JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Medium

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Application +4 | -1

The rate constant for the reaction

\mathrm{CO}_2+\mathrm{OH}^{-} \longrightarrow \mathrm{HCO}_3^{-}

in aqueous solution is

4.0 \times 10^3 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}

when

\left[\mathrm{CO}_2\right]=10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}

and

\left[\mathrm{OH}^{-}\right]=0.1 \mathrm{~mol} \mathrm{~L}^{-1}

Number of moles of

\mathrm{HCO}_3^{-}

formed per second is

x \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}

. What is the value of

x

View answer and explanation

Correct option: A

Option A (4) is correct because the rate of formation of HCO3- is calculated using the given rate constant and concentrations.

More Questions from CHEMICAL KINETICS

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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