Question

Consider the system:

A (aq) → B (aq)

at 283 K where ΔG^o_form A = -18.3 kJ/mol and ΔG^o_form B = -17.3 kJ/mol. Calculate the concentration of B at equilibrium when 2.21 mol of A and 1.71 mol of B are mixed in water to form exactly one liter of solution.

Answer 1

Consider the system:

A (aq) → B (aq)

at 283 K where ΔG^o_form A = -18.3 kJ/mol and ΔG^o_form B = -17.3 kJ/mol. Calculate the concentration of B at equilibrium when 2.21 mol of A and 1.71 mol of B are mixed in water to form exactly one liter of solution.

Ans :-

We know

ΔG⁰_rxn = ∑ Δ_fG⁰_Products - ∑ Δ_fG⁰_Rectants

ΔG⁰_rxn = - 17.3 KJ/mol + 18.3 KJ/mol

ΔG⁰_rxn = 1.0 KJ/mol

Also

ΔG⁰_rxn = - 2.303 RT log K

1.0 KJ/mol = - 2.303 x 8.314 x 10^-3 KJ^-1mol^-1 x 283 K log K

log K = - 0.1845

K = 10^{- 0.1845}

K = 0.654

ICE table is :

.................................A (aq) ---------------------------> B (aq)

Initial (I).....................2.21 M...................................1.71 M

Change (C)................ + y ........................................- y

Equilibrium (E).........(2.21+y) M.............................(1.71 - y) M

Expression of K is :

K = [B]/[A]

0.654 = (1.71-y) / (2.21+y)

1.445 + 0.654 y = 1.71 - y

1.654 y = 0.265

y = 0.16

So,

Equilibrium concentrations are :

[A] = 2.21+0.16 = 2.37 M

[B] = 1.71-0.16 = 1.55 M