Question
Mon December 17, 2012 By: Dipak Saikia

If a mixture is of 0.482 mole Nitrogen and 0.933 mol of oxygen is placed in a 10L reaction vessel and allowed to form N2O at a temperature, determine the composition of equilibrium mixture.

Expert Reply
Mon December 17, 2012

The given equation is:
2 N2 (g) + O2 (g) ? 2 N2O (g)
Therefore the equilibrium constant Kc, will be given as:
Kc = [N2O] 2 / [N22 [O2]

Initial molar concentrations are:
[N2] = 0.482 mol / 10.0 L = 0.0482 mol L-1
[O2] = 0.933 mol / 10.0 L = 0.0933 mol L-1
Let us assume that x mol of oxygen has reacted with 2x mol of N2 to form 2x mol of N2O, so that equilibrium concentrations are:
[N2] = 0.0482 – 2x
[O2] = 0.0933 – x
[N2O] = 2x

Therefore,
Kc = (2x) 2 / (0.0482 – 2x) 2 (0.0933 – x)
Since, Kc is very small (2.0 X 10-37), we can assume x to be infinitesimally small so that,
0.0482 – 2x ? 0.0482 and 0.0933 – x ? 0.0933
Kc = 4 x2 / (0.0482) 2 (0.0933)
i.e.
4x2 = (0.0482) 2 (0.0933) X (2.0 X 10-37)
4 x2 = 4.34 X 10-41

x2 = 4.34 X 10-41

Therefore x = 3.3 X 10-21

Thus, equilibrium concentrations are:
[N2] = 0.0482 – 2x ? 0.0482
[O2] = 0.0933 – x ? 0.0933
[N2O] = 2x = 2 X 3.3 X 10-21

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