(C) For a $1^{st}$ order reaction,the integrated rate equation is given by:$2.303 \log \left(\frac{[A]_0}{[A]_t}\right) = kt$Rearranging this equation

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(C) For a $1^{st}$ order reaction,the integrated rate equation is given by:$2.303 \log \left(\frac{[A]_0}{[A]_t}\right) = kt$Rearranging this equation,we get:$\log \left(\frac{[A]_0}{[A]_t}\right) = \left(\frac{k}{2.303}\right) t$Comparing this with the linear equation $y = mx + c$,where $y = \log \left(\frac{[A]_0}{[A]_t}\right)$,$x = t$,$m = \frac{k}{2.303}$,and $c$ is the intercept.Since there is no constant term added,the intercept $c = 0$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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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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A
$-\frac{k}{2.303}$
B
$-\log [A]_0$
C
zero
D
$\frac{2.303}{k}$

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Chemical Kinetics

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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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