Question

Reaction between \(N_2\) and \(O_2^–\) takes place as follows: 
\(2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)\)
If a mixture of \(0.482 \ mol\) \(N_2\) and \(0.933\  mol\) of \(O_2\) is placed in a \(10\  L\) reaction vessel and allowed to form \(N_2O\) at a temperature for which \(K_c = 2.0 × 10^{–37}\), determine the composition of equilibrium mixture.

Answer 1

Let the concentration of N₂O at equilibrium be x.
The given reaction is: 
                          \(2N_2(g) + O_2(g)  ↔ 2N_2O(g)\)
Initial conc.   \(0.482\ mol\)   \(0.933 \ mol\)    \(0\)
At equilibrium \((0.482 - x)\ mol\) \((1.933 - x)\ mol\)  \(x\) \(mol\)
Therefore, at equilibrium, in the 10 L vessel:
\([N_2] = \frac {0.482 - x }{10},\)

\([O_2] = \frac {0.933 - \frac x2}{10} ,\)

\([N2O] = \frac {x}{10}\)
The value of equilibrium constant i.e., \(K_c= 2.0 × 10^{-37}\) is very small. Therefore, the amount of N₂ and O₂ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N₂ and O₂.
Then, \([N_2] = \frac {0.482}{10} = 0.0482\  mol L^{-1 }\)

and \([O_2] = \frac {0.933}{10}= 0.0933 \ mol L^{-1}\)
Now,
\(K_c = \frac {[N_2O(g)]^2}{[N_2(g)]^2 [O_2(g)] }\)

⇒ \(2.0×10^{-37} = \frac {(\frac {x}{10})^2}{(0.0482)^2 (0.0933) }\)

⇒ \(\frac {x^2}{100} = 2.0×10^{-37}×(0.0482)^2×(0.0933)\)

⇒ \(x^2 = 43.35×10^{- 40}\)

 ⇒ \(x = 6.6×10^{-20}\)

\([N_2O] = \frac {x}{10}\)

\([N_2O] =\) \(\frac {6.6 × 10^{- 20}}{10}\)

\([N_2O]  = 6.6×10^{-21}\)