Question

3. Molecular bromine is 24% dissociated at 1600K and 1.00 bar in the equilibrium:

Br​​2 (g)-------- 2Br(g)

Calculate:

a) the equilibrium constant K at 1600K (8)

b) the standard Gibbs energy of reaction 4)

C) K at 2000K , given that ∆r H = +112kj/mol over the temperature range. (6)

Answer 1

Solution :-

Given data

Br​​2 (g)-------- 2Br(g)

Br2 dissociate 24 % at 1600 K

a)the equilibrium constant K at 1600K (8)

dissociation is 24 % that means if we assume 1 mole of the Br2 then

1 mol * 24 % / 100 % = 0.24 mol Br2 will dissociate

Therefore at equilibrium moles of Br2 remain = 1 mol – 0.24 mol = 0.76 mol

Moles of Br formed = 0.24 mol * 2 = 0.48 mol Br

Lets write the Kc equation

Kc= [Br]^2 / [Br2]

Kc =[0.48]^2 /[0.76]

Kc = 0.3032

b)the standard Gibbs energy of reaction 4)

formula to calculate the Gibbs energy is as follows

Delta Go = - RT ln K

Lets put the values in the formula

Delta Go = - 8.314 J per K mol * 1600 K * ln 0.3032

Delta Go = 15874 J

Delta Go = 15874 J * 1 kJ / 1000 J = 15.87 kJ

C) K at 2000K , given that ∆r H = +112kj/mol over the temperature range. (6)

Now lets calculate the equilibrium constant at 2000 K

ln k2/k1 = delta H / R [(1/T1)-(1/T2)]

ln k2 /0.3032 = 112000 J per mol / 8.314 J per K . mol * [(1/1600)-(1/2000)]

ln K2 / 0.3032 = 1.689

k2/0.3032 = antiln 1.689

k2/0.3032 = 5.414

K2 = 5.414 * 0.3032

K2 = 1.64

Therefore equilibrium constant at 2000 K = 1.64