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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.  

Graph 3.png

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