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The rate equation for a first order reaction is given by $[R]=[R]_0 e^{-k t}$. A straight line with positive slope is obtained by plotting. $[R]_0=$ initial concentration of reactant, $[R]=$ concentration of reactant at time, $t$
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Verified Answer
The correct answer is:
$\log \frac{[R]_0}{[R]}$ vs time
For the equation $[R]=[R]_0 e^{-k t}$
Taking log we get
$$
\begin{array}{rlrl}
& \ln [R] & =-k t+\operatorname{in}[R]_0 \\
& \therefore & \ln \frac{[R]}{[R]_0} & =-k t \\
& & & \\
k & =\frac{1}{t} \text { in } \frac{[R]_0}{[R]} \\
k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]}
\end{array}
$$
When graph is ploted of $\log \frac{[R]_0}{[R]}$ against $t$, we get a straight line with positive slope $=\frac{k}{2.303}$.
Taking log we get
$$
\begin{array}{rlrl}
& \ln [R] & =-k t+\operatorname{in}[R]_0 \\
& \therefore & \ln \frac{[R]}{[R]_0} & =-k t \\
& & & \\
k & =\frac{1}{t} \text { in } \frac{[R]_0}{[R]} \\
k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]}
\end{array}
$$
When graph is ploted of $\log \frac{[R]_0}{[R]}$ against $t$, we get a straight line with positive slope $=\frac{k}{2.303}$.
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