Ncert Exercises & Solutions

7.8 Reaction between N2 and O2– takes place as follows:

2N2 (g) + O2 (g) 2N2O (g)

If a mixture of 0.482 mol 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 the equilibrium mixture.

Let the concentration of

N_{2} O

at equilibrium be x.

The given reaction is :

2 N_{2 (g)} + O_{2 (g)} \leftrightarrow 2 N_{2} O_{(g)} Initial conc . 0.482 mol 0.933 mol 0 At equilibrium (0.48 - x) mol (1.933 - x / 2) mol x mol

Therefore, at equilibrium, in the 10L vessel:

[N_{2}] = \frac{0.482 - x}{10}, [O_{2}] = \frac{0.933 - 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,
$[N_{2}] = \frac{0.482}{10} = 0.482 m o l L^{- 1} a n d [O_{2}] = \frac{0.933}{10} = 0.0933 m o l L^{- 1}$
$N o w,$
$K_{c} = \frac{{[N_{2} O_{(g)}]}^{2}}{{[N_{2 (g)}]}^{2} [O_{2 (g)}]}$
$\Rightarrow 2.0 \times 10^{- 37} = \frac{{(\frac{x}{10})}^{2}}{{(0.0482)}^{2} (0.0933)}$
$\Rightarrow \frac{x^{2}}{100} = 2.0 \times 10^{- 37} \times {(0.0482)}^{2} \times 0.0933$
$\Rightarrow x^{2} = 43.35 \times 10^{- 40}$
$\Rightarrow x = 6.6 \times 10^{- 20}$
$[N_{2} O] = \frac{x}{10} = \frac{6.6 \times 10^{- 20}}{10}$
$= 6.6 \times 10^{- 21}$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions