Question

The standard free energy change for the reaction of 1.54 moles of CO2 (g) at 266 K, 1 atm would be favored under standard conditions at 266 K.

This reaction is

Assume that AH and

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2CO2(g)+5H2(g) + C₂H₂(g) + 4H2O(g)

AH-46.5 kJ and AS124.8 J/K

Answer 1

The reaction of 2CO2(g) + 5H2(g) to C2H2(g) and 4H2O(g) is favored at 266 K under standard conditions due to a negative enthalpy change and a positive entropy change, indicating a spontaneous process.

Step-by-step explanation:

The student question concerns whether the reaction of 2CO2(g) + 5H2(g) to form C2H2(g) and 4H2O(g) is favored under standard conditions at 266 K, given that the reaction has an enthalpy change (ΔH) of -46.5 kJ and an entropy change (ΔS) of +124.8 J/K. To determine if the reaction is favored, we calculate the standard free energy change (ΔG) using the equation ΔG = ΔH - TΔS. Since the enthalpy change is exothermic (negative ΔH) and the entropy increases (positive ΔS), we expect that this reaction would be favored at 266 K because both terms would lead to a negative ΔG, indicating a spontaneous process under standard conditions.