The rate equation for a first-order reaction | vedclass

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Question

(A) For a first-order reaction: $[R] = [R]_0 e^{-kt}$.Taking natural logarithm on both sides: $\ln [R] = \ln [R]_0 - kt$.Rearranging the terms: $\ln [

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$\log \frac{[R]_0}{[R]}$ vs $t$

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(A) For a first-order reaction: $[R] = [R]_0 e^{-kt}$.Taking natural logarithm on both sides: $\ln [R] = \ln [R]_0 - kt$.Rearranging the terms: $\ln [R]_0 - \ln [R] = kt$,which gives $\ln \frac{[R]_0}{[R]} = kt$.Converting to base $10$ logarithm: $2.303 \log \frac{[R]_0}{[R]} = kt$.Therefore,$\log \frac{[R]_0}{[R]} = \frac{k}{2.303} t$.Comparing this with the equation of a straight line $y = mx + c$,a plot of $\log \frac{[R]_0}{[R]}$ vs $t$ gives a straight line passing through the origin with a positive slope $m = \frac{k}{2.303}$.

The rate equation for a first-order reaction is given by $[R] = [R]_0 e^{-kt}$. $A$ straight line with a positive slope is obtained by plotting: ($[R]_0 =$ initial concentration of reactant,$[R] =$ concentration of reactant at time $t$)

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