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Using the following data at 298 K:Substance Df H° [kJ mol-1 ] S° [J mol-1 K -1 ]H 2 (g) 0 130.7CO (g) -110.5 197.7H 2 CO (g) -108.6 218.8(i) [2 marks]Calculate the standard Gibbs free energy change for the reaction at 298 K:H 2 (g) + CO (g) ® H 2 CO (g)

Question

Using the following data at 298 K:Substance Df H° [kJ mol-1 ] S° [J mol-1 K -1 ]H 2 (g) 0 130.7CO (g) -110.5 197.7H 2 CO (g) -108.6 218.8(i) [2 marks]Calculate the standard Gibbs free energy change for the reaction at 298 K:H 2 (g) + CO (g) ® H 2 CO (g)

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Solution

The standard Gibbs free energy change (ΔG°) for a reaction can be calculated using the equation:

ΔG° = ΔH° - TΔS°

where: ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change.

First, we need to calculate ΔH° and ΔS° for the reaction.

ΔH° for the reaction = ΔH° products - ΔH° reactants = (-108.6 kJ/mol) - [(0 kJ/mol) + (-110.5 kJ/mol)] = 1.9 kJ/mol

ΔS° for the reaction = ΔS° products - ΔS° reactants = (218.8 J/mol K) - [(130.7 J/mol K) + (197.7 J/mol K)] = -109.6 J/mol K

Now we can substitute these values into the equation for ΔG°:

ΔG° = ΔH° - TΔS° = (1.9 kJ/mol) - (298 K * -109.6 J/mol K) = 1.9 kJ/mol - (-32.7 kJ/mol) = 34.6 kJ/mol

So, the standard Gibbs free energy change for the reaction at 298 K is 34.6 kJ/mol.

This problem has been solved

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