Answers to Chapter 12 Study Questions

1. 1 mole $$\ce{CO2}$$

2. $$\mathrm{0.050 - 0.030\: moles/liter = 0.020\: moles/liter}$$

3. $$\mathrm{conc.\: of\: CO_2\: produced = conc.\: of\: CH_4\: used\: up}$$
$$\mathrm{conc.\: of\: CO_2\: after\: 20\: min\: = 0.050 - 0.020 = 0.030\: mol/L}$$

4. $$\mathrm{rate=\dfrac{\Delta [CO_2]}{\Delta time}}$$

5. $$\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=\dfrac{0.030-0.020\,mol/L}{20-10\,min}=\dfrac{0.0010\,mol/L}{min}}$$

6. $$\mathrm{conc.\: of\: CO_2\: after\: 30\: min = 0.050 - 0.015 = 0.035\: mol/L}$$
$$\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=\dfrac{0.035-0\,mol/L}{30-0\,min}=\dfrac{1.2\times10^{-3}\,mol/L}{min}}$$

7. $$\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=-\dfrac{\Delta [O_2]}{2\Delta t}}$$; $$\mathrm{\dfrac{\Delta [O_2]}{\Delta t}=-2\dfrac{\Delta[CO_2]}{\Delta time}}$$

8. $$\mathrm{rate=\dfrac{\Delta [O_2]}{\Delta t}=-2\dfrac{\Delta[CO_2]}{\Delta time}}$$        from e) $$\mathrm{\dfrac{\Delta [CO_2]}{\Delta time}=\dfrac{0.0010 \, mol/L}{min}}$$
$$\mathrm{\dfrac{\Delta [O_2]}{\Delta t} = -2(0.0010) = \dfrac{- 0.0020\: mol/L}{min}}$$

1. The mechanism, and therefore the number of steps in a reaction, are determined experimentally.
2. The second step; it's the slowest step.
3. $$\mathrm{rate = k \times (concentration\: NO_3) \times (concentration\: NO)}$$
4. rate increases by a factor of $$\mathrm{\dfrac{3}{2}}$$
$$\mathrm{\left(\dfrac{1}{2} \times 3 = \dfrac{3}{2}\right)}$$
5. the order with respect to $$\ce{NO}$$ is 1; the overall order is 2.

1.

1. concentration of reactants:  Reaction rate increases as concentration of reactants increases because number of collisions increases, making reaction more likely to occur.
2. surface area of reactants:  Rate increases as surface area of reactants increases because the greater the area of reactant exposed, the more likely are collisions that will result in product formation.
3. temperature:  As temperature increases, rate increases because at higher temperature, a greater proportion of reactant molecules have a kinetic energy greater than the activation energy so a greater proportion of collisions result in product formation.
4. catalyst:  Catalysts increase reaction rate by lowering the activation energy.
5. inhibitors:  Inhibitors decrease reaction rate by destroying a catalyst, reducing effective surface area or by using up reactant.

1. $$\ce{2 N2O(g) \rightarrow 2 N2(g) + O2(g)}$$
2. $$\ce{N2O(g) \rightarrow N2 + O}$$;    $$\ce{N2O(g) + O \rightarrow N2 + O2}$$
3. $$\ce{O}$$

1. $$\mathrm{rate = k [HgCl_2]^n \times [C_2O_4^{2-}]^m}$$

$$\mathrm{\dfrac{rate_2}{rate_1}=\dfrac{k[HgCl_2]_2^n}{k[HgCl_2]_1^n}=\left(\dfrac{[HgCl_2]_2}{[HgCl_2]_1}\right)^n}$$

$$\mathrm{\dfrac{2.6\times10^{-7}}{1.3\times10^{-7}}=\left(\dfrac{0.20}{0.10}\right)^n}$$

$$\mathrm{2 = 2^n}$$ ;   $$\mathrm{n = 1}$$

order with respect to $$\ce{HgCl2}$$ is 1.  (Rate is proportional to $$\ce{[HgCl2]}$$.)

$$\mathrm{\dfrac{rate_2}{rate_1}=\dfrac{k[C_2O_4^{2-}]_2^m}{k[C_2O_4^{2-}]_1^m}=\left(\dfrac{[C_2O_4^{2-}]_2}{[C_2O_4^{2-}]_1}\right)^m}$$

$$\mathrm{\dfrac{5.2\times10^{-7}}{1.3\times10^{-7}}=\left(\dfrac{0.20}{0.10}\right)^m}$$

$$\mathrm{4 = 2^m}$$ ;  $$\mathrm{m = 2}$$

order with respect to $$\ce{C_2O_4^{2-}}$$ is 2.  Overall order is 3.

1. $$\mathrm{rate = k [HgCl_2] [C_2O_4^{2-}]^2}$$
2. $$\mathrm{rate = k [HgCl_2] [C_2O_4^{2-}]^2}$$ ;  $$\mathrm{1.3 \times 10^{-7} = k (0.10)(0.10)^2}$$;  $$\mathrm{k = \dfrac{1.3 \times 10^{-7}}{0.001} = 1.3 \times 10^{-4}}$$
3. $$\mathrm{rate = 1.3 \times 10^{-4} \times [HgCl_2] [C_2O_4^{2-}]^2}$$; $$\mathrm{rate = 1.3 \times 10^{-4} \times (0.30) (0.30)^2 = \dfrac{3.5 \times 10^{-6}\:mol/L}{s}}$$