Question

The equilibrium constant (K_p) for the reaction below is 4.40 at 2000. K.

H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g)

Calculate

Δ

G

o

for the reaction.

kJ/mol

Calculate

Δ

G for the reaction when the partial pressures are

P_H2 = 0.22 atm,
P_CO2 = 0.72 atm,
P_H2O = 0.66 atm, and
P_CO = 1.16 atm.

Answer 1

if Kp = 4.4

then

Kp = Kc*(RT)^dn

dn = mol of product in gas - mol of reactants in gas

dn = 1+1 - (1+1) = 2-2 = 0

so

Kp = Kc

therefore

Kc = 4.4 as well

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kc = [H2O][CO] /([H2][CO2])

if thi is true, then

dG = -RT*ln(Kc)

dG = -8.314*298*ln(4.4)

dG = -3,670.7819 J/mol

dG = -3.67 kJ/mol

Now,

dG overall:

dG = dGº + RT*ln(Q)

Q = quotient

Q = P-H2O* P -CO /([P-H2][P-CO2])

Q = (0.66)(1.16)/((0.22*0.72))

Q = 4.8333

get:

dG = dGº + RT*ln(Q)

dG =  -3670.7819 + 8.314*298*ln(4.8333)

dG = 232.70 J/moll

dG = 0.232 kJ/mol