The following data were obtained during the first order thermal decomposition of $N_{2}O_{5(g)}$ at constant volume:
$2N_{2}O_{5(g)} \rightarrow 2N_{2}O_{4(g)} + O_{2(g)}$

$S.No.$	Time $/$ $s$	Total pressure $/$ $atm$
$1.$	$0$	$0.5$
$2.$	$100$	$0.512$

Calculate the rate constant.

(N/A) Let the initial pressure of $N_{2}O_{5(g)}$ be $P_0 = 0.5 \ atm$. Let the pressure of $N_{2}O_{5(g)}$ decrease by $2x \ atm$ at time $t$.
According to the stoichiometry: $2N_{2}O_{5(g)} \rightarrow 2N_{2}O_{4(g)} + O_{2(g)}$.

Time Phase	Reaction: $2N_{2}O_{5(g)} \rightarrow 2N_{2}O_{4(g)} + O_{2(g)}$
$t=0$	$0.5 \ atm \rightarrow 0 \ atm + 0 \ atm$
$t=100 \ s$	$(0.5 - 2x) \ atm \rightarrow 2x \ atm + x \ atm$

The total pressure $p_t$ at time $t$ is given by:
$p_t = p_{N_2O_5} + p_{N_2O_4} + p_{O_2} = (0.5 - 2x) + 2x + x = 0.5 + x$.
Thus,$x = p_t - 0.5$.
At $t = 100 \ s$,$p_t = 0.512 \ atm$,so $x = 0.512 - 0.5 = 0.012 \ atm$.
The partial pressure of $N_2O_5$ at $t = 100 \ s$ is:
$p_{N_2O_5} = 0.5 - 2x = 0.5 - 2(0.012) = 0.5 - 0.024 = 0.476 \ atm$.
For a first-order reaction,the rate constant $k$ is:
$k = \frac{2.303}{t} \log \frac{P_0}{p_{N_2O_5}} = \frac{2.303}{100} \log \frac{0.5}{0.476}$.
$k = \frac{2.303}{100} \log(1.0504) \approx \frac{2.303}{100} \times 0.02136 \approx 4.92 \times 10^{-4} \ s^{-1}$.