Question

Which of the following reactions will have the largest equilibrium constant (K) at 298 K?

a) Fe2O3(s) + 3 CO(g) ? 2 Fe(s) + 3 CO2(g) ?G° = -28.0 kJ
b) It is not possible to determine without more information.
c) 3 O2(g) ? 2 O3(g) ?G° = +326 kJ
d) CaCO3(s) ? CaO(s) + CO2(g) ?G° =+131.1 kJ
e) 2 Hg(g) + O2(g) ? 2 HgO(s) ?G° = -180.8 kJ

Answer 1

The largest equilibrium constant at 298 K corresponds to the reaction with the most negative standard free energy change. In this case, it is reaction e (2 Hg(g) + O2(g) → 2 HgO(s)) which has a ΔG° of -180.8 kJ.

Step-by-step explanation:

The equilibrium constant (K) is related to the standard free energy change (ΔG°) of a reaction through the equation: K = e^(-ΔG°/RT), where R is the universal gas constant and T is the temperature in Kelvin. A negative ΔG° indicates a spontaneous reaction and corresponds to a large K value, while a positive ΔG° points to a non-spontaneous reaction and results in a small K. Comparing the given reactions, reaction e, which is 2 Hg(g) + O2(g) → 2 HgO(s) with ΔG° = -180.8 kJ, will have the largest equilibrium constant at 298 K since it has the most negative ΔG° value.

Answer 2

The reaction with the largest equilibrium constant at 298 K is the one with the most negative ΔG°, which is 2 Hg(g) + O2(g) → 2 HgO(s) with ΔG° = -180.8 kJ.

Step-by-step explanation:

The reaction with the largest equilibrium constant (K) at 298 K can be determined using the relationship between the Gibbs free energy (ΔG°) and the equilibrium constant, which is given by the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. In this equation, a negative value of ΔG° corresponds to a positive value of ln(K), which indicates a large equilibrium constant K. Looking at the given reactions:

• Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔG° = -28.0 kJ
• 3 O2(g) → 2 O3(g) ΔG° = +326 kJ
• CaCO3(s) → CaO(s) + CO2(g) ΔG° = +131.1 kJ
• 2 Hg(g) + O2(g) → 2 HgO(s) ΔG° = -180.8 kJ

The reaction with the most negative ΔG° is the reaction 2 Hg(g) + O2(g) → 2 HgO(s) with ΔG° = -180.8 kJ, suggesting it would have the largest equilibrium constant at 298 K. Thus, the correct answer is:

e) 2 Hg(g) + O2(g) → 2 HgO(s) ΔG° = -180.8 kJ