A method for scrubbing CO2(g) from the air on a spacecraft is to allow CO2(g) to react with NaOH according
to the following (unbalanced reaction).
NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
Use the appropriate thermodynamic tables to calculate the enthalpy, entropy and standard Gibbs free energy
at 298K. (For the solid sodium carbonate, you need to use the following data: Molar heat capacity at constant
pressure 112.3 J/(mol K), Standard molar entropy 135 J/(mol K), Standard molar enthalpy of formation -1130.7
kJ/mol)
Any help to figure this out would be great I don't even know where to start
The problem doesn't say so but implies you want the dH, dS, dG FOR THE REACTION.
1. You have the equation, balanced it.
2. dHrxn = (n*dHf products) - (n*dHf reactants)
3. dSrxn = (n*dSf procuts) - (n*dSf reactants)
4. Then dGrxn = dHrxn - TdSrxn. Substitute and solve for dG.
Post your work if you get stuck.
The dHf and dSf you will need for Na2CO3 is listed in the problem. I presume the others are listed in your text/notes.
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.