Question

A $1d{m^3}$ of 2M $C{H_3}COOH$ is mixed with $1d{m^3}$ of 3M ethanol to form ester. The decrease in the initial rate if each solution is diluted with an equal volume of water would be:
A.2 times
B.4 times
C.6 times
D.5 times

Answer 1

For a reaction given by: $aA + bB \to cC + dD$
Where a, b, c, and d are the stoichiometric coefficients of the reactants or products, the rate equation for the reaction is given by:
Rate $\propto {[A]^x}{[B]^y} \to Rate = k{[A]^x}{[B]^y}$
When the equal volume of two solutions is mixed, concentration of the solutions reduces to half the initial value. Hence, the rate of reaction gets reduced to $\dfrac{1}{4}$ initial rate.

Complete step by step answer:
Given in the question is: 2 moles of acetic acid are present in 1 litre while 3 moles of ethanol are present in 1 litre. So, the rate for the reaction is
Esterification is a second order reaction $C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O$.
Let initial rate, ${r_1} = k \times \left[ a \right] \times \left[ b \right]$
${r_1} = k[C{H_3}COOH][{C_2}{H_5}OH]$
${r_1} = k \times 1 \times 1.5 = 1.5k$
When the solution is diluted with an equal amount of ${H_2}O$, concentration is reduced to half
Let final rate, ${r_2} = k \times \left[ {\dfrac{a}{2}} \right] \times \left[ {\dfrac{b}{2}} \right]$
${r_2} = k \times 0.5 \times 0.75 = 0.375k$
Hence, $\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{1.5k}}{{0.375k}} = 4$
$\dfrac{{{r_1}}}{{{r_2}}} = 4$

So, the rate is reduced by four times

Note: A rate law is an expression which relates the rate of a reaction to the rate constant and the concentrations of the reactants. A rate constant, k, is a proportionality constant for a given reaction. The general rate law is usually expressed as, Rate $= k{\left[ A \right]^s}{\left[ B \right]^t}$. The reaction rate is dependent on the concentration of the reactants as well as the rate constant. However, there are also other factors that can influence the rate of reaction. These factors include temperature and catalysts.