Use the standard free energies of formation (δgf°) of each substance to determine the value of equilibrium constant k at 25 °c for the following reaction. click here for a copy of the test 3 cover sheet.

co2(g) + 2 h2(g) ⇌ ch3oh(l)

δgf° kj/mol −394.4 0 −166.4

0

0.91

8.83 × 1039

1.13 × 10−40

k = 8,83x10³⁹

Explanation:

For the reaction:

CO₂(g) + 2H₂(g) ⇌ CH₃OH(l)

ΔG° = ΔGproducts - ΔGreactants

ΔG° = ΔGCH₃OH(l)  - (ΔGCO₂(g) + 2ΔGH₂(g)

As ΔGCH₃OH(l)  = -166,4 kJ/mol; ΔGCO₂(g) = -394,4 kJ/mol; ΔGH₂(g) = 0 kJ/mol

ΔG° = -166,4 kJ/mol - (-394,4 kJ/mol + 2×0 kJ/mol)

ΔG° = 228,0 kJ/mol

The ΔG could be defined as:

ΔG = -RT lnK

Where R is gas constant (8,314472x10⁻³ kJ/molK); Temperature is 25°C, 298,15K.

Replacing:

228,0 kJ/mol = 8,314472x10⁻³ kJ/molK×298,15K ln K

91,97 = ln k

K = 8,79x10³⁹ ≈ 8,83x10³⁹

I hope it helps!

56 moles of   lieter in chemistry