To calculate the thermodynamic properties (enthalpy, entropy, and standard Gibbs free energy) at 298K for the given reaction, you will need to use several thermodynamic equations and data. Here are the steps to solve this problem:
Step 1: Write the balanced equation
First, balance the reaction equation:
2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
Step 2: Determine the standard enthalpy change (∆H°rxn)
Use the standard enthalpy of formation (∆H°f) values for the reactants and products to calculate the enthalpy change:
∆H°rxn = ∑(∆H°f(products)) - ∑(∆H°f(reactants))
Given values:
- ∆H°f(Na2CO3) = -1130.7 kJ/mol
- ∆H°f(H2O) = -285.8 kJ/mol
∆H°rxn = (1 mol Na2CO3 * ∆H°f(Na2CO3)) + (1 mol H2O * ∆H°f(H2O)) - (2 mol NaOH * ∆H°f(NaOH)) - (∆H°f(CO2))
= (1 mol * -1130.7 kJ/mol) + (1 mol * -285.8 kJ/mol) - (2 mol * 0 kJ/mol) - (0 kJ/mol)
∆H°rxn = -1416.5 kJ/mol
Step 3: Determine the standard entropy change (∆S°rxn)
Use the standard molar entropy (∆S°) values for the reactants and products to calculate the entropy change:
∆S°rxn = ∑(∆S°(products)) - ∑(∆S°(reactants))
Given values:
- ∆S°(Na2CO3) = 135 J/(mol K)
- ∆S°(H2O) = 69.9 J/(mol K)
- ∆S°(CO2) = 213.7 J/(mol K)
∆S°rxn = (1 mol Na2CO3 * ∆S°(Na2CO3)) + (1 mol H2O * ∆S°(H2O)) - (2 mol NaOH * ∆S°(NaOH)) - (∆S°(CO2))
= (1 mol * 135 J/(mol K)) + (1 mol * 69.9 J/(mol K)) - (2 mol * 0 J/(mol K)) - (1 mol * 213.7 J/(mol K))
∆S°rxn = -8.8 J/(mol K)
Step 4: Calculate the standard Gibbs free energy change (∆G°rxn)
The standard Gibbs free energy change can be calculated using the equation:
∆G°rxn = ∆H°rxn - T∆S°rxn
Given:
T = 298K
∆G°rxn = -1416.5 kJ/mol - (298 K * (-8.8 J/(mol K)))
= -1416.5 kJ/mol - (-2614.4 kJ/mol)
∆G°rxn = 1197.9 kJ/mol
Therefore, at 298K, the enthalpy change (∆H°rxn) is -1416.5 kJ/mol, the entropy change (∆S°rxn) is -8.8 J/(mol K), and the standard Gibbs free energy change (∆G°rxn) is 1197.9 kJ/mol.
Note: The molar heat capacity for solid sodium carbonate is not used in this calculation, as it is not necessary for determining the thermodynamic properties of the reaction.