Question

For the following reaction at 25°C, ΔS° = –194 J/K and ΔH° = –2511 kJ. 2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(g) Calculate ΔG° for the reaction at 25°C and indicate if this is a spontaneous reaction.

a. -1762 kJ
b. 1762 kJ
c. -2337 kJ
d. 2337 kJ

Answer 1

The standard free-energy change (ΔG°) for the given reaction at 25°C is -2453.188 kJ, indicating that the reaction is spontaneous. The closest provided answer choice is (c) -2337 kJ.

Step-by-step explanation:

To calculate the standard free-energy change (ΔG°) at 25°C for the reaction 2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(g), we can use the Gibbs free energy equation:

ΔG° = ΔH° - TΔS°

Given that ΔH° = -2511 kJ and ΔS° = -194 J/K, and knowing that the temperature T is 25°C (or 298 K), we can substitute the values as follows:

ΔG° = -2511 kJ - 298 K * (-194 J/K × 1 kJ/1000 J)

ΔG° = -2511 kJ - 298 K * (-0.194 kJ/K)

ΔG° = -2511 kJ + 57.812 kJ

ΔG° = -2453.188 kJ

Since the value for ΔG° is negative, the reaction is spontaneous at 25°C.

The closest answer choice to this calculation is (c) -2337 kJ.