Question

Calculate ΔHrxn for the following reaction:
CaO(s)+CO2(g)→CaCO3(s)
Use the following reactions and given ΔH′s.
Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ
2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ

Express your answer using four significant figures.

Please help I keep getting all the wrong answers.

Answer 1

Use Hess's law.

Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ
2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ

We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation.

Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ
2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ

The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2.

2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ
2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ

Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.