7.8. Reaction between nitrogen and oxygen takes place as follows:
2N2?(g) + O2?(g) ? 2N2?O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the composition of the equilibrium mixture.
7.8. Reaction between nitrogen and oxygen takes place as follows:
2N2?(g) + O2?(g) ? 2N2?O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the composition of the equilibrium mixture.
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1 Answer
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Let x moles of N2 (g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is:
[N2? ]
[O2]
[N2? O]
Initial concentration
0.482? / 10
0.933/ 10
0
Conc. at equilibrium
(0.482 – x)? / 10
(0.933 – x/2) / 10
x/ 10
The value of equilibrium constant (2.0 x 10-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, value of x is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of chemical equilibrium, Kc = [N2O]2 / [N2]2 [O2]
= Kc = 2.0 x 10-37
= (x/10)2 / [ (0.482/ 10)2 x (0.933 / 10)2]
= 0.01 x2 / (2.1676 x 10-4)
x2= 4
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