Page 58 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

42 Chapter 4

EXAMPLES

Example No. I: Determine the concentration of all species at
equilibrium at 973 K, when 2 moles of 02(g)m in 4.0 dm3 are mixed
with 2.25 moles of N2(g) in 3.0 dm3, according to the equilibrium
reaction: 02(,) + N2(,) .C 2NO(,), given that Kc = 4.13 x lo-’. Is
this an example of a homogeneous or a heterogeneous equilibrium?
Evaluate Kp, given that R = 0.08314 dm3 bar IC-’ mol-’.

1. Read the question carefully.
2. Identify the species involved: 02(g), N2(g) and NO(,). All species
are in the gaseous phase; therefore this is a homogeneous
equilibrium (one single phase).
3. The balanced chemical equation is already given in the question:
02(g) + N2(g) 2NO(,).

4. K, = ~OI~/{[N~~[O,~) lo-’.
4.13
x
=
5. Kp = p(N0)2/(p(N2)p(02)} = K, (RT)*”,where Av = (2) - (2) = 0.
Hence Kp =K,(RT)O = 4.13 x lo-’, since from the rules of
indices, xo = 1.
6. Initial concentration of O~(S> 2 mol in 4.0 dm3 + 0.5 mol in
=
1 dm3, i.e. 0.5 M
Initial concentration of N2(,) = 2.25 mol in 3.0 dm3 + 0.75 mol
in 1 dm3, i.e. 0.75 M
2
02(t3) + - 2NO(f3)
Initial conc./M 0.5 0.75 0
Change --x -X + 2x
Final conc./M 0.5 - x 0.75 -X 0 + 2x

7. K,=~O]2/{[N2J[02]} =4.13~ (2~)~/{(0.5 -x)(0.75 -x)}.
8. Since K, << 1 (i.e. the equilibrium lies to the left), assume
xaO.5 and x <0.75+ K, = (4x2)/{(0.5) x (0.75)) =4.13 x lo-’.
Hence, 2 = 3.8719 x lO-’Oandx = 1.97 x lo-’.
9. (x/0.5)% = (1.97 x 10-’/0.5)% = 0.00394% and (x/0.75)%
= (1.97 x 10-’/0.75)% = 0.00263% i.e., both are <5%.
Hence the assumptions made were valid and it is not necessary
to solve a quadratic equation in this question.
10. [OZ(~)] 0.5 M; [Nz(~)] x 0.75 M (both slightly reduced);
M
= 2x = 2 x (1.97 x lo-’) = 3.94 x lO-’M.
11. K~ = 4.13 x