Question

Consider the following reaction:

A + B + C right arrow D

The rate law for this reaction is as follows:

Rate = k time fraction [A][C]^2 over [B]^1/2

Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 1.12×10-2 M/s.


What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled?

Rate 2 = ? M/s

Answer 1

Answer: The rate of the reaction when concentrations are changed is $5.17\times 10^{-2}M/s$

Explanation:

For the given chemical reaction:

$A+B+C\rightarrow D$

The given rate law of the reaction follows:

$\text{Rate}_1=k\frac{[A][C]^2}{[B]^{1/2}}$

We are given:

$\text{Rate}_1=1.12\times 10^{-2}M/s$

When the concentrations are changed:

New concentration of A = 2A  (Concentration is doubled)

New concentration of C = 2C   (Concentration is doubled)

New concentration of B = 3B     (Concentration is tripled)

The new rate law expression becomes:

$\text{Rate}_2=k\frac{[2A][2C]^2}{[3B]^{1/2}}\\\\\text{Rate}_2=\frac{2\times 2^2}{3^{1/2}}\times (k\frac{[A][C]^2}{[B]^{1/2}})\\\\\text{Rate}_2=4.62\times (\text{Rate}_1)\\\\\text{Rate}_2=4.62\times 1.12\times 10^{-2}\\\\\text{Rate}_2=5.17\times 10^{-2}M/s$

Hence, the rate of the reaction when concentrations are changed is $5.17\times 10^{-2}M/s$