Question

For the reaction 2A (g) + B (aq) + 3 C (`) −−)−−* D (s) + 3 E (g), the concentrations at equilibrium are found to be

PA = 2.8 × 103 Pa [B] = 1.2 × 10−2 M [C] = 12.8 M PE = 2.6 × 104 Torr

(a) Find the numerical value of the equilibrium constant.

(b) Calculate ∆G◦ .

(c) Is the reaction spontaneous as written?

Answer 1

a)

in equilibrium

Kp = P-E^3 / (P-A)^2 *[B][C]

Pv = nRT

n/V = P/(RT)

for A

M = P/(RT) --> (2.8*10^3)/(101325) / (0.082*298) = 0.001130 M of A

for E

M = P/(RT) --> (2.6*10^4)/(760) / (0.082*298) = 1.40 M of E

then

Keq = [E]^3 /([A]^2 * [B][ C][^3

Keq = (1.40 ^3) / ((0.001130 ^2)(1.2*10^-2)(12.8)^3) = 85391.779

b)

dG° = -RT*ln(Keq)

dG° = -8.314*298*ln(85391.779)

dG° = -28132.842 J/mol

dG°= -28.13 kJ/mol

c)

since dG < 0, and K> 1

this favours products strongly, i.e. it is spontaneous