Dissociation of a gas A2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A2(g) 2A(g) The standard Gibbs energy of formation of the involved substances has been provided below: arraycc Substance & Gf /kJ mol-1 \ A2 & -100.00 \ A & -50.832 array The degree of dissociation of A2(g) is given by (x 10-2)1/2 where x = . (Nearest integer). [ Given : R = 8 J mol-1 K-1, 2 = 0.3010, 3 = 0.48 ] Assume degree of dissociation is not negligible.
Answer
33
Solution
r G = 2 Gf (A) - Gf (A2)
= 2(-50.832) - (-100.00) = -1.664 kJ mol-1 = -1664 J mol-1
r G = -2.303 RT Kp
Kp = 16642.303 8 300 = 16645527.2 = 0.3010
Kp = 2
Let 1 mol of A2 dissociate to extent α:
A2: 1 - , A: 2 , total = 1 +
Kp = pA2pA2 = ( 2 1+ )2 P2 ( 1- 1+ )P = 4 2 P1- 2
With P = 1 bar:
4 21- 2 = 2
4 2 = 2 - 2 2
2 = 13 = 33.3 10-2
= (33.3 10-2)1/2
x 33
Answer: 33