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

##### 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, Answered by Pragya Singh | 1 year ago

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