Question 15.1: Determining a Rate Law The reaction between bromate ions and...
Determining a Rate Law
The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the following equation:
BrO_{3}^{-} (aq) + 5 Br^{-} (aq) + 6 H^{+} (aq) → 3Br _{2} (l) + 3 H_{2} O (l)
Table 15.5 gives the results of four experiments involving this reaction. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant.
Table 15.5
The Results from Four Experiments to Study the Reaction
BrO_{3}^{-} (aq) + 5 Br^{-} (aq) + 6 H^{+} (aq) → 3Br _{2} (l) + 3 H_{2} O (l)
|
Experiment |
Initial Concentration of BrO_{3}^{-} (mol / L) |
Initial Concentration of Br^{-} (mol / L) |
Initial Concentration of H^{+} (mol / L) |
Measured Initial Rate (mol L^{-1} s ^{-1}) |
| 1 | 0.10 | 0.10 | 0.10 | 8.0 × 10 ^ {-4} |
| 2 | 0.20 | 0.10 | 0.10 | 1.6 × 10 ^ {-3} |
| 3 | 0.20 | 0.20 | 0.10 | 3.2 × 10 ^ {-3} |
| 4 | 0.10 | 0.10 | 0.20 | 3.2 × 10 ^ {-3} |
The general form of the rate law for this reaction is
Rate = k [BrO_{3}^{-} ] ^{n} [ Br^{-} ] ^ {m} [H^{+} ]^{p}
We can determine the values of n, m, and p by comparing the rates from the various experiments. To determine the value of n, we use the results from Experiments 1 and 2, in which only [BrO_{3}^{-} ] changes:
\frac{Rate 2}{Rate 1} = \frac{ 1.6 × 10 ^ {-3} mol L^{-1} s ^{-1}}{ 8.0 × 10 ^ {-4} mol L^{-1} s ^{-1}}
= \frac{k (0.20 mol / L) ^{n} (0.10 mol / L) ^ {m}(0.10 mol / L)^{p}}{k (0.10 mol / L) ^{n} (0.10 mol / L) ^ {m}(0.10 mol / L)^{p}}
2.0 = ( \frac{(0.20 mol / L) }{(0.10 mol / L) } ) ^{n} = (2.0) ^{n}
Thus n is equal to 1.
To determine the value of m, we use the results from Experiments 2 and 3, in which only [ Br^{-} ] changes:
\frac{Rate 3}{Rate 2} = \frac{ 3.2 × 10 ^ {-3} mol L^{-1} s ^{-1}}{ 1.6 × 10 ^ {-3} mol L^{-1} s ^{-1}}
= \frac{k (0.20 mol / L) ^{n} (0.20 mol / L) ^ {m}(0.10 mol / L)^{p}}{k (0.20 mol / L) ^{n} (0.10 mol / L) ^ {m}(0.10 mol / L)^{p}}
2.0 = ( \frac{(0.20 mol / L) }{(0.10 mol / L) } ) ^{m} = (2.0) ^{m}
Thus m is equal to 1.
To determine the value of p, we use the results from Experiments 1 and 4, in which [BrO_{3}^{-} ] and [ Br^{-} ] are constant but [H^{+} ] changes:
\frac{Rate 4}{Rate 1} = \frac{ 3.2 × 10 ^ {-3} mol L^{-1} s ^{-1}}{ 8.0 × 10 ^ {-4} mol L^{-1} s ^{-1}}
= \frac{k (0.10 mol / L) ^{n} (0.10 mol / L) ^ {m}(0.20 mol / L)^{p}}{k (0.10 mol / L) ^{n} (0.10 mol / L) ^ {m}(0.10 mol / L)^{p}}
4.0 = ( \frac{(0.20 mol / L) }{(0.10 mol / L) } ) ^{p}
4.0 = (2.0) ^{p} = (2.0 )²
Thus p is equal to 2.
The rate of this reaction is first order in BrO_{3}^{-} and Br^{-} and second order in H^{+} . The overall reaction order is n + m + p = 4.
The rate law can now be written:
Rate = k [BrO_{3}^{-} ] [ Br^{-} ] [H^{+} ]²
The value of the rate constant k can be calculated from the results of any of the four experiments. For Experiment 1 the initial rate is 8.0 × 10 ^ {-4} mol L^{-1} s ^{-1} , and [BrO_{3}^{-} ] = 0.100 M , [ Br^{-} ] = 0.100 M , and [H^{+} ] = 0.100 M. Using these values in the rate law gives
8.00 × 10 ^ {-4} mol L^{-1} s ^{-1} = k (0.10 mol / L) (0.10 mol / L) (0.10 mol / L)²
8.00 × 10 ^ {-4} mol L^{-1} s ^{-1} = k (1.0 × 10 ^ {-4} mol^{4} / L ^{4})
k = \frac{ 8.0 × 10 ^ {-4} mol L^{-1} s ^{-1} }{1.0 × 10 ^ {-4} mol^{4} / L ^{4}}
= 8.00 L³ mol^{-3} s ^{-1}
Check: Verify that the same value of k can be obtained from the results of the other experiments.