The rate constant for a first-order reaction is $2.2 \times10^{-5} \mathrm{~s}^{-1}$ at $320^{\circ} \mathrm{C}$. Calculate the time in which the reaction is $15 \%$ complete.

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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The rate constant for a first-order reaction is

2.2 \times10^{-5} \mathrm{~s}^{-1}

320^{\circ} \mathrm{C}

. Calculate the time in which the reaction is

15 \%

complete.

View answer and explanation

Correct option: A

7391 \mathrm{~sec}

Option A (7391 sec) is correct because it is the time calculated using the first-order kinetics equation for 15% completion.

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\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}

\begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

k

(Initial\mathrm{ }conc.\mathrm{ } = a)

C_0 \sqrt{T}

C_0

T

100 \mathrm{~kJ/mol}.

75 \%

\log _{10} \frac{K_2}{K_1}

27^{\circ} \mathrm{C}

2.303 \times 8.314=19.147