Chemistry – Physical chemistry | e-Consult

Answer:

To calculate ΔG, we use the equation: ΔG = ΔH - TΔS. We are given ΔH = -890 kJ/mol and T = 298 K. We need to determine the entropy change (ΔS) for the reaction.

The change in moles of gas is: Products: 1 mole CO₂ + 2 moles H₂O = 3 moles. Reactants: 1 mole CH₄ + 2 moles O₂ = 3 moles. The change in the number of moles of gas is zero. However, we must consider the state of the reactants and products. The reaction involves a change from a gas (reactants) to gases (products). The increase in the number of gas molecules generally leads to an increase in entropy. Therefore, ΔS is positive. A reasonable estimate for ΔS would be around 85 J/mol·K. (Note: This value is an estimation and could be obtained from a table of standard entropy values).

Now we can calculate ΔG:

ΔG = ΔH - TΔS = -890 kJ/mol - (298 K)(85 J/mol·K) = -890000 J/mol - (298 K)(85 J/mol·K) = -890000 J/mol - 25830 J/mol = -915830 J/mol = -915.83 kJ/mol

The magnitude of ΔG is -915.83 kJ/mol, which is a large negative value. The sign of ΔG is negative, indicating that the reaction is spontaneous at 298 K. A negative ΔG means that the reaction will proceed in the forward direction without requiring any external energy input. The larger the magnitude of the negative ΔG, the more spontaneous the reaction is. Therefore, this reaction is highly spontaneous at 298 K.