The reaction N2g-O2g- 2NOg contributes to air pollution whenever a fuel is burnt in air at a high temperature- At 1500K- equilibrium constant K for it is 1-0- 10-5- Suppose in a case -N2-0-80 mol L-1 and -O2-0-20 mol L-1 before any reaction occurs- Calculate the equilibrium concentrations of the product after the mixture has heated to 1500K-

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The reaction ...

Question

The reaction $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500K, equilibrium constant K for it is $1.0 \times 10^{- 5}$ . Suppose in a case $[N_{2}] = 0.80 m o l L^{- 1}$ and $[O_{2}] = 0.20 m o l L^{- 1}$ before any reaction occurs. Calculate the equilibrium concentrations of the product after the mixture has heated to 1500K.

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Solution

$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$

$\begin{matrix} Intitial conc. & 0.80 & 0.20 & O At equilibrium & 0.80 - x & 0.20 - x & 2 x \end{matrix}$
Now,
$K = \frac{[N O]^{2}}{[O_{2}] [N_{2}]} = \frac{(2 x)^{2}}{(0.8 - x) (0.2 - x)} = 10^{- 5}$

$\Rightarrow \frac{4 x^{2}}{0.16 - 1 x + x^{2}} = 10^{- 5}$

$\Rightarrow 400000 x^{2} = x^{2} - x + 0.16$

$\Rightarrow 399999 x^{2} + x - 0.16 = 0$

$b^{2} - 4 a c = (1)^{2} - 4 (399999) (- 0.16) = 1 + 255999.36 = 256000.36$

$\Rightarrow \sqrt{b^{2} - 4 a c} = \sqrt{256000.36} = 505.96$

$x = \frac{- 1 \pm 505.96}{2 \times 399999} = \frac{504.96}{2 \times 399999}$

$= 0.63 \times 10^{- 3}$

$[N O] = 2 x = 2 \times 0.63 \times 10^{- 3} = 1.26 \times 10^{- 3} m o l / l i t r e$

Equilibrium concentration of the product NO is $1.26 \times 10^{- 3} m o l e / l i t r e$

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