The equilibrium constant at 100 K for the reaction with ΔH° = 10 kJ/mol and ΔS° = -100 J/K is 1.23 x 10^5.
The given reaction is an endothermic reaction, so the equilibrium constant, Keq, will increase with increasing temperature.
To calculate the equilibrium constant at 100 K, we can use the following equation:
ΔG° = -RT ln(Keq)
where ΔG° is the standard Gibbs free energy change, R is the ideal gas constant (8.314 J/mol·K), T is the temperature in Kelvin (100 K), and Keq is the equilibrium constant.
Given that ΔH° = 10 kJ/mol and ΔS° = -100 J/K, we can calculate ΔG° using the following equation:
ΔG° = ΔH° - TΔS°
Plugging in the values, we get:
ΔG° = (10 kJ/mol) - (100 K)(-100 J/mol·K) = 11 kJ/mol
Now we can solve for Keq:
Keq = e^(-ΔG°/RT)
Plugging in the values, we get:
Keq = e^(-(11 kJ/mol)/(8.314 J/mol·K)(100 K)) = 1.23 x 10^5
Therefore, the equilibrium constant at 100 K for the reaction with ΔH° = 10 kJ/mol and ΔS° = -100 J/K is 1.23 x 10^5.
Question
Calculate the equilibrium constant at 40 K for a reaction with ΔH∘=10 kJ and ΔS∘=−100 J/K. (Don't round until the end. Using the exponent enlarges any round-off error.) Hint given in feedback. Answer: Calculate the equilibrium constant at 100 K for the thermodynamic data in the previous question. Notice that Keq is larger at the larger temperature for an endothermic reaction.