For the thermal decomposition of at constant volume, the following table can be formed, for the reaction mentioned below: [2025]

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Solution

Q.
For the thermal decomposition of $N_{2} O_{5} (g)$ at constant volume, the following table can be formed, for the reaction mentioned below:

$2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)$

Sr. No. Times Total Pressure (atm)

1 0 0.6

2 100 $' x'$

$x$ = _____ $\times 10^{- 3}$ atm (nearest integer)

Given: Rate constant for the reaction is $4.606 \times 10^{- 2} s^{- 1}$ . [2025]

Ans.
(897)

$\begin{matrix} 2 N_{2} O_{5} (g) & ⟶ & 2 N_{2} O_{4} (g) & + & O_{2} (g) \\ t = 0 & p ° & 0 & 0 \\ t = 100 s & p ° - 2 p & 2 p & p \end{matrix}$

Total pressure at $t = 0$ is $p ° + 0 + 0 = p ° = 0.6$

Total pressure at $t = 100 s$ is $(p ° - 2 p) + 2 p + p = p ° + p = x$

From above two equations, $p = x - 0.6$

For first order reaction, rate constant is given as:

$k = \frac{2.303}{t} \log \frac{P_{N_{2} O_{5}} (t = 0)}{P_{N_{2} O_{5}} (t = t)}$

$4.606 \times 10^{- 2} = \frac{2.303}{100} \log \frac{0.6}{0.6 - 2 (x - 0.6)}$

$4.606 \times 10^{- 2} = \frac{2.303}{100} \log \frac{0.6}{1.8 - 2 x}$

$\log \frac{0.6}{1.8 - 2 x} = 2$

$\frac{0.6}{1.8 - 2 x} = 10^{2} = 100$

$1.8 - 2 x = 0.006$

$x = 0.897 atm = 897 \times 10^{- 3} atm$

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