Question

The reaction of methylacetate with water is shown below:

CH3CO2CH3 (aq) + H2O (l) ? CH3CO2H (aq) + CH3OH (aq)

The rate of the reaction is given by the following rate law: rate = k[H+][CH3CO2CH3] Consider one liter of solution that is 0.15 M CH3CO2CH3 (aq) and 0.015 M H+ (aq) at 25?C (a) For each change listed below state whether the rate of reaction increase, decreases or remains the same. Why?

(i) methylacetate (CH3CO2CH3) is added to the solution

2. The temperature is lowered

3 Water is added to the solution

For those 3 cases does the K increase or remain the same?

Answer 1

The balanced reaction is

CH3CO2CH3 (aq) + H2O (l) = CH3CO2H (aq) + CH3OH (aq)

rate of reaction = k[H+][CH3CO2CH3]

Ans 1

Part a

methylacetate (CH3CO2CH3) is added to the solution

Increases the rate of reaction because the rate is first order with respect to CH3CO2CH3.

Part b

The temperature is lowered

Decreases the rate of reaction because rate is proportional to exp(-1/T)

Part c

Water is added to the solution

Decreases the rate of reaction because the more water added the more solution will dilute.

Ans 2

Higher the reaction cate, higher the rate constant because rate of reaction is directly proportional to rate constant.

Part a

methylacetate (CH3CO2CH3) is added to the solution

K will increase because of Increases the rate of reaction because the rate is first order with respect to CH3CO2CH3.

Part b

The temperature is lowered

K will decrease because of Decreases the rate of reaction because rate is proportional to exp(-1/T)

Part c

Water is added to the solution

K will decrease because Decreases the rate of reaction because the more water added the more solution will dilute.