JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Analyzing +4 | -1

Rate constant

(\mathrm{K})

varies with temperature as given by equation

\log \mathrm{K}\left(\mathrm{min}^{-1}\right)=5-\frac{2000}{\mathrm{T}}

Consider the following about this equation
(I) Pre exponential factor is

10^5

.
(II)

\mathrm{E}_\alpha

9.212 \mathrm{~kcal}

.
(III) Variation of

\log \mathrm{K}

with

1 / \mathrm{T}

is linear.
Select the correct statement.

View answer and explanation

Correct option: A

I, II and III

Option A (I, II and III) is correct because all three statements are true based on the Arrhenius equation and its logarithmic form.

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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