(C) The reaction is $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$.Initial pressure $(t=0)$: $P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$.At $t = 20 \ min$:

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(C) The reaction is $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$.Initial pressure $(t=0)$: $P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$.At $t = 20 \ min$: $(80-x) + x + x = 120 \ mm \implies 80 + x = 120 \implies x = 40 \ mm$.The pressure of $C_2H_4O$ remaining is $80 - 40 = 40 \ mm$.Since the pressure reduced to half ($80 \ mm$ to $40 \ mm$) in $20 \ minutes$,the half-life $t_{1/2}$ is $20 \ min$.Thus,option $C$ is correct.

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

Calculate the half-life of the first-order reaction $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$. If the initial pressure of $C_2H_4O_{(g)}$ is $80 \ mm$ and the total pressure at the end of $20 \ minutes$ is $120 \ mm$,find the half-life in $\text{min}$.

A
$40$
B
$120$
C
$20$
D
$80$

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Chemical Kinetics

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First Order reaction

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Calculate the half-life of the first-order reaction $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$. If the initial pressure of $C_2H_4O_{(g)}$ is $80 \ mm$ and the total pressure at the end of $20 \ minutes$ is $120 \ mm$,find the half-life in $\text{min}$.

Similar Questions

For a first-order reaction,the concentration decreases from $800 \ mol/dm^3$ to $50 \ mol/dm^3$ in $2 \times 10^4 \ s$. What is the rate constant of the reaction?

Which is a correct integrated rate equation for a first-order reaction?

The half-life period for a first-order reaction is . . . . . . .

$75\%$ of a first-order reaction was completed in $32 \text{ minutes}$. When was $50\%$ of the reaction completed? (in $\text{minutes}$)

For a first-order reaction,if $75\%$ of the reaction is completed in $15 \ min$,then the time required for $90\%$ completion of the reaction will be ........... $min$.

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