Answers to Chapter 12 Study Questions
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 11188

 1 mole \(\ce{CO2}\)
 \(\mathrm{0.050  0.030\: moles/liter = 0.020\: moles/liter}\)
 \(\mathrm{conc.\: of\: CO_2\: produced = conc.\: of\: CH_4\: used\: up}\)
\(\mathrm{conc.\: of\: CO_2\: after\: 20\: min\: = 0.050  0.020 = 0.030\: mol/L}\)
 \(\mathrm{rate=\dfrac{\Delta [CO_2]}{\Delta time}}\)
 \(\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=\dfrac{0.0300.020\,mol/L}{2010\,min}=\dfrac{0.0010\,mol/L}{min}}\)
 \(\mathrm{conc.\: of\: CO_2\: after\: 30\: min = 0.050  0.015 = 0.035\: mol/L}\)
\(\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=\dfrac{0.0350\,mol/L}{300\,min}=\dfrac{1.2\times10^{3}\,mol/L}{min}}\)
 \(\mathrm{rate=\dfrac{\Delta[CO_2]}{\Delta time}=\dfrac{\Delta [O_2]}{2\Delta t}}\); \(\mathrm{\dfrac{\Delta [O_2]}{\Delta t}=2\dfrac{\Delta[CO_2]}{\Delta time}}\)
 \(\mathrm{rate=\dfrac{\Delta [O_2]}{\Delta t}=2\dfrac{\Delta[CO_2]}{\Delta time}}\) 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 mole \(\ce{CO2}\)

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

 concentration of reactants: Reaction rate increases as concentration of reactants increases because number of collisions increases, making reaction more likely to occur.
 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.
 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.
 catalyst: Catalysts increase reaction rate by lowering the activation energy.
 inhibitors: Inhibitors decrease reaction rate by destroying a catalyst, reducing effective surface area or by using up reactant.

 \(\ce{2 N2O(g) \rightarrow 2 N2(g) + O2(g)}\)
 \(\ce{N2O(g) \rightarrow N2 + O}\); \(\ce{N2O(g) + O \rightarrow N2 + O2}\)
 \(\ce{O}\)

 \(\mathrm{rate = k [HgCl_2]^n \times [C_2O_4^{2}]^m}\)
\(\mathrm{\dfrac{rate_2}{rate_1}=\dfrac{k[HgCl_2]_2^n}{k[HgCl_2]_1^n}=\left(\dfrac{[HgCl_2]_2}{[HgCl_2]_1}\right)^n}\)
\(\mathrm{\dfrac{2.6\times10^{7}}{1.3\times10^{7}}=\left(\dfrac{0.20}{0.10}\right)^n}\)
\(\mathrm{2 = 2^n}\) ; \(\mathrm{n = 1}\)
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.
 \(\mathrm{rate = k [HgCl_2] [C_2O_4^{2}]^2}\)
 \(\mathrm{rate = k [HgCl_2] [C_2O_4^{2}]^2}\) ; \(\mathrm{1.3 \times 10^{7} = k (0.10)(0.10)^2}\); \(\mathrm{k = \dfrac{1.3 \times 10^{7}}{0.001} = 1.3 \times 10^{4}}\)
 \(\mathrm{rate = 1.3 \times 10^{4} \times [HgCl_2] [C_2O_4^{2}]^2}\); \(\mathrm{rate = 1.3 \times 10^{4} \times (0.30) (0.30)^2 = \dfrac{3.5 \times 10^{6}\:mol/L}{s}}\)