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The reaction between \mathrm{N}_{2} and \mathrm{O}_{2} takes place as follows: 2 N_{2}(g)+O_{2} \rightleftharpoons 2 N_{2} O(g) If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form \mathrm{N}_{2} \mathrm{O} at a temperature for which \mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}, determine the composition of the equilibrium solution.
CBSE | Chemistry | Class 11 | Equilibrium | NCERT
The reaction between \mathrm{N}_{2} and \mathrm{O}_{2} takes place as follows: 2 N_{2}(g)+O_{2} \rightleftharpoons 2 N_{2} O(g) If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form \mathrm{N}_{2} \mathrm{O} at a temperature for which \mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}, determine the composition of the equilibrium solution.

Answer:

Assuming the concentration of \mathrm{N}_{2} \mathrm{O} at equilibrium be \mathrm{x}

The given reaction is:

2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \quad \rightleftharpoons \quad 2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})

Given, intial mol.= 0.482 \mathrm{~mol} of N_2 &

0.933 \mathrm{~mol} of O_2

At equilibrium, the moles will be following of Nitrogen and oxygen, respectively, (0.482-\mathrm{x}) \mathrm{mol}

(1.933-\mathrm{x}) \mathrm{mol}

The moles product is, \mathrm{x} \mathrm{mol}
Thus, concentration of each will be,

\left[N_{2}\right]=\frac{0.482-x}{10} ,\left[O_{2}\right]=\frac{0.933-\frac{\pi}{2}}{10},\left[N_{2} O\right]=\frac{x}{10}

The equilibrium constant has a value that is exceedingly tiny. This implies that only tiny quantities are involved. Then,

\left[N_{2}\right]=\frac{0.482}{10}=0.0482 \mathrm{~mol} \mathrm{~L}^{-1} and \left[O_{2}\right]=\frac{0.933}{10}=0.0933 \mathrm{~mol} L^{-1}

Now substutiting the value,

K_{c}=\frac{\left[N_{2} O_{(g)}\right]^{2}}{\left[N_{2(g)}\left[O_{2(g)}\right]\right.}

\Rightarrow 2.0 \times 10^{-37}=\frac{\left(\frac{x}{2}\right)^{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}\left[N_{2} O\right]=\frac{x}{10}=\frac{6.6 \times 10^{-20}}{10}

=6.6 \times 10^{-21}

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