Q39 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry

For the reaction $2\text{NOBr}_{(g)} \longrightarrow 2\text{NO}_{(g)} + \text{Br}_{2(g)}$, rate law is $r = k[\text{NOBr}]^2$. Rate constant is $1.62\ \text{M s}^{-1}$ and concentration of NOBr is $5 \times 10^{-3}\ \text{M}$,
What is rate of reaction?

  • Correct answer
    A
    $4.05 \times 10^{-5}\ \text{M s}^{-1}$
  • B
    $6.48 \times 10^{-6}\ \text{M s}^{-1}$
  • C
    $2.12 \times 10^{-5}\ \text{M s}^{-1}$
  • D
    $8.63 \times 10^{-6}\ \text{M s}^{-1}$

Solution

Given rate law is $r = k[\text{NOBr}]^2$

Substituting the given values:

$r = 1.62 \times (5 \times 10^{-3})^2$

$r = 1.62 \times 25 \times 10^{-6}$

$r = 40.5 \times 10^{-6}$

$r = 4.05 \times 10^{-5}\ \text{M s}^{-1}$

Answer: $4.05 \times 10^{-5}\ \text{M s}^{-1}$

Hence, the correct answer is option: 'A'.

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