JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Medium

1 of 1

Application +4 | -1

For the reaction,

C{l_2} + 2{l^ - } \to {I_2} + 2C{l^ - }

the initial concentration of

{I^ - }

was

0.20\;mol\;{L^{ - 1}}

and the concentration after 20 min was

0.18\;mol\;{L^{ - 1}}

.Then the rate of formation of

{I_2}

mol\;{L^{ - 1}}

would be

View answer and explanation

Correct option: B

5 \times {10^{ - 4}}

Option B (

5 \times {10^{ - 4}}

) is correct because it is calculated using the rate of reaction formula considering the stoichiometry of the reaction.

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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