At 500,^oC,cyclopropane isomerizes to propene | vedclass

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(B) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -kt$. Given: Initial concentration $[A]_0 = 0.05\, M$. Rate cons

Vedclass · Accepted Answer

$0.015$

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(B) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -kt$. Given: Initial concentration $[A]_0 = 0.05\, M$. Rate constant $k = 6.7 	imes 10^{-4}\,s^{-1}$. Time $t = 30\, min = 30 	imes 60\, s = 1800\, s$. Substituting the values: $\ln([A]_t / 0.05) = -(6.7 	imes 10^{-4}\,s^{-1}) 	imes (1800\, s)$. $\ln([A]_t / 0.05) = -1.206$. Taking the exponential on both sides: $[A]_t / 0.05 = e^{-1.206} \approx 0.299$. $[A]_t = 0.05 	imes 0.299 \approx 0.01495\, M \approx 0.015\, M$.

At $500\,^oC$,cyclopropane isomerizes to propene. This reaction is first order with a rate constant of $6.7 \times 10^{-4}\,s^{-1}$. If the initial concentration of cyclopropane is $0.05\, M$,what will be the molarity of cyclopropane after $30\, min$ (in $, M$)?

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