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.

Asked by Pragya Singh | 1 year ago | 80

Let the concentration of N_{2}O at equilibrium be x.

The given reaction is:

2N_{2}(g) + O_{2}(g) ⇌ 2N_{2}O(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,

Answered by Pragya Singh | 1 year ago

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