Question

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

Answer 1

To find the equilibrium constant (K) at 25 °C, calculate the standard Gibbs free energy change (ΔG°) using the given ΔGf° values and apply the equation ΔG° = -RTlnK. Substitute the values into the expression and solve for K.

Step-by-step explanation:

To calculate the equilibrium constant (K) at 25 °C using the standard free energies of formation (ΔGf°), the following equation is used:

ΔG° = -RTlnK
where ΔG° is the standard Gibbs free energy change for the reaction, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.

The standard Gibbs free energy change (ΔG°) is calculated by subtracting the sum of ΔGf° of the reactants from the sum of ΔGf° of the products:

ΔG° = Σ ΔGf° (products) - Σ ΔGf° (reactants)

For the reaction CO2(g) + 2 H2(g) ⇌ CH3OH(l), the ΔGf° values are:

• CO2(g): -394.4 kJ/mol
• H2(g): 0 kJ/mol (since elemental hydrogen is in its standard state)
• CH3OH(l): -166.4 kJ/mol

Plugging in the values, we get:

ΔG° = [-166.4 kJ/mol] - [(-394.4 kJ/mol) + 2(0 kJ/mol)] = 228 kJ/mol

Next, we convert ΔG° to Joules (1 kJ = 1000 J) and T to Kelvin (25 °C = 298 K). Then we solve for K:

ΔG° = 228 kJ/mol × 1000 J/kJ = 228000 J/mol

K = e^{-ΔG°/RT} = e^{-228000 J/mol / (8.314 J/mol·K × 298 K)}

By calculating this expression, we will obtain the value of K.

Answer 2

k = 8,83x10³⁹

Step-by-step 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!