JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Analyzing +4 | -1

For a reaction of the type

A + B \to x

products, it is observed that doubling the concentration of

A

causes the reaction rate

({k_1})

to be four times as great but doubling the amount of

B

does not affect the rate

({k_2})

. The rate equation is

View answer and explanation

Correct option: B

k = {k_1}{k_2}

Option B is correct because the rate is proportional to [A]² and independent of [B], so the rate law is k[A]², which can be represented as k = k1*k2.

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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