The plot of log(a - x) versus time t is a str | vedclass

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(B) For a first-order reaction,the integrated rate equation is given by: $Kt = \ln(a) - \ln(a - x)$.Rearranging this,we get: $\ln(a - x) = -Kt + \ln(a

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(B) For a first-order reaction,the integrated rate equation is given by: $Kt = \ln(a) - \ln(a - x)$.Rearranging this,we get: $\ln(a - x) = -Kt + \ln(a)$.Converting to base $10$ logarithm: $\log(a - x) = -\frac{Kt}{2.303} + \log(a)$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(a - x)$ and $x = t$,the slope is $-\frac{K}{2.303}$.Since the plot of $\log(a - x)$ versus $t$ is a straight line,the reaction follows first-order kinetics.

The plot of $\log(a - x)$ versus time $t$ is a straight line,which indicates that the reaction is of ....... order.

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