Ncert Exercises & Solutions

Compounds A and B react according to the following chemical equation.

A(g) + 2B(g) $\overset{}{\to}$ 2C(g)

The concentration of either A or B was changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. The following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment	Initial concentration of [A]/mol L^–1	Initial concentration of [B]/moI L^–1	Initial rate (mol L^–1 s^–1)
1. 2. 3.	0.30 0.30 0.60	0.30 0.60 0.30	0.10 0.40 0.20

Rate = k {[A]}^{2} [B]

Rate = k [A] {[B]}^{2}

Rate = k [A] [B]

Rate = k {[A]}^{2} {[B]}^{0}

HINT: Find out the effect of change in rate by changing concentrations of reactant.

Step 1:

The rate of reaction is the change in concentration of reactant with respect to time.

$\begin{aligned} r & = k [A]^{x} [B]^{γ} \\ \frac{Rate of \exp . 1}{Rate of \exp . 2} & = \frac{[0 .30]^{x} [0 .30]^{y}}{[0 .30]^{x} [0 .60]^{y}} \\ \frac{0 .10}{0 .40} & = \frac{[0 .30]^{γ}}{[0 .60]^{y}} \\ \frac{1}{4} & = {[\frac{1}{2}]}^{γ} \\ {[\frac{1}{2}]}^{2} & = {[\frac{1}{2}]}^{γ} \\ y & = 2 \end{aligned}$
$\begin{aligned} \end{aligned}$

Step 2:

$\begin{aligned} \frac{Rate of \exp . 1}{Rate of \exp . 3} & = \frac{[0 .30]^{x} [0 .30]^{y}}{[0 .60]^{x} [0 .30]^{y}} \\ \frac{0 .10}{0 .20} & = {[\frac{0 .30}{0 .60}]}^{x} {[\frac{0 .30}{0 .30}]}^{y} \\ \frac{1}{2} & = {[\frac{1}{2}]}^{x} [1]^{y} \\ \frac{1}{2} & = {[\frac{1}{2}]}^{x} \\ i . e ., x & = 1 \\ ∴ Rate & = k [A]^{x} [B]^{y} \\ Rate & = k [A]^{1} [B]^{2} \end{aligned}$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions