The half-life of a first-order reaction is 85 | vedclass

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Question

(A) Step $1$: Calculate the rate constant $k$ for the first-order reaction using $k = \frac{0.693}{t_{1/2}}$.$k = \frac{0.693}{850 	ext{ s}} \approx

Vedclass · Accepted Answer

$0.023 	ext{ mol dm}^{-3}$

Vedclass · Answer

(A) Step $1$: Calculate the rate constant $k$ for the first-order reaction using $k = \frac{0.693}{t_{1/2}}$.$k = \frac{0.693}{850 	ext{ s}} \approx 8.153 	imes 10^{-4} 	ext{ s}^{-1}$.Step $2$: Use the integrated rate law for a first-order reaction: $[R] = [R]_0 e^{-kt}$.$[R] = 0.06 	imes e^{-(8.153 	imes 10^{-4} 	imes 1200)}$.Step $3$: Calculate the exponent: $-(8.153 	imes 10^{-4} 	imes 1200) \approx -0.97836$.Step $4$: Calculate the final concentration: $[R] = 0.06 	imes e^{-0.97836} \approx 0.06 	imes 0.3759 \approx 0.02255 	ext{ mol dm}^{-3}$.Rounding to two significant figures, we get $0.023 	ext{ mol dm}^{-3}$. Thus, option $A$ is correct.

The half-life of a first-order reaction is $850 \text{ s}$. The initial concentration of the reactant is $[R]_0 = 0.06 \text{ mol dm}^{-3}$. What concentration $[R]$ would remain after $t = 1200 \text{ s}$?

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