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(D) For a first-order reaction,the integrated rate equation is given by $k = \frac{1}{t} \ln \frac{[R]_0}{[R]_t}$.At two different times $t_1$ and $t_

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$I: F, II: F$

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(D) For a first-order reaction,the integrated rate equation is given by $k = \frac{1}{t} \ln \frac{[R]_0}{[R]_t}$.At two different times $t_1$ and $t_2$,we have:$k = \frac{1}{t_1} \ln \frac{[R]_0}{[R]_1} \implies \ln [R]_1 = \ln [R]_0 - kt_1$$k = \frac{1}{t_2} \ln \frac{[R]_0}{[R]_2} \implies \ln [R]_2 = \ln [R]_0 - kt_2$Subtracting the two equations: $\ln [R]_1 - \ln [R]_2 = -kt_1 - (-kt_2) = k(t_2 - t_1)$.Thus,$k = \frac{1}{t_2 - t_1} \ln \frac{[R]_1}{[R]_2} = -\frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2} = \frac{1}{t_1 - t_2} \ln \frac{[R]_2}{[R]_1}$.Comparing this with the given statements:Statement $I$: $k = \frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2}$ is False because the denominator should be $(t_2 - t_1)$ or the expression should be $-\frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2}$.Statement $II$: $k = -\frac{1}{t_1 - t_2} \ln \frac{[R]_2}{[R]_1} = \frac{1}{t_2 - t_1} \ln \frac{[R]_2}{[R]_1}$,which is also False based on the standard derivation.Therefore,both statements are False.

Identify True $(T)$ and False $(F)$ statements of the following for a first-order reaction $R \rightarrow P$.
Statement $I$: $k = \frac{1}{(t_1 - t_2)} \ln \frac{[R]_1}{[R]_2}$
Statement $II$: $k = -\frac{1}{(t_1 - t_2)} \ln \frac{[R]_2}{[R]_1}$

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Identify True $(T)$ and False $(F)$ statements of the following for a first-order reaction $R \rightarrow P$.Statement $I$: $k = \frac{1}{(t_1 - t_2)} \ln \frac{[R]_1}{[R]_2}$Statement $II$: $k = -\frac{1}{(t_1 - t_2)} \ln \frac{[R]_2}{[R]_1}$