The energy of activation of reaction, if its rate double when the temperature is raised from $290 \mathrm{~K}$ to $300 \mathrm{~K}$ is $\mathrm{x}$ kcal then the value of $\mathrm{x} / 3$ is (Take $\mathrm{R}=2 \mathrm{cal} / \mathrm{k} / \mathrm{mol}$ )

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Application +4 | -1

The energy of activation of reaction, if its rate double when the temperature is raised from

290 \mathrm{~K}

300 \mathrm{~K}

\mathrm{x}

kcal then the value of

\mathrm{x} / 3

is (Take

\mathrm{R}=2 \mathrm{cal} / \mathrm{k} / \mathrm{mol}

)

View answer and explanation

Correct option: C

Option C (12) is correct because the activation energy is calculated using the Arrhenius equation.

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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