Use the table below to answer the questions that follow.
Thermodynamic Quantities for Selected Substances at 298.15 K (25°C)
-The equilibrium constant for a reaction is 0.48 at 25 °C. What is the value of ΔG° (kJ/mol) at this temperature?

Question

Use the table below to answer the questions that follow.
Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) 
-The equilibrium constant for a reaction is 0.48 at 25 °C. What is the value of ΔG° (kJ/mol) at this temperature?

Quizplus · Accepted Answer

We can use ΔG° = -RT ln(K) where R is the gas constant, T is temperature in Kelvin and K is the equilibrium constant. Substituting the given values, we get: 
ΔG° = - (8.314 J/mol-K × 298.15 K) × ln(0.48) = -18.2 kJ/mol
Rounding to one decimal place gives us ΔG° = -18.2 ≈ - 18 kJ/mol, which is closest to choice C, 1.8 kJ/mol.

Use the Table Below to Answer the Questions That Follow