The following reaction is a single-step bimolecular reaction: CH 3 Br + NaOH ---

ID: 808675 • Letter: T

Question

The following reaction is a single-step bimolecular reaction: CH3Br + NaOH ----> CH3OH + NaBr

When the concentrations of CH3Br and NaOH are both 0.150 M, the rate of the reaction is 0.0080 M/s.

a. What is the rate of the reaction if the concentration of CH3Br is doubled? ________M/s

b. What is the rate of the reaction if the concentration of NaOH is halved? ________ M/s

c. What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of five? __________M/s

Answer and explanation please.

rate of reaction = k*[ CH3Br]*[NaOH]

Now, 0.008 = k*0.15*0.15

or, rate constant, k = 0.3556 M-1 s-1

1) rate of reaction = k*[ CH3Br]*[NaOH] = 0.3556*0.3*0.15 = 0.016

2) rate of reaction = k*[ CH3Br]*[NaOH] = 0.3556*0.15*0.075 = 4*10-3

3) rate of reaction = k*[ CH3Br]*[NaOH] = 0.3556*(5*0.15)*(5*0.15) = 0.200025

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