Question

In: Chemistry

Kp has the value 1.00x10^-5 for the equilibrium CO2(g) + H2(g) <--> CO(g) + H2O(g) at...

Kp has the value 1.00x10^-5 for the equilibrium

CO2(g) + H2(g) <--> CO(g) + H2O(g)

at 25oC and ΔSo is -41.9 J/K (ΔHo and ΔSo do not change much with temperature). One mole of CO, 2 moles of H2, and 3 moles of CO2 are introduced into a 5-liter flask at 25oC. Calculate

(a) ΔGo at 25oC,

(b) the equilibrium pressure,

(c) the moles of each species present at equilibrium,

(d) Kp at 100 0C. The mixture obeys the ideal equation of state.

Solutions

Expert Solution

(a) dHo at 25 oC = dHo(products) - dHo(reactants)

= (-110.5 - 241.818) - (-393.509)

= 41.191 kJ/mol

dGo = dHo - TdSo

        = 41.191 - (298 x -0.0419)

        = 53.68 kJ/mol

(b) The equilibrium pressure P = nRT/V

= 6 x 0.08205 x 298/5 = 29.34 atm

(c) initial concentrations,

[CO] = 1/5 = 0.2 M

[H2] = 2/5 = 0.4 M

[CO2] = 3/5 = 0.6 M

at equilibrium let x be the amount the reactant has gone under conversion then,

Kp = 1 x 10^-5 = [CO][H2O]/[CO2][H2]

1 x 10^-5 = (0.2+x)(x)/(0.4-x)(0.6-x)

1 x 10^-5 = 0.2x + x^2/0.24 - x + x^2

2.4 x 10^-6 - 1 x 10^-5x + 1 x 10^-5x^2 = 0.2x + x^2

x^2 + 0.2x - 2.4 x 10^-6 = 0

x = 1.2 x 10^-5 M

Equilibrium concentrations,

[H2O] = 1.2 x 10^-5 M ; 6 x 10^-5 mols

[CO] = 0.2 M ; 1 mols

[H2] = 0.4 M ; 2 mols

[CO2] = 0.6 M ; 3 mols

Total = 6 mols

(d) Kp at 100 oC = 373 K

dGo = -RTlnKp

53.68 x 1000 = -8.314 x 373 ln Kp

Kp = 3.04 x 10^-8

queen_honey_blossom answered 1 hour ago
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