JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Application +4 | -1

The rate law for a reaction between the substances

A

and

B

is given by

rate = k{\left[ A \right]^n}{\left[ B \right]^m}

. On doubling the concentration of

A

and halving the concentration of

B

, the ratio of the new rate to the earlier rate of the reaction will be as

View answer and explanation

Correct option: D

{2^{\left( {n - m} \right)}}

Option D (

{2^{\left( {n - m} \right)}}

) is correct because the new rate is the original rate multiplied by

2^n

(due to A) and

(1/2)^m

(due to B), resulting in a factor of

2^{n-m}

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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