JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Application +4 | -1

The slope of Arrhenius plot

\left( {\ln \,k\,\mathrm{v/s}\frac{1}{T}} \right)

of first order reaction is

- 5 \times {10^3}K.

The value of

{E_a}

of the reaction is.
Choose the correct option for your answer.
[Given

R = 8.314\mathrm{ J}{\mathrm{K}^{ - 1}}\mathrm{ mo}{\mathrm{l}^{ - 1}}

]

View answer and explanation

Correct option: C

41.5\,\mathrm{KJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}

Option C (41.5 kJ/mol) is correct because it is the activation energy calculated from the slope of the Arrhenius plot.

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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