Consider the following reaction in a sealed vessel at equilibrium with concentrations of N2 = 3.0 10-3 M, O2 = 4.2 10-3 M and NO = 2.8 10-3 M. 2NO(g) N2(g) + O2(g) If 0.1 mol L-1 of NO(g) is taken in a closed vessel, what will be degree of dissociation ( ) of NO(g) at equilibrium?
- 0.00889
- 0.0889
- 0.8889
- 0.717
Answer
(4) 0.717
Solution
Kc = [N2][O2][NO]2 = (3.0 10-3)(4.2 10-3)(2.8 10-3)2
= 12.67.84 = 1.607
Starting with 0.1 M NO: at equilibrium [NO] = 0.1(1- ), [N2] = [O2] = 0.05
Kc = (0.05 )2[0.1(1- )]2 = ( 2(1- ) )2
2(1- ) = 1.607 = 1.268
= 2.536(1- )
= 2.5363.536 = 0.717
Answer: (4)