For the thermal decomposition of N_2O_{5(g)} | vedclass

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(D) The reaction is $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$.For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{

Vedclass · Accepted Answer

$900$

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(D) The reaction is $2N_2O_{5(g)} ightarrow 2N_2O_{4(g)} + O_{2(g)}$.For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is the initial pressure of $N_2O_5$ and $P_t$ is the pressure at time $t$.Given $k = 4.606 	imes 10^{-2} \ s^{-1}$ and $t = 100 \ s$.$4.606 	imes 10^{-2} = \frac{2.303}{100} \log \frac{0.6}{P_{N_2O_5}}$$2 = \log \frac{0.6}{P_{N_2O_5}} \Rightarrow \frac{0.6}{P_{N_2O_5}} = 10^2 = 100$.$P_{N_2O_5} = \frac{0.6}{100} = 0.006 \ atm$.Let $x$ be the pressure of $N_2O_5$ decomposed. Then $0.6 - x = 0.006 \Rightarrow x = 0.594 \ atm$.The total pressure $P_{total} = (0.6 - x) + x + \frac{x}{2} = 0.6 + \frac{x}{2}$.$P_{total} = 0.6 + \frac{0.594}{2} = 0.6 + 0.297 = 0.897 \ atm$.$P_{total} = 897 	imes 10^{-3} \ atm$.Rounding to the nearest integer,$X = 897$.

$S.NO$	$Time/s$	Total pressure $(atm)$
$1.$	$0$	$0.6$
$2.$	$100$	$X$

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