Question 12.2: Look at the second reaction in Table 12.2, shown in progress......
Look at the second reaction in Table 12.2, shown in progress in Figure 12.4. What is the order of the reaction with respect to each of the reactants, and what is the overall reaction order?
STRATEGY
To find the reaction order with respect to each reactant, look at the exponents in the rate law, not the coefficients in the balanced chemical equation, and then sum the expo-nents to obtain the overall reaction order.
TABLE 12.2 Balanced Chemical Equations and Experimentally Determined Rate Laws for Some Reactions
| Reaction | Rate Law |
| (CH_{3})_{3}CBr (aq) + H_{2}O (aq) \longrightarrow (CH_{3})_{3}COH (aq) + H^{+} (aq) + Br^{-} (aq) | Rate = k[(CH_{3})_{3}CBr] |
| HCO_{2}H (aq) + Br_{2} (aq) \longrightarrow 2 H^{+} (aq) + 2 Br^{-} (aq) + CO_{2} (g) | Rate = k[Br_{2}] |
| BrO^{-}_{3} (aq) + 5 Br^{-} (aq) + 6 H^{+} (aq) \longrightarrow 3 Br_{2} (aq) + 3 H_{2}O (l) | Rate = k[BrO^{-}_{3}][Br^{-}][H^{+}]^{2} |
| H_{2} (g) + I_{2} (g) \longrightarrow 2 HI (g) | Rate = k[H_{2}][I_{2}] |
| CH_{3}CHO (g) \longrightarrow CH_{4} (g) + CO (g) | Rate = k[CH_{3}CHO]^{3/2} |
The rate law for the second reaction in Table 12.2 is
\text{Rate} = k [Br_{2}]Because HCO_{2}H (formic acid) does not appear in the rate law, the rate is independent of the concentration of HCO_{2}H ,and the reaction is zeroth order in HCO_{2}H . Because the exponent on [Br_{2}] (understood) is 1, the reaction is first order in Br_{2} . The reaction is first order overall because the sum of the exponents is 1.