At 25degreeC, the pressure of water vapor over an equilibrium mixture of CuSO_4

Question

At 25degreeC, the pressure of water vapor over an equilibrium mixture of CuSO_4 5H_2O(s) and CuSO_4 3H_2O(s) is 7.8 torr. Calculate delta_rG and delta_rGdegree for CuSO_4 5H_2O(s) rightarrow CuSO_4O 3H_2O(S) + 2H_2O(g) 2. Calculate delta_rG at 25degreeC for H_2O(1) rightarrow H^+(aq) + OH*(aq) given |H^+| = |OH*| = 1 times 10^-7 M. 3. At 400degreeC and 350 bar, a 1:3 mixture of nitrogen and hydrogen reacts to form an equilibrium mixture containing x(NH_3) = 0.50. Assuming ideal gas behavior, calculate K for N_2(g) + 3H_2(g) rightarrow 2NH_3(g) 4. Calculate K at 298K for the equation trans-1,2-dichloroethene rightarrow cis-l,2-dichloroethene given delta_fG^theta = 27.34 kj/mol for trans-l,2-dichloroethene and delta_fG^theta - 22.11 kj/mol for cis-1,2-dichloroethene. Assuming the activities of the isomers to be given by the respective concentrations, what is the ratio of the concentrations of the cis isomer to the trans isomer at equilibrium?

Explanation / Answer

1) For the equilibrium:

Kp = p2H2O

Kp = 7.8 X 7.8 = 60.84

rG = 0 (for equilibrium)

rG0 = -RTlnKp = -8.314 x 298 X ln60.84 = 10180.3 Joules = 10.18 KJ / mole

2) Again rG = 0 for equilbirum reaction

rG0 = -RT lnKc

Kc = [H+] [ OH-] = 10^-14

rG0 = 79881 J / mole

3) Let the intial concentration of N2 = 1 mole and H2 = 3moles

Mole fraction of NH3 = 0.5

Let the amount of NH3 formed = 2x

so [N2] = 1-x

[H2] = 3-3x

Total moles = 2x + 1-x + 3-3x = 4-2x

Mole fraction of NH3 = 0.5 = 2x / 4 - 2x

2 - x = 2x

2 = 3x

x =0.67 mole

so Kc = [NH3]^2 / [N2] [ H2]^3

Kc = 0.67 X 0.67 / (1-0.67)(3-3X0.67)

Kc = 0.449 / 0.33 X 0.99

Kc = 1.38

4) The Delta G of of reaction = Delta G of products - Delta G of reactants

rG0 = rG0 products - rG0 reactants

rG0 = 22.11 - 27.34 = -5.23 KJ / mole

rG0 = -RT lnK

-5.23 = -8.314 X 298 X ln K

on calculating

K = 1.002

K = [Cis-1,2dichloroethene] / [Trans-1,2-dichloroethene]

So the ratio will be almost 1:1

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