Question

Methanol, CH 3 OH CH3OH , is formed from methane and water in a two‑step process. Step 1: CH4 ( g ) + H2O ( g ) ⟶ CO ( g ) + 3H2 ( g ) Δ S ∘ = 214.7 J / K Step 2: CO ( g ) + 2 H 2 ( g ) ⟶ CH 3 OH ( l ) Δ S ∘ = − 332.3 J / K Calculate Δ H∘ and ΔG∘ at 298 K for step 1.

Answer 1

Answer:

ΔH° = 206.1 kJ

ΔG° = 142.1 kJ

Explanation:

Let's consider the first step in the synthesis of methanol.

Step 1: CH₄(g) + H₂O(g) ⟶ CO(g) + 3 H₂(g) ΔS° = 214.7 J / K

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(p) are the standard enthalpies of formation of reactants and products

ΔH° = [1 mol × ΔH°f(CO(g)) + 3 mol × ΔH°f(H₂(g))] - [1 mol × ΔH°f(CH₄(g)) + 1 mol × ΔH°f(H₂O(g))]

ΔH° = [1 mol × (-110.5 kJ/mol) + 3 mol × (0 kJ/mol)] - [1 mol × (-74.81 kJ/mol) + 1 mol × (-241.8 kJ/mol)]

ΔH° = 206.1 kJ

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

ΔG° = 206.1 kJ - 298 K × (214.7 × 10⁻³ kJ/K)

ΔG° = 142.1 kJ