Question

Question 7.8 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 × 10–37, determine the composition of equilibrium mixture.

Class XI Equilibrium Page 225

Answer 1

k is very small, which means negligible amounts of N2 and O2 react .
it means conc. of reactants at intial ≈ conc . at equilibrium .
so,
conc. of N2 = [ N2 ] = (0.482 - 2x)/10 = 0.482/10
= 0.0482
conc. of O2 = [O2 ] = (0.933 - x)/10 = 0.933/10
= 0.0933

Kc = [N2O]²/[N2]² [O2]
2 × 10^-37 = [2x/10]²/(0.0482)² (0.0933)
2 × 10^-37 = 4x²/100(0.0482)² (0.0933)
x² = 10.837 × 10^-40
take square root
x = 3.292 × 10^-20

hence, conc. of [N2O] = 2x/10 = 2 × 3.92 × 10^-20/10 = 6.58 × 10^-21 M

Answer 2

Answer:

2 N2(g) +O2(g) <=>2N2O(g)

0.482 0.933 0

(0.482-2x) (0.933-x) 2x

(0.482-2x) /10 (0.933-x) /10 2x/10

k is very small, which means negligible amounts of N2 and O2 react.

[N2]=(0.482-2x) /10 = 0.482/10 = 0.0482 mol L-¹

[O2] = (0.933-x) /10 = 0.933/10 = 0.0933 mol L-¹

Kc = [N2O]²/[N2]² [O2]²

=> 2×10^-37 = [2x/10]² / [0.0482]² [0.0933]

=> 2×10^-37 = 4x² / 100 (0.0482) ² (0.0933)

=> x² = 10.837×10^-40

=> x = √ 10.837×10^-40

=> x = 3.292×10^-20

Now,

[N2O] = 2x / 10

= 2×3.92×10^-20 / 10 = 6.58×10^-21 mol L-¹