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Derivation of Kp and Kc

Reaction betw...

Question

Reaction between $N_{2}$ and $O_{2}$ takes place as follows:
$2 N_{2} (g) + O_{2} (g) \leftrightarrow 2 N_{2} O (g)$
If a mixture of 0.482 mol of $N_{2}$ and 0.933 mol of $O_{2}$ is placed in a 10 L reaction vessel and allowed to form $N_{2} O$ at a temperature for which $K_{c} = 2.0 \times 10^{- 37}$ , determine the composition of equilibrium mixture.

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Solution

Let the concentration of $N_{2} O$ at equilibrium be x.
The given reaction is:
$\begin{matrix} 2 N_{2 (g)} & + & O_{2 (g)} & \leftrightarrow & 2 N_{2} O_{(g)} I n i t i a l c o n c . & 0.482 m o l & 0.933 m o l & 0 A t e q u i l i b r i u m & (0.482 - x) m o l & (1.933 - x) m o l & x m o l \end{matrix}$
Therefore, at equilibrium, in the 10 L vessel:
$[N_{2}] = \frac{0.482 - x}{10}, [O_{2}] = \frac{0.933 - \frac{x}{2}}{10}, [N_{2} O] = \frac{x}{10}$
The value of equilibrium constant i.e. $K_{c} = 2.0 \times 10^{- 37}$ is very small. Therefore, the amount of $N_{2}$ and $O_{2}$ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of $N_{2}$ and $O_{2}$ . Then,
$[N_{2}] = \frac{0.482}{10} = 0.0482 m o l L^{- 1} a n d [O_{2}] = \frac{0.933}{10} = 0.0933 m o l L^{- 1}$
Now,
$K_{c} = \frac{[N_{2} O_{(g)}]^{2}}{[N_{2 (g)}]^{[} O_{2 (g)}]} \Rightarrow 2.0 \times 10^{- 37} = \frac{{(\frac{x}{10})}^{2}}{(0.0482)^{2} (0.0933)} \Rightarrow \frac{x^{2}}{100} = 2.0 \times 10^{- 37} \times (0.0482)^{2} \times (0.0933) \Rightarrow x^{2} = 43.35 \times 10^{- 40}$

$[N_{2} O] =\Rightarrow x = 6.6 \times 10^{- 20} m o l L^{- 1}$

$[N_{2}] = 0.0482 m o l L^{- 1} a n d [O_{2}] = 0.0933 m o l L^{- 1}$

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Derivation of Kp and Kc

Standard XII Chemistry

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Reaction between N2 and O2 takes place as follows: 2N2(g) +O2(g) ↔2N2O(g) If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc=2.0×10−37, determine the composition of equilibrium mixture.