Question

Given, 100 ${ cm }^{ 3 }$ of 1 M ${ CH }_{ 3 }COOH$ was mixed with 100 ${ cm }^{ 3 }$ of 2 M ${ CH }_{ 3 }OH$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water will be?
(a) 2 times
(b) 4 times
(c) 0.5 times
(d) 0.25 times

Answer 1

Hint: The rate of esterification reaction generally depends upon the concentration of both the acid and the alcohol. Hence the order of the esterification reaction is 2.

Complete answer:
The formation of an ester is a second order reaction i.e. the rate of the reaction will depend upon the concentration of the acid (acetic acid) and the alcohol (methanol). Let the rate constant be k, hence:
${ Rate }_{ 1 }=k\left[ { CH }_{ 3 }COOH \right] \left[ { CH }_{ 3 }OH \right] $
Let the concentration of the acetic acid be a and the concentration of the methanol be b. Then the above equation will become:
${ Rate }_{ 1 }=kab$....(1)
Now if acetic acid solution and methanol are diluted by the same volume then their respective concentration will become half of their initial value i.e. $\cfrac { a }{ 2 } $ and $\cfrac { b }{ 2 } $ respectively. Hence the rate equation will become:
${ Rate }_{ 2 }=k\times \cfrac { a }{ 2 } \times \cfrac { b }{ 2 } $...(2)
Now, we will divide equation (2) by equation (1):
$\cfrac { { Rate }_{ 2 } }{ { Rate }_{ 1 } } =\cfrac { k\times \cfrac { a }{ 2 } \times \cfrac { b }{ 2 } }{ kab } =\cfrac { 1 }{ 4 } =0.25$

Hence the correct answer is (d) 0.25 times.

Note:
Always remember that for writing the equation for the rate of a reaction we should always know the order of the reaction. If there are a number of reactions in a reaction mechanism then the rate of the reaction will depend upon the reactants which are involved in the rate determining step. It should be noted that the slowest step is often regarded as the rate determining step.