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(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$:At $t=0$,pressure of $A$ is $P_0$,and total pressure is $P_0$.At time $t$,let the pressu

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$A, D$

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(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$:At $t=0$,pressure of $A$ is $P_0$,and total pressure is $P_0$.At time $t$,let the pressure of $A$ reacted be $x$. Then $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.Total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.Thus,$x = \frac{P_t - P_0}{2}$.Partial pressure of $A$ at time $t$ is $P_A = P_0 - \frac{P_t - P_0}{2} = \frac{3P_0 - P_t}{2}$.For a first order reaction,$k = \frac{1}{t} \ln \left( \frac{P_0}{P_A} \right) = \frac{1}{t} \ln \left( \frac{P_0}{(3P_0 - P_t)/2} \right) = \frac{1}{t} \ln \left( \frac{2P_0}{3P_0 - P_t} \right)$.Rearranging gives $\ln(3P_0 - P_t) = \ln(2P_0) - kt$. This represents a straight line with a negative slope $(-k)$,matching graph $A$.For a first order reaction,the rate constant $k$ is independent of the initial concentration $[A]_0$,matching graph $D$.Also,$t_{1/3}$ (time for $A$ to become $1/3$ of its initial value) is $\frac{\ln 3}{k}$,which is independent of $[A]_0$.

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