Question

The reaction \[C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH\] is allowed to take place with initial concentrations of 0.2 mole/lit of each reactant. If the reaction mixture is diluted with water so that the initial concentration of each reactant becomes 0.1mole/lit, the rate of the reaction will be:
A. 1/8th of the original rate
B. 1/4th of the original rate
C. 1/2th of the original rate
D. Same as the original rate.

Answer 1

Hint: Rate of reaction is the change in concentration of reactants and products in unit time. Rate of reaction with respect to products is the increase in the concentration of products in unit time, Rate of reaction with respect to reactants is the decrease in the concentration of reactants in unit time.

Complete step by step answer:
According to the equation \[C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH\]rate of reaction depends on the concentration of both $C{H_3}COO{C_2}{H_5}$ and NaOH, hence it is a second order reaction.
Thus, rate law is given as $[C{H_3}COO{C_2}{H_5}]^{1}[NaOH]^{1}$, order =1+1=2.
The initial concentration of the reactant is 0.2 mol/lit.
$r_1= K[0.2]^2 = 0.4 K $
Concentration of the reactant when it is reduced to 0.1 mole/lit
$r_2= K[0.1] = 0.1 K$
$\dfrac{{r2}}{{r1}} = \dfrac{{k(0.1)}}{{k(0.4)}} = \dfrac{1}{{0.25}} = 0.25,i.e,\dfrac{1}{4}th$

Hence the correct option is option B.

Note:
Order of reaction, it is the sum of the power of the molar concentration terms of the reactants expressed in rate law expression, it can also be predicted by rate of reaction: in a reaction, if the rate of reaction depends on any one of the species then it is the first-order reaction. If it depends on two species it is a second-order reaction.