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(D) For a first-order reaction,the concentration at time $t$ is given by $[C]_t = [C]_0 e^{-kt},$ where $k = \frac{\ln 2}{t_{1/2}}.$Given $[A]_0 = [B]

Vedclass · Accepted Answer

$900$

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(D) For a first-order reaction,the concentration at time $t$ is given by $[C]_t = [C]_0 e^{-kt},$ where $k = \frac{\ln 2}{t_{1/2}}.$Given $[A]_0 = [B]_0,$ we want to find $t$ such that $[A]_t = 4[B]_t.$Substituting the expressions: $[A]_0 e^{-(\ln 2 / 300)t} = 4[B]_0 e^{-(\ln 2 / 180)t}.$Since $[A]_0 = [B]_0,$ we have $e^{-(\ln 2 / 300)t} = 4 e^{-(\ln 2 / 180)t}.$Rearranging gives $e^{(\frac{\ln 2}{180} - \frac{\ln 2}{300})t} = 4.$Taking the natural logarithm on both sides: $(\frac{\ln 2}{180} - \frac{\ln 2}{300})t = \ln 4 = 2 \ln 2.$Dividing by $\ln 2$: $(\frac{1}{180} - \frac{1}{300})t = 2.$Solving for $t$: $(\frac{300 - 180}{180 	imes 300})t = 2 \Rightarrow \frac{120}{54000}t = 2.$$t = \frac{2 	imes 54000}{120} = \frac{108000}{120} = 900 \ s.$

Similar Questions

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