For the first-order gaseous reaction A_{(g)} | vedclass

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Question

(A) For the reaction $A_{(g)} 	o 2B_{(g)} + C_{(g)}$:At $t=0$,pressure is $P_0, 0, 0$ for $A, B, C$ respectively.At time $t$,let the pressure of $A$

Vedclass · Accepted Answer

$K = \frac{1}{t} \ln \left( \frac{2P_0}{3P_0 - P_t} \right)$

Vedclass · Answer

(A) For the reaction $A_{(g)} 	o 2B_{(g)} + C_{(g)}$:At $t=0$,pressure is $P_0, 0, 0$ for $A, B, C$ respectively.At time $t$,let the pressure of $A$ decrease by $x$. Then $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.The total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.From this,$2x = P_t - P_0$,so $x = \frac{P_t - P_0}{2}$.The pressure of $A$ at time $t$ is $P_A = P_0 - x = P_0 - \frac{P_t - P_0}{2} = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$.For a first-order reaction,$K = \frac{1}{t} \ln \left( \frac{P_0}{P_A} ight) = \frac{1}{t} \ln \left( \frac{P_0}{(3P_0 - P_t)/2} ight) = \frac{1}{t} \ln \left( \frac{2P_0}{3P_0 - P_t} ight)$.

For the first-order gaseous reaction $A_{(g)} \to 2B_{(g)} + C_{(g)}$,the expression for the rate constant $K$ in terms of initial pressure $P_0$ and total pressure $P_t$ at time $t$ is:

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