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First Order Reaction

The decomposi...

Question

The decomposition of $N_{2} O_{5}$ according to the equation $2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
is a first order reaction. After 2 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 500 mm Hg. On complete decomposition, the total pressure is 800 mm Hg. If the rate constant of the reaction is $y m i n^{- 1}$ then, what is the value of y? (given log 2 = 0.3, log 3 = 0.5)

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Solution

The given reaction is
$2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$
$\begin{matrix} p^{0} & 0 & 0 & i n i t i a l l y p^{0} - 2 p & 4 p & p & a t 2 m i n 0 & 2 p^{0} & \frac{p^{0}}{2} & a f t e r l o n g t i m e \end{matrix}$
Given:
$p^{0} + 3 p = 500 m m H g$ and
$2.5 p^{0} = 800 m m H g$
$\Rightarrow p^{0} = 320 m m H g$ and p = $60 m m H g$
So, $k = \frac{2.303}{2} l o g \frac{p^{0}}{p^{0} - 2 p} = \frac{2.303}{2} l o g \frac{320}{200} = \frac{2.303}{2} l o g 1.6 = \frac{2.303}{2} (4 \times l o g 2 - 1) = 0.2303$

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