1. ⇒ (MHT CET 2023 12th May Evening Shift )

The rate law for the reaction $A + B \to C$ at $25^{\circ} C$ is given by rate $= k [A] [B]^{2}$ . Calculate the rate of reaction if rate constant at same temperature is $6.25 {mol}^{- 2} {dm}^{6} s^{- 1} [[A] = 1 M, [B] = 0.2 M]$

A. $0.25 mol {dm}^{- 3} s^{- 1}$

B. $0.5 mol {dm}^{- 3} s^{- 1}$

C. $0.75 mol {dm}^{- 3} s^{- 1}$

D. $1.25 mol {dm}^{- 3} s^{- 1}$

Correct Option is (A)

$\begin{aligned} Rate & = k [A] [B]^{2} \\ = 6.25 {mol}^{- 2} {dm}^{6} s^{- 1} \times 1 mol {dm}^{- 3} \\ \times 0.2 mol {dm}^{- 3} \times 0.2 mol {dm}^{- 3} \\ = 0.25 mol {dm}^{- 3} s^{- 1} \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

Identify rate law expression for $2 {NO}_{(g)} + {Cl}_{2 (g)} \to 2 {NOCl}_{(g)}$ if the reaction is second order in $NO$ and first order in ${Cl}_{2}$ .

A. Rate $= k [NO]^{2} [{Cl}_{2}]$

B. Rate $= k [NO] [{Cl}_{2}]$

C. $Rate = k [NO]^{2}$

D. Rate $= k [{Cl}_{2}]$

Correct Option is (A)

Currently no explanation available

3. ⇒ (MHT CET 2023 11th May Evening Shift )

For a reaction $A + B \to$ product, if $[A]$ is doubled keeping $[B]$ constant, the rate of reaction doubles. Calculate the order of reaction with respect to A.

A. 0

B. 1/2

C. 1

D. 2

Correct Option is (C)

$\begin{array}{ll} A + B ⟶ Product \\ Rate = k [A]^{x} [B]^{y} \\ {Rate}_{1} = k [2 A]^{x} [B]^{y} \\ ∴ & \frac{{Rate}_{1}}{Rate} = \frac{k 2^{x} [A]^{x} [B]^{y}}{k [A]^{x} [B]^{y}} \end{array}$

$\begin{array}{ll} ∴ & \frac{2 \times Rate}{Rate} = 2^{x} \\ ∴ & 2^{x} = 2 \\ ∴ & x = 1 \end{array}$

$∴$ Order of reaction with respect to A = 1

4. ⇒ (MHT CET 2023 11th May Morning Shift )

What is half life time of a first order reaction if initial conc. of reactant is $0.01 mol L^{- 1}$ and rate of reaction is $0.00352 mol L^{- 1}$ minute $^{- 1}$ ?

A. 1.969 minute

B. 16.69 minute

C. 16.69 minute

D. 19.69 minute

Correct Option is (A)

For a first order reaction, rate $= k [A]$

$∴ 0.00352 mol L^{- 1}$ minute $^{- 1} = k \times 0.01 mol L^{- 1}$

$k = 0.352$ minute $^{- 1}$

Half-life period, $(t_{1 / 2}) = \frac{0.693}{k} = \frac{0.693}{0.352} = 1.969$ minute

5. ⇒ (MHT CET 2023 10th May Evening Shift )

Calculate the rate constant for the first order reaction, $A \to B$ if the rate of reaction is $5.4 \times 10^{- 6} mol {dm}^{- 3} s^{- 1}$ and $[A] = 0.3 M$ .

A. $1.8 \times 10^{- 5} s^{- 1}$

B. $1.5 \times 10^{- 5} s^{- 1}$

C. $2.1 \times 10^{- 5} s^{- 1}$

D. $2.4 \times 10^{- 5} s^{- 1}$

Correct Option is (A)

For the first order reaction, $A ⟶ B$ ,

$\begin{aligned} Rate = k [A] \\ \begin{aligned} ∴ k & = \frac{Rate}{[A]} = \frac{5.4 \times 10^{- 6} mol {dm}^{- 3} s^{- 1}}{0.3 mol {dm}^{- 3}} \\ = 1.8 \times 10^{- 5} s^{- 1} \end{aligned} \end{aligned}$

⇒Chemical Kinetics