po. ((1-2). P) (³(1-za). P.) ³ 4 2a 4 2a Simplification gives: 2α 4-2α P 4-2α 3 (+²2 F-)² (1-²8 - F) (+-²a - F-)³ = 16a²(2-a)³² pm² 4-2α 3(1-a) 27(1-a)* P2¹
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