The decomposition reaction 2N_2O_{5(g)} xrigh | vedclass

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(B) For the reaction $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$ at constant $V$ and $T$: At $t = 0$,$P_{N_2O_5} = 1 \ atm$,$P_{N_2O_4} = 0$,$P

Vedclass · Accepted Answer

$2.30$

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(B) For the reaction $2N_2O_{5(g)} ightarrow 2N_2O_{4(g)} + O_{2(g)}$ at constant $V$ and $T$: At $t = 0$,$P_{N_2O_5} = 1 \ atm$,$P_{N_2O_4} = 0$,$P_{O_2} = 0$. At $t = Y 	imes 10^3 \ s$,let the pressure of $O_2$ formed be $P$. Then $P_{N_2O_5} = 1 - 2P$,$P_{N_2O_4} = 2P$,and $P_{O_2} = P$. The total pressure $P_T = (1 - 2P) + 2P + P = 1 + P = 1.45 \ atm$. Thus,$P = 0.45 \ atm$. The initial pressure of $N_2O_5$ is $P_0 = 1 \ atm$ and the pressure at time $t$ is $P_t = 1 - 2P = 1 - 2(0.45) = 0.1 \ atm$. For a first-order reaction,$k = \frac{2.303}{t} \log \left(\frac{P_0}{P_t}ight)$. Substituting the values: $5 	imes 10^{-4} = \frac{2.303}{Y 	imes 10^3} \log \left(\frac{1}{0.1}ight)$. $5 	imes 10^{-4} = \frac{2.303}{Y 	imes 10^3} 	imes 1$. $0.5 = \frac{2.303}{Y} \implies Y = \frac{2.303}{0.5} = 4.606$. Given the options provided,there is a discrepancy in the calculation or the provided options. Based on standard calculation,$Y = 4.606$.

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