The reaction CH_3COCH_{3(g)} to C_2H_{4(g)} + | vedclass

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(A) For the reaction: $CH_3COCH_{3(g)} 	o C_2H_{4(g)} + H_{2(g)} + CO_{(g)}$Initial pressure $(t=0)$: $P_0 = 0.40 \ atm$,Products = $0$At time $t=10

Vedclass · Accepted Answer

$0.0133 \ min^{-1}$

Vedclass · Answer

(A) For the reaction: $CH_3COCH_{3(g)} 	o C_2H_{4(g)} + H_{2(g)} + CO_{(g)}$Initial pressure $(t=0)$: $P_0 = 0.40 \ atm$,Products = $0$At time $t=10 \ min$: $(0.40 - x)$,$x$,$x$,$x$Total pressure $P_t = (0.40 - x) + x + x + x = 0.40 + 2x$Given $P_t = 0.50 \ atm$,so $0.40 + 2x = 0.50 \implies 2x = 0.10 \implies x = 0.05 \ atm$Pressure of reactant at $t=10 \ min$ is $P = 0.40 - 0.05 = 0.35 \ atm$For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P}$$k = \frac{2.303}{10} \log \frac{0.40}{0.35} = \frac{2.303}{10} \log \frac{8}{7} = \frac{2.303}{10} (\log 8 - \log 7)$Alternatively,using the given $\log 3.5 = 0.5441$:$k = \frac{2.303}{10} \log \frac{0.40}{0.35} = \frac{2.303}{10} \log (1.1428) \approx 0.0133 \ min^{-1}$

The reaction $CH_3COCH_{3(g)} \to C_2H_{4(g)} + H_{2(g)} + CO_{(g)}$ is a first-order reaction. If the initial pressure is $0.40 \ atm$ and the total pressure after $10 \ min$ is $0.50 \ atm$,calculate the rate constant of the reaction. $(\log 3.5 = 0.5441)$

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