Calculate the half-life of the first-order re | vedclass

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

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$20$

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(C) The reaction is $C_2H_4O_{(g)} \longrightarrow CH_{4(g)} + CO_{(g)}$.Initial pressure $(t=0)$: $P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$.At time $t=20 \ min$: $P_{C_2H_4O} = 80-x$,$P_{CH_4} = x$,$P_{CO} = x$.Total pressure $P_t = (80-x) + x + x = 80+x = 120 \ mm$.So,$x = 40 \ mm$.Pressure of reactant remaining $a-x = 80-40 = 40 \ mm$.Rate constant $k = \frac{2.303}{t} \log \frac{P_0}{P_t} = \frac{2.303}{20} \log \frac{80}{40} = \frac{2.303}{20} \log 2$.Since $\log 2 \approx 0.3010$,$k = \frac{2.303 	imes 0.3010}{20}$.Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693 	imes 20}{2.303 	imes 0.3010}$.Since $0.693 \approx 2.303 	imes 0.3010$,$t_{1/2} = 20 \ min$.

Calculate the half-life of the first-order reaction $C_2H_4O_{(g)} \longrightarrow CH_{4(g)} + CO_{(g)}$. 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 $\min$.

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