1. Chapter 13 Chemical Equilibrium

2. Copyright © Houghton Mifflin Company. All rights reserved.13–2 Figure 13.1: A molecular representation of the reaction 2NO 2 (g) N 2 O4(g) over time in a closed vessel.

3. Copyright © Houghton Mifflin Company. All rights reserved.13–3 Figure 13.2: The changes in concentrations with time for the reaction H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) when equimolar quantities of H 2 O(g) and CO(g) are mixed.

4. Copyright © Houghton Mifflin Company. All rights reserved.13–4 Figure 13.3: (a) H 2 O and CO are mixed in equal numbers and begin to react (b) to form CO2 and H2. After time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules then remain constant over time (d).

5. Copyright © Houghton Mifflin Company. All rights reserved.13–5 Figure 13.4: The changes with time in the rates of forward and reverse reactions for H 2 O(g) + CO(g)H 2 (g) + CO 2 (g) when equimolar quantities of H 2 O(g) and CO(g) are mixed. The rates do not change in the same way with time because the forward reaction has a much larger rate constant than the reverse reaction.

6. Copyright © Houghton Mifflin Company. All rights reserved.13–6 Figure 13.5: A concentration profile for the reaction N 2 (g) + 3H 2 (g)2NH 3 (g) when only N 2 (g) and H 2 (g) are mixed initially.

7. Copyright © Houghton Mifflin Company. All rights reserved.13–7

8. Copyright © Houghton Mifflin Company. All rights reserved.13–8 Figure 13.6: The position of the equilibrium CaCO 3 (s) CaO(s) + CO 2 (g) does not depend on the amounts of CaCO 3 (s) and CaO(s) present.

9. Copyright © Houghton Mifflin Company. All rights reserved.13–9 Hydrated copper (II) sulfate on the left. Water applied to anhydrous copper (II) sulfate, on the right, forms the hydrated compound.

10. Figure 13.7: (a) A physical analogy illustrating the difference between thermodynamic and kinetic stabilities. (b) The reactants H 2 and O 2 have a strong tendency to form H 2 O.

11. Copyright © Houghton Mifflin Company. All rights reserved.13–11 Apollo II lunar landing module at Tranquility Base, 1969.

12. Copyright © Houghton Mifflin Company. All rights reserved.13–12 Procedure for Solving Equilibrium Problems

13. Copyright © Houghton Mifflin Company. All rights reserved.13–13

14. Copyright © Houghton Mifflin Company. All rights reserved.13–14 Figure 13.8: (a) The initial equilibrium mixture of N 2, H 2, and NH 3. (b) Addition of N 2.(c) The new equilibrium position for the system containing more N 2 (due to addition of N 2 ), less H 2, and more NH 3 than in (a).

15. Copyright © Houghton Mifflin Company. All rights reserved.13–15 Figure 13.9: (a) A mixture of NH 3 (g), N 2 (g), and H 2 (g) at equilibrium. (b) The volume is suddenly decreased. (c) The new equilibrium position for the system containing more NH 3 and less N 2 and H 2. The reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) shifts to the right (toward the side with fewer molecules) when the container volume is decreased.

16. Figure 13.10: (a) Brown NO 2 (g) and colorless N 2 O 4 (g) in equilibrium in a syringe. (b) The volume is suddenly decreased, giving a greater concentration of both N 2 O 4 and NO 2 (indicated by the darker brown color). (c) A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts the brown NO 2 (g) to colorless N 2 O 4 (g) as predicted by Le Châtelier’s principle, since in the equilibrium 2NO 2 (g) ∆ N 2 O 4 (g) the product side has the smaller number of molecules.

17. Copyright © Houghton Mifflin Company. All rights reserved.13–17 Shifting the N 2 O 4 (g) 2NO 2 (g) equilibrium by changing the temperature. (a) At 100ºC the flask is definitely reddish brown due to a large amount of NO 2 present. (b) At 0ºC the equilibrium is shifted toward colorless N 2 O 4 (g).

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19. Copyright © Houghton Mifflin Company. All rights reserved.13–19