JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Application +4 | -1

If the rate constant of the reaction,

X \rightarrow Y

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

and initial concentration of

X

3 \mathrm{~M}

then the concentration of

{Y}

after 100 minutes will be

View answer and explanation

Correct option: C

1.5 \mathrm{~M}

Option C (

1.5 \mathrm{~M}

) is correct because it's calculated by multiplying the rate constant by the time to find the concentration of Y formed in a zero-order reaction.

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