Question

∆g° = 78.5 kj/mol for the reaction 3 a (g) b (g) →2 c (g). at 1120 k, the partial pressures of a, b, and c are 11.5 atm, 8.60 atm, and 0.510 atm respectively. what is the free energy for this reaction?

Answer 1

To find the free energy for a reaction, we can use the formula ΔG° = ΔH° - TΔS°. By calculating the values of ΔH° and ΔS° using the given data, we can then substitute these values into the formula to find the free energy change.

Step-by-step explanation:

The standard free energy change (ΔG°) can be calculated using the formula ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change.

In this case, the given values are ΔG° = 78.5 kJ/mol, T = 1120 K, and the partial pressures of a, b, and c are 11.5 atm, 8.60 atm, and 0.510 atm respectively.

To find ΔH°, we need to use the equation ΔG° = ΔH° - TΔS°. Rearranging the equation, we find ΔH° = ΔG° + TΔS°. Substituting the given values, we get ΔH° = 78.5 kJ/mol + 1120 K × ΔS°.

The value of ΔS° can be calculated using the ideal gas law: ΔS° = R × ln(Q), where R is the ideal gas constant and Q is the reaction quotient. Since the reaction is a → b, the reaction quotient is Q = (p_b^2)/(p_a^3). Plugging in the given values, we find Q = (8.60^2)/(11.5^3).

Now, we can substitute the calculated ΔS° value into the equation for ΔH°: ΔH° = 78.5 kJ/mol + 1120 K × ΔS°. Finally, we can calculate ΔG° using the formula ΔG° = ΔH° - TΔS°, substituting the values we have obtained.