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(D) For a first-order reaction,the half-life is given by $t_{50\%} = \frac{\ln 2}{k}$.For $t_{87.5\%}$,the amount remaining is $100 - 87.5 = 12.5\%$,w

Vedclass · Accepted Answer

$t_{90\%} = 4 \cdot t_{50\%}$

Vedclass · Answer

(D) For a first-order reaction,the half-life is given by $t_{50\%} = \frac{\ln 2}{k}$.For $t_{87.5\%}$,the amount remaining is $100 - 87.5 = 12.5\%$,which is $(1/8)$ of the initial amount. Thus,$t_{87.5\%} = \frac{\ln 8}{k} = \frac{3 \ln 2}{k} = 3 \cdot t_{50\%}$.For $t_{90\%}$,the amount remaining is $100 - 90 = 10\%$,which is $(1/10)$ of the initial amount. Thus,$t_{90\%} = \frac{\ln 10}{k} \approx \frac{2.303}{k} \approx 3.32 \cdot t_{50\%}$.Therefore,the statement $t_{90\%} = 4 \cdot t_{50\%}$ is $INCORRECT$.

For a first order gas phase elementary reaction $A_{(g)} \xrightarrow{k} 2B_{(g)}$,which of the following is $INCORRECT$?

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