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(B) For a first-order reaction,the integrated rate equation is given by: $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ Rearranging this equation to the

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$\frac{K}{2.303}$

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(B) For a first-order reaction,the integrated rate equation is given by: $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ Rearranging this equation to the form $y = mx + c$: $\log \frac{[R]_0}{[R]} = \frac{K}{2.303} 	imes t$ Comparing this with the equation of a straight line $y = mx$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,and the slope $m = \frac{K}{2.303}$. Therefore,the slope of the graph is $\frac{K}{2.303}$.

What is the value of the slope when a graph is plotted of $\log \frac{[R]_0}{[R]}$ versus $t$ (time) for a first-order reaction?

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