Answers to Chapter 12 Study Questions
1. a) 1 mole CO2
b) 0.050 - 0.030 moles/liter = 0.020 moles/liter
c) conc. of CO2 produced = conc. of CH4 used up
conc. of CO2 after 20 min = 0.050 - 0.020 = 0.030 mol/L
f) conc. of CO2 after 30 min = 0.050 - 0.015 = 0.035 mol/L
= -2(0.0010) = - 0.0020 mol/L/min
2 a) The mechanism, and therefore the number of steps in a reaction, are determined experimentally.
b) The second step; it's the slowest step.
c) rate = k x (concentration NO3) x (concentration NO)
d) rate increases by a factor of (1/2 x 3 = 3/2) 3/2
e) the order with respect to NO is 1; the overall order is 2.
4. 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.
5. a) 2 N2O(g) ® 2 N2(g) + O2(g)
b) N2O(g) ® N2 + O; N2O(g) + O ® N2 + O2
6. a) rate = k [HgCl2]n x [C2O42¯]m
2 = 2n ; n = 1
order with respect to HgCl2 is 1. (Rate is proportional to [HgCl2].)
4 = 2m ; m = 2
order with respect to C2O42- is 2. Overall order is 3.
b) rate = k [HgCl2] [C2O42-]2
c) rate = k [HgCl2] [C2O42-]2 ; 1.3 x 10-7 = k (0.10)(0.10)2; k = 1.3 x 10-7/0.001 = 1.3 x 10-4
d) rate = 1.3 x 10-4 x [HgCl2] [C2O42-]2; rate = 1.3 x 10-4 x (0.30) (0.30)2 = 3.5 x 10-6 mol/L/s