End of Chapter Material

1. Use bond energies to estimate the energy change of this reaction.

   C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

   Note: C₃H₈ is made up of two C-C bonds and 8 C-H bonds.

   Bond Energies

   C−H = 413 kJ/mol

   C−C = 348 kJ/mol

   O=O = 495 kJ/mol

   O-H = 463 kJ/mol

2. Use bond energies to estimate the energy change of this reaction.

   N₂H₄ + O₂ → N₂ + 2H₂

   Bond Energies

   N-H = 391 kJ/mol

   O=O = 495 kJ/mol

   N=N = 945 kJ/mol

   O-H = 463 kJ/mol

3. Is the following reaction endothermic or exothermic?

   PCl₃ (g) + Cl₂ (g) ⇄ PCl₅ (g) + 60 kJ

4. Is the following reaction endothermic or exothermic?

   N₂O₄ (g) + 57 kJ ⇄ 2NO₂ (g)

5. Determine how much energy is given off when 67.9 g of H₂ reacts in the following thermochemical equation:
6. Determine how much energy is given off when 75.3 g of NH₃ is produced in the following thermochemical equation:
7. Does entropy increase or decrease for the following reaction?

   PCl₃ (g) + Cl₂ (g) ⇄ PCl₅(g) + 60 kJ

8. Does entropy increase or decrease for the following reaction?

   N₂O₄ (g) + 57 kJ ⇄ 2NO₂ (g)

9. Does a very large K_eq favor the reactants or the products? Explain your answer.
10. Is the K_eq for reactions that favor reactants large or small? Explain your answer.
11. Write the equilibrium expression for:  2 HF (g) ⇄ H₂(g) + F₂(g)
12. Write the equilibrium expression for:  4 NH₃ (g) +7 O₂ (g) ⇌ 4 NO₂ (g) + 6 H₂O (g)
13. Write the equilibrium expression for:  2 PbO (s) + O₂ (g) ⇄ 2 PbO₂ (s)
14. How will the equation shift if more HF (g) is added?  2 HF (g) ⇄ H₂(g) + F₂(g)
15. How will the equation shift if more F₂ (g) is added?  2 HF (g) ⇄ H₂(g) + F₂(g)
16. How will the equation shift if the pressure is increased?  2 HF (g) ⇄ H₂(g) + F₂(g)
17. How will the equation shift if the pressure is reduced?  2 HF (g) ⇄ H₂(g) + F₂(g)
18. How will the equation shift if the temperature is increased? N₂O₄ (g) + 57 kJ ⇄ 2NO₂ (g)
19. How will the equation shift if the temperature is decreased? N₂O₄ (g) + 57 kJ ⇄ 2NO₂ (g)

1. −2,023 kJ

3. Exothermic

5. -1030 kJ

7.. Entropy decreases as 2 moles of gas go to 1 mole of gas

9. Favors products because the numerator of the ratio for the Keq is larger than the denominator

11. $K_{\text{eq}}={\displaystyle \frac{[\mathrm{H}{\phantom{A}}_2][\mathrm{F}{\phantom{A}}_2]}{[\mathrm{HF}{]}^2}}$

13.$K_{\text{eq}}={\displaystyle \frac{1}{[\mathrm{O}{\phantom{A}}_2]}}$

15. The reaction will shift to the left

17. There will be no impact on the reaction since there are 2 moles of gas on both sides of the equation

19. The reaction will shift to the left