The reaction between \( N_2\) and \( O_2 \)takes place as follows:
\( 2N_{2}(g)+O_{2} ⇌ 2N_{2}O(g) \)
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 \( N_2O\) at a temperature for which
\( Kc =2.0\times 10^{-37}\), determine the composition of the equilibrium solution.
1 Answer
Solution :-
Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g) + O2(g) ⇌ 2N2O(g)
Initial conc. 0.482 mol 0.933 mol 0
At equilibrium(0.482-x)mol (1.933-x)mol x mol
\( [N_{2}]=\dfrac{0.482-x}{10}\cdot [O_{2}]=\dfrac{0.933-\dfrac{x}{2}}{10},\)
\( [N_{2}O]=\dfrac{x}{10} \)
The value of equilibrium constant is extremely small. This means that only small amounts . Then,
Now,
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