Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

Let the order of reaction with respect to A is x and with respect to B is y. Thus, rate= k[A]^x [B]^y
For the given cases,
(I) rate= k(0.1)^x (0.1)^y=6.0x10^-3
(II) rate= k(0.3)^x (0.2)^y=7.2x10^-2
(III) rate= k(0.3)^x (0.40)^y=2.88x10^-1
(IV) rate= k(0.4)^x (0.1)^y=2.40x10^-2

On dividing eq. (II) by (III), we get
(0.3/0.3)^x (0.2/0.4)^y = 7.2X10^-2/2.88X10^-1
or (1/2)^y=1/4
or (1/2)^y=(1/2)²
          y=2

Thus, rate law is, rate= k[A]¹[B]² or =k[A][B]²

For the reaction,

A+B$\to$Products

On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore 

       Rate$\propto$[A]¹        ....(i)

Let initially rate law is

      Rate = k[A][B]^y  ...(ii)

If concentration of A and B both are doubled, the rate gets changed by a factor of 8.

         8 x Rate = k[2A][2B]^y   ...(iii) [._.. Rate$\propto$

Key Idea Rate of disappearance of reactant

= - 1/stoichiometry coefficient of the reactant x change in conc of reactant / time taken

Rate of appearance of product

= 1/stoichiometry coefficient  x change in conc of product / time taken

For the reaction, 

N₂+3H₂ $\to$ 2NH₃

Rate = -[N₂]/dt = -1/3d[H₂]/dt = +1/2d[NH₃]/dt

or -1/3d[H₂]/dt = +1/2d[NH₃]/dt

-d[H₂]/dt = 3/2 x 2x10^-4 molL^-1s^-1

=3x10^-4 molL^-1s^-1

Key Idea: Rate of appearance/disappearance 

= +1/stoichiometric coefficient x [reactant or product ] / time taken 

For reacton, 

BrO^-₃(aq) + 5Br^-(aq) + 6H⁺ $\to$3Br₂(l) + 3H₂O(l)

Rate of appearance of bromine (Br₂) = +1/3 d[Br₂]/dt

Rate of diappearance of bromide ion (Br^-) = -1/5d[Br^-]/dt

or d[Br₂]/dt = (-3/5)d[Br^-]/dt

(d) Key Idea: The Arrhenius equation is represented as

                  $\mathrm{k} = {\mathrm{Ae}}^{\raisebox{1ex}{$-{\mathrm{E}}_{\mathrm{a}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{RT}$}\right.}$

In the given equations, first take log and then compare them.

                  $$\begin{gathered} {\mathrm{k}}_1 = {10}^{16}{\mathrm{e}}^{\raisebox{1ex}{$-2000$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.} \\ {\mathrm{k}}_2 = {10}^{15}{\mathrm{e}}^{\raisebox{1ex}{$-1000$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.} \end{gathered}$$

On taking log, we get

$$\begin{gathered} \log {\mathrm{k}}_1 = \log {10}^{16} - \frac{2000}{2.303\mathrm{T}} ...\left(\mathrm{i}\right) \\ \log {\mathrm{k}}_2 = \log {10}^{15} - \frac{1000}{2.303\mathrm{T}} ...\left(\mathrm{ii}\right) \\ \because {\mathrm{k}}_1={\mathrm{k}}_2 \\ \mathrm{Hence}, \mathrm{from} \mathrm{EQS} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{T} = \frac{1000}{2.303}\mathrm{k} \end{gathered}$$

(a) Key Idea : By comparing the rate andconcentration, the order of the reaction can be
calculated.
Let the rate of the reaction Wrt [CH₃COCH_3], [Br₂] and [H⁺] are x, y and z respectively. Thus,

        Rate $\propto [{\mathrm{CH}}_3{\mathrm{COCH}}_3{]}^{\mathrm{x}} [{\mathrm{Br}}_2{]}^{\mathrm{y}}{\left[{\mathrm{H}}^{+}\right]}^{\mathrm{z}}$

     5.7 x 10^-5 = [0.30]x [0.05]y [0.05]z    ...(i)
     5.7 x 10^-5 = [0.30]x [0.10]y [0.05]z    ...(ii)
     1.2 × 10^-4 = [0.30]x [0.10]y [0.10]z   ...(iii)
     3.1 x 10-4 = [0.40]x [0.05]y [0.20]z   ...(iv)
From Eqs (i) and (ii)

                            y = 0
From eqs (ii) and (iii)

                            z = 1
from eqs (i) and (iv)
                            x = 1

Thus, rate law $\propto$ [CH₃COCH₃] [H⁺]
                       = k [CH₃COCH₃] [H⁺]

(d) For first order reaction.

$$\begin{gathered} k=\frac{2.303}{t}\log \left(\frac{\mathrm{a}}{\left(\mathrm{a}-\mathrm{x}\right)}\right) \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2.303}{k}\log \left(\frac{\mathrm{a}}{\mathrm{a}-{\displaystyle \frac{\mathrm{a}}{2}}}\right) \\ = \frac{2.303}{k}\log \left(2\right) \\ \mathrm{or} \frac{\ln 2}{k} \end{gathered}$$

For the reaction 2A + B → 3C + D

The reaction rate is written as follows
The reaction rate w.r.t. A = $-\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$
The reaction rate w.r.t B = $-\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}$
The reaction rate w.r.t. C = $+\frac{1}{3}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}$
The reaction rate w.r.t. D = $\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}$
Hence, the answer (a) is not correct expression for represent the rate of reaction.

For the reaction

 N₂(g) + 3H₂(g) $\to$2NH₃(g)

The rate of reaction w.r.t N₂ = -d[N₂]/dt

The rate of reaction w.r.t H₂ = -1/3.d[H₂]/dt

The rate of reaction w.r.t NH₃ = =1/2.d[NH₃]/dt

Hence, at a fixed time

           -d[N₂]/dt = -1/3. d[H₂]/dt

                          = +1/2.d[NH₃]/dt

or          =d[NH₃]/dt = -2/3.d[H₂]/dt

or                            = -2d[N₂]/dt