# If the initial partial pressures are PPH3 = 0.0260 atm, PP2 = 0.871 atm, PH2 = 0.517 atm, calculate Qp and determine which direction the reaction proceeds in to reach equilibrium.

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Chemistry 122 (General College Chemistry I)-‐ Fall 2018 Discussion Worksheet Week 10 (Coordinated by Dr. R. Hatfield)

Topics: equilibrium state and the equilibrium constant (K), reaction quotient (Q), expressing equilibrium with pressure terms-‐ relationship between Kc and Kp, determining reaction direction, solving a variety of equilibrium problems, Le Chatelier’s principle Relevant Chapters in Silberberg, 8e: Chapter 17 Sections 1-‐6, pp. 747-‐780. Student Directions: Try all problems with your best effort and bring your work to discussion. Use your text as a resource for any constants or relationships needed. You will have to work ahead. When topics are completely covered in lecture, go back and redo any problems you did not understand. See pages 782-‐783 of your text for a summary list of key equations and relationships. Basic Math: Review the general form for quadratic equations and solving quadratic equations. Mon., Tues., and Wed. Discussions Only: Complete up through #5 for credit this week.

1. The reaction for the formation of ammonia is shown as:

N!(g) + 3 H!(g) 2NH!(g) K!= ? a) Write the equilibrium constant expression (K!) for this reaction.

b) One equation for the breakdown of ammonia is shown as: NH!(g) !! N!(g) +

! ! H!(g) K! ’ = ?

Manipulate the reaction for ammonia formation, as necessary, to express K! ’ in terms of K! . (Hint: A summary table for ways of writing and calculating K is found on page 755 of your text).

c) Calculate the value of K! at 500. K for the formation of ammonia in part a) using the following measured concentrations for the equilibrium mixture: [N!] = 3.0 x 10-‐2 M; [H!] = 3.7 x 10-‐2 M; [NH!] = 1.6 x 10-‐2 M. [1.7 x 102]

d) Next, calculate the value of K! for the formation of ammonia at 500. K. [0.10]

e) Using the calculated value of K! at 500. K from part a as well as the expression you

derived for K! ’ in part b, calculate the value of K! ’. [7.7 x 10-‐2]

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2. Equilibrium terminology. When solving equilibrium problems, K and Q are used. Define Q. When does K = Q?

3. Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, Q!:

a) KNO!(s) KNO!(s) + O!(g)

b) HCl(g) + O!(g) H!O(l) + Cl!(g)

4. Calculate the value of the equilibrium constant at 427 °C for the reaction:

Na!O (s) +!! O! (g) Na!O!(s) Kc = ? [4000] Equilibrium constants for the following reactions are given at 427 °C: Na!O(s) 2 Na(l) + !! O!(g) Kc = 2 x 10-‐25 Na!O!(s) 2 Na(l) + O!(g) Kc = 5 x 10-‐29

5. Phosphine (PH!) decomposes at higher temperatures as follows:

2 PH! (𝑔) P! (g) + 3 H! (g) Kp = 398 at 873 K

a) If the initial partial pressures are PPH3 = 0.0260 atm, PP2 = 0.871 atm, PH2 = 0.517 atm, calculate Qp and determine which direction the reaction proceeds in to reach equilibrium. [178]

b) When a mixture of PH!, P!, and H! reaches equilibrium at 873 K, PP2 = 0.412 atm and PH2 = 0.822 atm. Find PPH3. [0.0240 atm]

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6. Iodine gas and hydrogen gas are sealed in a flask and heated to 600. K. The initial partial pressures of iodine and hydrogen gas at 600. K before they react are: PI2 =3.00 atm and PH2 =2.00 atm. No other reactions take place. Determine the equilibrium partial pressures of I!,H!, and HI at 600. K. [I! = 1.13 atm]

[H! = 0.134 atm] I!(g) + H!(g) 2 HI(g) K! = 92 at 600. K [HI = 3.73 atm]

7. In the decomposition of hydrogen sulfide: 2H!S(g) 2H!(g) + S!(g) Kc = 9.30 x 10-‐8 at 700 °C

If 0.51 mol H!S is placed in a 3.0 liter container, what is the equilibrium concentration of H!(g) at 700 °C? Can the small x approximation be used here when solving for x? [1.8 x 10-‐3 M]

8. The following reaction is endothermic: C(s) + H!O(g) CO (g) + H!(𝑔)

For a reaction mixture at equilibrium, how would the following changes affect the equilibrium (shift right, shift left, no change) and the equilibrium concentration of CO (g) (increase, decrease, no change)? Change Effect of equilibrium Effect on concentration of CO Decreasing the temperature Adding C to the mixture Adding H!O to the mixture Adding a catalyst Adding H! to the mixture Lowering the volume Adding an inert gas