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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 32 \ min$,so $k = \frac{0.693}{32} \ min

Vedclass · Accepted Answer

$7.1 	imes 10^{13} \ molecules/L$

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 32 \ min$,so $k = \frac{0.693}{32} \ min^{-1} = 0.021656 \ min^{-1}$.The time elapsed is $1.5 \ hr = 90 \ min$.The concentration at time $t$ is given by $[A]_t = [A]_0 	imes e^{-kt}$.$[A]_t = 5.0 	imes 10^{14} 	imes e^{-(0.021656 	imes 90)}$.$[A]_t = 5.0 	imes 10^{14} 	imes e^{-1.949}$.$[A]_t = 5.0 	imes 10^{14} 	imes 0.1424 = 7.12 	imes 10^{13} \ molecules/L$.Thus,the correct option is $C$.

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