We are given with following information about concentration of reactant with initial rate of reaction. $\begin{array}{*{20}{c}} {{\text{Initial concentration}}}&{{\text{Initial rate}}} \\ {0.005{\text{ M}}}&{7.5 \times {{10}^{ - 4}}} \\ {0.02{\text{ M}}}&{3 \times {{10}^{ - 3}}} \end{array}$ Find out order of reaction with respect to that reactant.

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Medium

1 of 1

Application +4 | -1

We are given with following information about concentration of reactant with initial rate of reaction.

\begin{array}{cccccccccccccccccccc} {{\text{Initial concentration}}}&{{\text{Initial rate}}} \\ {0.005{\text{ M}}}&{7.5 \times {{10}^{ - 4}}} \\ {0.02{\text{ M}}}&{3 \times {{10}^{ - 3}}} \end{array}

Find out order of reaction with respect to that reactant.

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Correct option: A

The correct answer is 1 because the order of the reaction is determined by comparing the change in rate with the change in concentration.

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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