Page 59 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

Equilibrium I 43

Example No. 2: When 1 M H2(g) and 1 M 12(g) come to equilibrium
at 730 K, determine the concentration of each substance at
equilibrium, if K, = 48.9. Determine the value of Kp given that R
= 0.08314 dm3 bar K-' mol-'.

1. Read the question carefully.
2. Species involved: H2(g), 12!g) and HI(g), i.e. all gases + homo-
geneous equilibrium (one single phase).
3. No balanced chemical equation is given. Therefore:
(a) identify the reactants: H2(g) and 12(g);
(b) identify the product: HI(g);
(c) write a chemical equation for the equilibrium reaction and
balance:
H2(g) + I2(g) == 2HI(g).
4. K, = [HI]2/{~2][12]) = 48.9.
5. Kp = P(HI)~/(~(H~)~(I~)) = K, (RqA"g, where Avg = (2) - (2) =
0.
Hence Kp = Kc(Rq0 = 48.9.
6.
- 2H4,).
A
Initial conc./M 1 1 0
Change -X --x + 2x
FinalconcJM 1 - x 1 -x 0 + 2x

7. K, = [HI]2/{[H2][12]) = 48.9 = (2~)~/(1 - x)(l - X) = (2~)~/
(1 - x)2.
8. In this example, if you assume x << 1 + K, = 4x2 = 48.9.
Hence, x2 = 12.225 and x = 3.496. Since x > 1, the assump-
tion made is not valid (as expected since Kc >> 1, i.e. the
equilibrium lies towards the products), and a quadratic equa-
tion must be solved. Returning to the original expression of K,
in step 7: (2~)~/(1 - x)~ 48.9 . . . (t).
=
Therefore: 4x2 = 48.9(x2 + 1 - 2x) = 48.9~~ 97.8~ + 48.9
-
=+ 44.9~~ 97.8~ + 48.9 = 0
-
This equation is in the form ax2 + bx + c = 0, with solution,
d
-b f n
X=
2a
where a = 44.9, b = -97.8 and c = 48.9.