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(A) $1$. Consider the reaction: $A(g) 	o B(g) + C(g)$.$2$. At $t = 0$, the initial pressure of $A$ is $p_i$, and the pressures of $B$ and $C$ are $0$

Vedclass · Accepted Answer

$\frac{1}{t} \ln \frac{p_i}{2p_i - p_t}$

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(A) $1$. Consider the reaction: $A(g) 	o B(g) + C(g)$.$2$. At $t = 0$, the initial pressure of $A$ is $p_i$, and the pressures of $B$ and $C$ are $0$.$3$. At time $t$, let $x$ be the decrease in pressure of $A$. The pressures are: $P_A = p_i - x$, $P_B = x$, and $P_C = x$.$4$. The total pressure $p_t$ at time $t$ is given by: $p_t = (p_i - x) + x + x = p_i + x$.$5$. From this, we find $x = p_t - p_i$.$6$. The partial pressure of $A$ at time $t$ is $P_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$.$7$. For a first-order reaction, the rate constant $k$ is given by: $k = \frac{1}{t} \ln \frac{p_i}{P_A}$.$8$. Substituting $P_A$, we get: $k = \frac{1}{t} \ln \frac{p_i}{2p_i - p_t}$.

For a first-order gas-phase reaction $A(g) \to B(g) + C(g)$, let $p_i$ be the initial pressure of gas $A$ and $p_t$ be the total pressure of the reaction mixture at time $t$. The expression for the rate constant $(k)$ is:

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