Question

Enter your answer in the provided box. In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a gas and can therefore be separated from solid impurities: Ni(s) + 4CO(g) ⇌ Ni(CO)4(g) Given that the standard free energies of formation of CO(g) and Ni(CO)4(g) are −137.3 and −587.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 70.0°C. Assume that ΔG o f is temperature-independent.

Answer 1

Answer:

The equilibrium constant is  6.6 * 10^5

Explanation:

Step 1: Data given

The standard free energies of formation of CO(g) = -137.3 kJ/mol

The standard free energies of formation of Ni(CO)4(g) = −587.4 kJ/mol,

Temperature = 70.0 °C

Step 2: The balanced equation

Ni(s) + 4CO(g) ⇌ Ni(CO)4(g)

For 1 mol Ni we need 4 moles CO to produce 1 mol Ni(CO)4

Step 3: Calculate Gibbs free energy

Gibbs free energy=  standard free energy of formation of products - standard free energy of formation of reactants

Gibbs free energy =  standard free energy of formation(Ni(CO)4) - standard free energy of formation(Ni(s)) +4Gf(CO) )

Gibbs free energy =  -587.4 -(4*-137.3)

Gibbs free energy = -587. 4 +549.2

Gibbs free energy = -38.2 kJ/mol

Step 4: Calculate the equilibrium cosntant K

Δ G ∘  = -RT ln K

-38.2 kJ/mol = -8.314 * 343 * ln K

ln K = 13.4

K = e^13.4

K = 6.6 * 10^5

The equilibrium constant is  6.6 * 10^5

Answer 2

Answer:

656055.062

Explanation:

Data:

- gas constant, R = 8.3144 J/(mol*K)

- temperature, T = 70 °C or 70+273 = 343 K

The standard free energy of reaction is:

ΔG = G of formation of Ni(CO)4(g) - 4 * G of formation of CO(g) - G of formation of Ni(s)

ΔG = −587.4 - 4*(-137.3) - 0 = -38.2  kJ/mol or -38200 J/mol

The relationship between the standard free energy of reaction (ΔG) and the  the equilibrium constant (Kp) is:

ΔG = -R*T*ln(Kp)

-38200/(-8.3144*343) = ln(Kp)

e^13.394 = Kp

656055.062 = Kp