Calculate the temperature (K) at which the equation as written is at equilibrium. Express your answer to three significant figures.
CaO(s) + CO2(g) → CaCO3(s)
∆G°f (kJ/mol)
CaO(s)-604.0
CO2(g)-394.4
CaCO3(s)-1128.8
∆H°f (kJ/mol)
CaO(s)-635.1
CO2(g)-393.5
CaCO3(s)-1206.9
∆H°f (kJ/mol)
CaO(s)39.7
CO2(g)213.7
CaCO3(s)92.9
Here are the values I got:
∆G°rxn = -130.4 kJ
∆H°rxn = -178.3 kJ
∆S°rxn = -160.5 kJ
Solving for T,
T= (∆H-∆G)/∆S
plugging in the values I get: 298K
however the answer returned states that it's 1110K.
You have made two errors.
a. dS is given in J/mol and not kJ/mol so change that to 0.1605 kJ/mol
b. Equilibrium is when dG = 0
dG = dH - TdS
0 = dH - TdS
TdS = dH and
T = dH/dS = -178.3/-0.1605 = 1110.9 K so to three s.f. that would be 1.11E1 K.
To calculate the temperature at which the equation is at equilibrium, you need to use the equation:
∆G°rxn = ∆H°rxn - T∆S°rxn
where:
∆G°rxn is the standard Gibbs free energy change for the reaction,
∆H°rxn is the standard enthalpy change for the reaction,
T is the temperature in Kelvin,
∆S°rxn is the standard entropy change for the reaction.
You have already determined the values for ∆G°rxn and ∆H°rxn. Now, you need to use the known value for ∆S°rxn to solve for T.
Given that ∆G°rxn = -130.4 kJ/mol and ∆H°rxn = -178.3 kJ/mol, you can substitute these values into the equation:
-130.4 kJ/mol = -178.3 kJ/mol - T∆S°rxn
Now, you need the value of ∆S°rxn to proceed. Unfortunately, it seems that you have not provided the correct value for ∆S°rxn in your question. In order to calculate the temperature at which the equation is at equilibrium, you need the correct value for ∆S°rxn.
Please check your data and provide the correct value for ∆S°rxn so that we can calculate the equilibrium temperature accurately.