The enthalpy change was determined for the
following reactions;
Ca (s) + CO2 (g) + ½ O2 (g ) → CaCO3 (s)
ΔH = -812.8 kJ/mo l
Ca (s) + ½ O2 (g) → CaO (s)
ΔH = -634.9 kJ/mo l
What is the enthalpy change for the following reaction?
CaCO3 (s) → CaO (s) + CO2 (g)
Add equation 1 to the reverse of equation 2 to obtain the reaction you want. dH for the new reaction is dH for equation 1 + the negative dH (which will make is positive) for equation 2.
To find the enthalpy change for the given reaction, we can use Hess's Law. Hess's Law states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual reactions that make up the overall reaction.
In this case, we need to use the enthalpy changes of the reactions given to calculate the enthalpy change for the desired reaction.
Firstly, let's take the reverse of the first equation:
CaCO3 (s) → Ca (s) + CO2 (g)
Since the given enthalpy change is -812.8 kJ/mol, the enthalpy change for this reverse reaction would be +812.8 kJ/mol.
Next, we need to use the second equation:
Ca (s) + ½ O2 (g) → CaO (s)
The given enthalpy change is -634.9 kJ/mol.
Now, we add both equations together:
CaCO3 (s) → Ca (s) + CO2 (g)
Ca (s) + ½ O2 (g) → CaO (s)
The Ca (s) on both sides cancels out, leaving us with:
CaCO3 (s) → CaO (s) + CO2 (g)
Now, we add the enthalpy changes of the two equations:
+812.8 kJ/mol + (-634.9 kJ/mol)
The result is:
ΔH = 177.9 kJ/mol
Therefore, the enthalpy change for the reaction CaCO3 (s) → CaO (s) + CO2 (g) is 177.9 kJ/mol.