Question

Consider the following reaction: Br2(l) + 2I-(aq) ---> 2Br-(aq) + I2(s)

a.Calculate ΔGorxn

b.Calculate K

Answer 1

It is much easier to do this via Ecell (reduction potentials) than dG information

first, find the half cells

I2(s) + 2 e− ⇌ 2 I− +0.54

Br2(l) + 2 e− ⇌ 2 Br− +1.066

Clearly, I2 is being formed, i.e. reduced so invert it

2 I−⇌ I2(s) + 2 e− -0.54

Br2(l) + 2 e− ⇌ 2 Br− +1.066

Add all

Br2(l) + 2 e− +  2 I− ⇌ I2(s) + 2 e− + 2 Br− E = -0.54 +1.066

cancel common terms

Br2(l) + 2 I− ⇌ I2(s) + 2Br− E = 0.526 V

now,

relate dG and Ecell via

dG = -n*F*Ecell

where n = moles of e-, F = faraday constant 96500 C/mol, Ecell =0.526 V

dG = -2*96500*0.526

dG = -101,518 J/mol

dG = -101.518 kJ/mol

b)

for K, we can always relate dG and K, free energy and equilibrium conditions via:

dG = -RT*ln(K)

solve for K

K = exp(-dG/(RT))

K = exp(--101518/(8.314*298))

K = 6.2391*10^17

this is clearly favoured toward products