A and B decompose via first order kinetics wi | vedclass

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(B) Given $(t_{1/2})_A = 54 \, min$ and $(t_{1/2})_B = 18 \, min$. Starting with equimolar concentrations,let $[A]_0 = [B]_0 = x$. For first order kin

Vedclass · Accepted Answer

$108$

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(B) Given $(t_{1/2})_A = 54 \, min$ and $(t_{1/2})_B = 18 \, min$. Starting with equimolar concentrations,let $[A]_0 = [B]_0 = x$. For first order kinetics,the concentration at time $t$ is given by $[A]_t = [A]_0 	imes (1/2)^n$,where $n = t / t_{1/2}$. We want $[A]_t = 16 	imes [B]_t$. Substituting the expressions: $x 	imes (1/2)^{t/54} = 16 	imes x 	imes (1/2)^{t/18}$. $(1/2)^{t/54} = 2^4 	imes (1/2)^{t/18}$. $(1/2)^{t/54} = (1/2)^{-4} 	imes (1/2)^{t/18}$. $(1/2)^{t/54} = (1/2)^{(t/18) - 4}$. Equating the exponents: $t/54 = (t/18) - 4$. $4 = t/18 - t/54 = (3t - t) / 54 = 2t / 54 = t / 27$. $t = 4 	imes 27 = 108 \, min$.

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