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Question

Reaction between N2 and O2 takes place as follows:
2N2(g) +O2(g) 2N2O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc=2.0×1037, determine the composition of equilibrium mixture.

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Solution

Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g)+O2(g)2N2O(g)Initial conc.0.482 mol0.933 mol0At equilibrium(0.482x)mol(1.933x) molx mol
Therefore, at equilibrium, in the 10 L vessel:
[N2]=0.482x10,[O2]=0.933x210,[N2O]=x10
The value of equilibrium constant i.e. Kc=2.0×1037 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2. Then,
[N2]=0.48210=0.0482 mol L1 and [O2]=0.93310=0.0933 mol L1
Now,
Kc=[N2O(g)]2[N2(g)][O2(g)]2.0×1037=(x10)2(0.0482)2(0.0933)x2100=2.0×1037×(0.0482)2×(0.0933)x2=43.35×1040


[N2O]=x=6.6×1020 mol L1

[N2]=0.0482 mol L1 and [O2]=0.0933 mol L1


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