T= 298 K,
BrCl(g) = 1.40 mol
Volume =12.0 L
ΔG0[Br2(g)] = 3.11 kJ/mol,
ΔG0[BrCl(g)] =−0.98 kJ/mol.]
Pls correct the equation it must be as below
BrCl(g)⇌1/2Br_2(g) + 1/2Cl_2(g)
Or
2BrCl(g)⇌Br_2(g) + Cl_2(g)
Part A
Calculate ∆ G
∆ G = 1/2 x ∆ G (Br_2) - ∆ G (BrCl)
∆ G = ½ x (3110)- (-980)
∆ G = 2535 J
∆ G = - RT lnKp
lnKp = -∆ G/RT
(R = 8.314 J/mol and T = 298K)
lnKp = (-2535)/(8.134 x 298) = -1.023
Kp = 0.3595
Given BrCl(g)= 1.25 moles in beginning
2BrCl(g) ------> Br2(g) + Cl_2(g)
I 1.25............................ 0 ............................... 0
C -2x ......................... +x ........................... +x
E (1.25-2x) ........................ x ...................... x
Kp = [Br_2(g)] [Cl_2(g)] /[BrCl(g)]2
Kp = x2 / (1.25 - 2x)2
K p = x/(1.25 – 2x)
K p = 0.3595
(0.3595) x (1.25 – 2x) = x
x = 0.4493 – 0.719x
x + 0.719x = 0.4493
1.719x = 0.4493
x = 0.5033/1.719
x = 0.261
BrCl = 1.25 -2x
BrCl = 1.25 – (2 x 0.261)
BrCl = 1.25 – 0.522
BrCl = 0.727 moles
Moles of BrCl present at equillibrium = 0.727 moles
Moles of Br2 at equllibrium = x = 0.261 moles
Moles of Cl2 at equllibrium = x = 0.261 moles