# 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

CH = 413 kJ/mol

CC = 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 Keq favor the reactants or the products? Explain your answer.
10. Is the Keq 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. $K_{\text{eq}}=\dfrac{[\ce{H2}][\ce{F2}]}{[\ce{HF}]^2}$

13.$K_{\text{eq}}=\dfrac{\ce{1}}{[\ce{O2}]}$

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