JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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For the reaction

2 \mathrm{N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2

rate & rate constant are

1.02 \times 10^{-4}

and

3.4 \times 10^{-5} \mathrm{sec}^{-1}

respectively then concentration of

\mathrm{N}_2 \mathrm{O}_5

at that time will be:

View answer and explanation

Correct option: B

Option B (3) is correct because the concentration of N2O5 is calculated directly from the rate law, given the rate and rate constant.

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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