For a first-order reaction at 100,^oC,the rat | vedclass

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(D) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -Kt$ Given: $K = 1.5 	imes 10^{-2} \, s^{-1}$ $t = 10 \, min =

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$10^{-2} \, mol \, L^{-1}$

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(D) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -Kt$ Given: $K = 1.5 	imes 10^{-2} \, s^{-1}$ $t = 10 \, min = 600 \, s$ $[A]_0 = 100 \, mol \, L^{-1}$ Substituting the values: $\ln([A]_t / 100) = -(1.5 	imes 10^{-2} \, s^{-1}) 	imes (600 \, s)$ $\ln([A]_t / 100) = -9$ $[A]_t / 100 = e^{-9}$ $[A]_t = 100 	imes e^{-9} \approx 100 	imes 1.23 	imes 10^{-4} \approx 1.23 	imes 10^{-2} \, mol \, L^{-1}$ Since the calculated value is approximately $10^{-2} \, mol \, L^{-1}$,the correct option is $D$.

For a first-order reaction at $100\,^oC$,the rate constant is $K = 1.5 \times 10^{-2} \, s^{-1}$. If the initial concentration of the reactant is $100 \, mol \, L^{-1}$,what will be the concentration of the reactant after $10 \, min$?

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