Dissociation of N_2O_5 dissolved in CCl_4 at | vedclass

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Question

(A) For the reaction $N_2O_{5(soln)} 	o 2NO_{2(soln)} + \frac{1}{2}O_{2(g)}$,let the initial concentration of $N_2O_5$ be $[A]_0 = 0.25 \ mol \ L^{-1

Vedclass · Accepted Answer

$1022$

Vedclass · Answer

(A) For the reaction $N_2O_{5(soln)} 	o 2NO_{2(soln)} + \frac{1}{2}O_{2(g)}$,let the initial concentration of $N_2O_5$ be $[A]_0 = 0.25 \ mol \ L^{-1}$.Let $x$ be the amount of $N_2O_5$ dissociated at time $t$.According to stoichiometry,$[NO_2] = 2x = 0.20 \ mol \ L^{-1}$,so $x = 0.10 \ mol \ L^{-1}$.The concentration of $N_2O_5$ remaining at time $t$ is $[A]_t = [A]_0 - x = 0.25 - 0.10 = 0.15 \ mol \ L^{-1}$.For a first-order reaction,$k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.Substituting the values: $5.0 	imes 10^{-4} = \frac{2.303}{t} \log(\frac{0.25}{0.15})$.$t = \frac{2.303}{5.0 	imes 10^{-4}} \log(1.666) \approx 4606 	imes 0.2218 \approx 1022 \ s$.

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