Nitrous acid (HNO2) is a weak acid that dissociates in water in the following manner: HNO2(aq) + H2O(l) → NO2−(aq) + H3O+(aq) At a temperature of 298.15 K, the acid-dissociation constant (Ka) for nitrous acid is 5.6 × 10−4. (a) Calculate the change in standard free energy (ΔG°) for this equilibrium reaction. kJ/mol (b) What is the value of ΔG at equilibrium? ΔG > 0 ΔG = 0 ΔG < 0 (c) If the reaction were spontaneous in the forward direction, what type of value would you expect for ΔG? ΔG > 0 ΔG = 0 ΔG < 0
The Correct Answer and Explanation is:
Let’s work through the problem step by step.
(a) Calculate the change in standard free energy (ΔG°) for the equilibrium reaction.
The relationship between the standard free energy change (ΔG°) and the acid dissociation constant (Ka) is given by the following equation: ΔG∘=−RTln(Ka)\Delta G^\circ = -RT \ln(K_a)ΔG∘=−RTln(Ka)
Where:
- RRR is the universal gas constant, 8.314 J/mol\cdotpK8.314 \, \text{J/mol·K}8.314J/mol\cdotpK,
- TTT is the temperature in Kelvin, 298.15 K298.15 \, \text{K}298.15K,
- KaK_aKa is the acid dissociation constant, 5.6×10−45.6 \times 10^{-4}5.6×10−4.
First, substitute the values into the equation: ΔG∘=−(8.314 J/mol\cdotpK)(298.15 K)ln(5.6×10−4)\Delta G^\circ = -(8.314 \, \text{J/mol·K}) (298.15 \, \text{K}) \ln(5.6 \times 10^{-4})ΔG∘=−(8.314J/mol\cdotpK)(298.15K)ln(5.6×10−4)
Now calculate: ln(5.6×10−4)=−7.478\ln(5.6 \times 10^{-4}) = -7.478ln(5.6×10−4)=−7.478 ΔG∘=−(8.314)(298.15)(−7.478)\Delta G^\circ = -(8.314) (298.15) (-7.478)ΔG∘=−(8.314)(298.15)(−7.478) ΔG∘=18465.8 J/mol≈18.47 kJ/mol\Delta G^\circ = 18465.8 \, \text{J/mol} \approx 18.47 \, \text{kJ/mol}ΔG∘=18465.8J/mol≈18.47kJ/mol
Thus, the change in standard free energy is: ΔG∘≈+18.47 kJ/mol\Delta G^\circ \approx +18.47 \, \text{kJ/mol}ΔG∘≈+18.47kJ/mol
(b) What is the value of ΔG at equilibrium?
At equilibrium, the change in free energy (ΔG\Delta GΔG) is equal to zero because the system has reached a state where the forward and reverse reactions occur at the same rate. This is true for any system at equilibrium, whether it’s a chemical reaction or a physical process.
Thus: ΔG=0\Delta G = 0ΔG=0
(c) If the reaction were spontaneous in the forward direction, what type of value would you expect for ΔG?
A spontaneous reaction occurs when the change in free energy (ΔG\Delta GΔG) is negative. If a reaction is spontaneous in the forward direction, it indicates that the products are thermodynamically more stable than the reactants, meaning that the free energy decreases as the reaction proceeds.
So, if the reaction were spontaneous, you would expect: ΔG<0\Delta G < 0ΔG<0
Summary:
- (a) ΔG∘≈+18.47 kJ/mol\Delta G^\circ \approx +18.47 \, \text{kJ/mol}ΔG∘≈+18.47kJ/mol.
- (b) ΔG=0\Delta G = 0ΔG=0 at equilibrium.
- (c) If the reaction were spontaneous, ΔG<0\Delta G < 0ΔG<0.
