For reaction $A \to B,$ the rate constant ${k_1} = {A_1}{e^{ - E{a_1}/RT}}$ and for the reaction $X \to Y,$ the rate constant${k_2} = {A_2}{e^{ - E{a_2}/RT}}.$ If ${A_1} = {10^8},{A_2} = {10^{10}}$ and $E{a_1} = 600\;{\rm{cal}}/{\rm{mol}},\;E{a_2} = 1800\;{\rm{cal}}/{\rm{mol}},$ then the temperature at which ${k_1} = {k_2}\;is\;\left( {R = 2\;{\rm{cal}}/{\rm{k }}\;{\rm{mol}}} \right)$

$\frac{{300}}{{2.303}}{\rm{K}}$

$\frac{{200}}{{3.506}}{\rm{K}}$

$\frac{{400}}{{6.506}}{\rm{K}}$

$\frac{{200}}{{5.204}}{\rm{K}}$

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Application +4 | -1

For reaction

A \to B,

the rate constant

{k_1} = {A_1}{e^{ - E{a_1}/RT}}

and for the reaction

X \to Y,

the rate constant

{k_2} = {A_2}{e^{ - E{a_2}/RT}}.

{A_1} = {10^8},{A_2} = {10^{10}}

and

E{a_1} = 600\;\mathrm{cal}/\mathrm{mol},\;E{a_2} = 1800\;\mathrm{cal}/\mathrm{mol},

then the temperature at which

{k_1} = {k_2}\;is\;\left( {R = 2\;\mathrm{cal}/\mathrm{k }\;\mathrm{mol}} \right)

View answer and explanation

Correct option: B

\frac{{300}}{{2.303}}\mathrm{K}

Option A (

\frac{{200}}{{3.506}}\mathrm{K}

) is incorrect. Option B (

\frac{{300}}{{2.303}}\mathrm{K}

) is correct because equating the rate constants and solving for T using the Arrhenius equation leads to this result.

Practice All Chapter MCQs →

\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