Question

A sample of air consisting of ${{N}_{2}}$ and ${{O}_{2}}$ was heated to 2500K until the equilibrium ${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}$ was established with an equilibrium constant ${{K}_{c}}=2.1\times {{10}^{-3}}$. At equilibrium, the mole% of NO was 1.8. Estimate the initial composition of air in mole fraction of ${{N}_{2}}$ and ${{O}_{2}}$.

Answer 1

Hint: Take the reaction and assume some arbitrary ‘a’, ‘b’ moles of reactants are present at time, t=0. Then try to solve using equilibrium constant concept and derive the mole fraction formula, $\dfrac{a}{a+b}$ to obtain a mole fraction of one of the reactants. Then, a second mole fraction can just be obtained by subtracting 1 by mole fraction of another reactant.

Complete answer:
Let’s have a look at the reaction,
${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}$
At equilibrium, the mole% of NO was 1.8. That means, the mole fraction of NO is 0.018.

	${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}$
Initially (t=0)	a	b	0
At equilibrium	(a-x)	(b-x)	2x

Now, let’s calculate equilibrium constant, ${{K}_{c}}$
${{K}_{c}}=\dfrac{{{\left[ NO \right]}^{2}}}{\left[ {{N}_{2}} \right]\left[ {{O}_{2}} \right]}=\dfrac{{{(2x)}^{2}}}{(a-x)(b-x)}=2.1\times {{10}^{-3}}$
But, from the definition of mole fraction we have,
Mole fraction of NO $=\frac{2x}{a-x+b-x+2x}=\dfrac{2x}{a+b}=0.018$
$\therefore x=0.009(a+b)$
Substituting the value of x in the equation ${{K}_{c}}=\dfrac{{{(2x)}^{2}}}{(a-x)(b-x)}=2.1\times {{10}^{-3}}$ and solving it we get,
$\dfrac{a}{a+b}=0.79$
Mole fraction of ${{N}_{2}}=\dfrac{a}{a-x+b-x+2x}=\dfrac{a}{a+b}=0.79$
We know that the sum of mole fractions of all the components is equal to unity.
Using this relation, mole fraction of oxygen, ${{O}_{2}}=1-0.79=0.21$
Therefore, the initial composition of air in mole fraction of ${{N}_{2}}$and ${{O}_{2}}$ was estimated and found to be 0.79 and 0.21 respectively.

Note:
Remember the sum of mole fractions of all components in the system is always equal to unity. This relation can be used to avoid lengthy calculation of mole fraction of second component and simplifying the answer.