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!

b

explanation:

mitosis

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.