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(C) For the reaction $C_2H_4O_{(g)} 	o CH_{4(g)} + CO_{(g)}$:At $t=0$,pressure of $C_2H_4O = P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$. Total pressu

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$20$

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(C) For the reaction $C_2H_4O_{(g)} 	o CH_{4(g)} + CO_{(g)}$:At $t=0$,pressure of $C_2H_4O = P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$. Total pressure $P_t = 80 \ mm$.At $t=20 \ min$,pressure of $C_2H_4O = 80-x$,$P_{CH_4} = x$,$P_{CO} = x$. Total pressure $P_t = (80-x) + x + x = 80+x = 120 \ mm$.Thus,$x = 40 \ mm$.The pressure of reactant remaining is $P_0 - x = 80 - 40 = 40 \ mm$.Using the first-order rate constant formula $k = \frac{2.303}{t} \log \frac{P_0}{P_0-x} = \frac{2.303}{20} \log \frac{80}{40} = \frac{2.303}{20} \log 2$.Since $\log 2 \approx 0.3010$,$k = \frac{2.303 	imes 0.3010}{20} \approx \frac{0.693}{20} \ min^{-1}$.The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{0.693 	imes 20}{0.693} = 20 \ min$.

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 $120 \ mm$. (in $min$)

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