(A) The integrated rate equation for a first-order reaction is given by: $\ln \frac{[R]_0}{[R]} = Kt$. Converting to base $10$ logarithm: $\log \frac{

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$\log \frac{[R]_0}{[R]} \text{ vs } t \text{ (Time)}$

(A) The integrated rate equation for a first-order reaction is given by: $\ln \frac{[R]_0}{[R]} = Kt$. Converting to base $10$ logarithm: $\log \frac{[R]_0}{[R]} = \frac{Kt}{2.303}$. Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$ and $x = t$,the slope $m$ is equal to $\frac{K}{2.303}$. Therefore,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ gives a slope of $\frac{K}{2.303}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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For which of the following graphs of a first-order reaction will the value of the slope be $\frac{K}{2.303}$?

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A
$\log \frac{[R]_0}{[R]} \text{ vs } t \text{ (Time)}$
B
$\log \frac{[R]}{[R]_0} \text{ vs } t \text{ (Time)}$
C
$\ln \frac{[R]_0}{[R]} \text{ vs } t \text{ (Time)}$
D
$\ln \frac{[R]}{[R]_0} \text{ vs } t \text{ (Time)}$

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For which of the following graphs of a first-order reaction will the value of the slope be $\frac{K}{2.303}$?

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For a first order reaction,the intercept of the graph between $\log \left(\frac{[A]_0}{[A]_t}\right)$ ($Y$-axis) and time ($X$-axis) is equal to

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Identify True $(T)$ and False $(F)$ statements for the first order reaction $R \rightarrow P$ given below:
$I. \ k = \frac{1}{t} \ln \frac{[R]_0}{[R]}$
$II. \ k = \frac{1}{t} \ln \frac{[R]}{[R]_0}$