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(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.The time $t$ required for a reaction to complete $90 \%$

Vedclass · Accepted Answer

$3.32$

Vedclass · Answer

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.The time $t$ required for a reaction to complete $90 \%$ is given by $t_{90\%} = \frac{2.303}{K} \log \left( \frac{100}{100 - 90} ight) = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.Substituting $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}$ into the equation:$t_{90\%} = \frac{2.303}{0.693} 	imes t_{1/2}$.Since $2.303 \approx \ln 10$ and $0.693 \approx \ln 2$,we have $t_{90\%} = \frac{\ln 10}{\ln 2} 	imes t_{1/2} = \frac{2.303}{2.303 	imes 0.3010} 	imes t_{1/2} = \frac{1}{0.3010} 	imes t_{1/2} \approx 3.32 	imes t_{1/2}$.Therefore,$x = 3.32$.

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

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