At 298 K . for a first order reaction $(\mathrm{A} \rightarrow \mathrm{P})$ the following graph is obtained.…

At 298 K . for a first order reaction $(\mathrm{A} \rightarrow \mathrm{P})$ the following graph is obtained. The rate constant (in $\mathrm{s}^{-1}$ ) and initial concentration (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of ' A ' are respectively $(y$-axis $=\ln (\mathrm{a}-\mathrm{x}) ; x$-axis $=$ time in sec$)$
  1. $2.303 ; 10^{-1}$
  2. $10^{-2} ; 2.303$
  3. $10^{-1} ; 10^{-2}$
  4. $10^{-2} ; 10^{-1}$

Solution

From first order reaction, $\begin{aligned} & \ln \frac{a}{a-x}=K t \\ & \ln a-\ln (a-x)=K t\end{aligned}$ $\begin{aligned} & \ln (a-x)=\ln a-K t \\ & y=c+m x\end{aligned}$ $\begin{aligned} & \mathrm{c}(\text { intercept })=\ln \mathrm{a} \\ & \mathrm{m}(\text { slope })=-\mathrm{K} \\ & \therefore \text { given, }\end{aligned}$ $\mathrm{c}=\ln \mathrm{a}$ or, $\ln \mathrm{a}=-2.303$ $\begin{aligned} & 2.303 \log a=-2.303 \\ & \log a=-1 \\ & \therefore a=10^{-1}\end{aligned}$ again, $\begin{gathered}-\mathrm{K}=-(10)^{-2} \\ \mathrm{~K}=10^{-2}\end{gathered}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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