End of Chapter Material
Additional Exercises
- 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
- 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
- Is the following reaction endothermic or exothermic?
PCl3 (g) + Cl2 (g) ⇄ PCl5 (g) + 60 kJ
- Is the following reaction endothermic or exothermic?
N2O4 (g) + 57 kJ ⇄ 2NO2 (g)
- Determine how much energy is given off when 67.9 g of H2 reacts in the following thermochemical equation:
- Determine how much energy is given off when 75.3 g of NH3 is produced in the following thermochemical equation:
- Does entropy increase or decrease for the following reaction?
PCl3 (g) + Cl2 (g) ⇄ PCl5(g) + 60 kJ
- Does entropy increase or decrease for the following reaction?
N2O4 (g) + 57 kJ ⇄ 2NO2 (g)
- Does a very large Keq favor the reactants or the products? Explain your answer.
- Is the Keq for reactions that favor reactants large or small? Explain your answer.
- Write the equilibrium expression for: 2 HF (g) ⇄ H2(g) + F2(g)
- Write the equilibrium expression for: 4 NH3 (g) +7 O2 (g) ⇌ 4 NO2 (g) + 6 H2O (g)
- Write the equilibrium expression for: 2 PbO (s) + O2 (g) ⇄ 2 PbO2 (s)
- How will the equation shift if more HF (g) is added? 2 HF (g) ⇄ H2(g) + F2(g)
- How will the equation shift if more F2 (g) is added? 2 HF (g) ⇄ H2(g) + F2(g)
- How will the equation shift if the pressure is increased? 2 HF (g) ⇄ H2(g) + F2(g)
- How will the equation shift if the pressure is reduced? 2 HF (g) ⇄ H2(g) + F2(g)
- How will the equation shift if the temperature is increased? N2O4 (g) + 57 kJ ⇄ 2NO2 (g)
- How will the equation shift if the temperature is decreased? N2O4 (g) + 57 kJ ⇄ 2NO2 (g)
Answers
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. [latex]K_{\text{eq}}=\dfrac{[\ce{H2}][\ce{F2}]}{[\ce{HF}]^2}[/latex]
13.[latex]K_{\text{eq}}=\dfrac{\ce{1}}{[\ce{O2}]}[/latex]
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