Trimethyl bromomethane reacts with sodium hydroxide to give trimethyl methanol and sodium bromide according to the reaction shown below.
(CH3)3CBr + NaOH → (CH3)3COH + NaBr
A student studying this reaction produces the following data.
| Trial | [NaOH]t=0 | [(CH3)3CBr]t=0 |
Rate t=0 (M/s) |
|
1 |
0.1 M | 0.5 M | 5 x 10-4 |
|
2 |
0.2 M | 1.0 M |
1 x 10-3 |
| 3 | 0.1 M | 4.0 M |
4 x 10-3 |
What is the rate law for the reaction above?
A. Rate = k[A]
B. Rate = k[(CH3)3CBr]
C. Rate = k[A][(CH3)3CBr]
D. Rate = k[A][(CH3)3CBr]2
Click for Explanation
To determine the rate law, you must compare trials to see how different initial reactant concentrations affect the initial rate of the reaction. Comparing trial 1 and trial 3, the initial rate in trial 3 increased by a factor of 8. Since [NaOH] is constant and [(CH3)3CBr] increases by a factor of 8, the reaction rate is first order with respect to (CH3)3CBr.
Comparing trial 1 and trial 2, the initial rate of trial 2 increases by a factor 2. Since [(CH3)3CBr] also increases by a factor of 2 and the reaction rate is first order with respect to (CH3)3CBr, the increase in the reaction rate is due to [(CH3)3CBr] alone. Thus, the reaction is zero order with respect to NaOH. Thus, the rate law can be described by answer choice B.
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