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(C) For a first order reaction,the integrated rate equation is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.At half-life,$t = t_{1/2}$ and $x = \

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$\frac{0.693}{k}$

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(C) For a first order reaction,the integrated rate equation is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.At half-life,$t = t_{1/2}$ and $x = \frac{a}{2}$,so $a-x = \frac{a}{2}$.Substituting these values: $k = \frac{2.303}{t_{1/2}} \log \frac{a}{a/2} = \frac{2.303}{t_{1/2}} \log 2$.Since $\log 2 \approx 0.3010$,we get $k = \frac{2.303 	imes 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}$.Therefore,$t_{1/2} = \frac{0.693}{k}$.

For the first order reaction with rate constant $k$,which expression gives the half-life period? (Initial concentration $= a$)

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