I need some help on working out the enthalpy change of formation of CaCO3. Below I have posted a picture of the question incl. relevant data. Any help please?
Working out Enthalpy Change of Formation Watch
- 3 followers
- 2 badges
- Thread Starter
- 10-03-2013 12:42
- 3 followers
- 1 badge
- 10-03-2013 13:23
Ok so we know the principle of these diagrams is that whatever route taken the enthalpy change is always the same. When the diagram is drawn like this it's easy to see that the energy change going down the left hand side is going to be the same as the energy change going down the right hand side.
The right hand side is broken into 3 separate energy changes that we have to add up to find the total energy change for the right hand side. Looking carefully at the chemicals on each level we can see that the first energy change is representing the formation of CaCl2, which you know is -426.5 KJmol-1 from the first equation on the sheet.
The next arrow we can see is representative of the formation of H2O which we know from the data is equal to -286KJmol-1.
The final arrow on the right hand side is representative of the formation of CO2, which again we can tell from the data to be equal to -394KJmol-1.
Now we can calculate the total enthalpy change for the right hand side (which is equivalent to the enthalpy change of formation of CaCl2, CO2 and H2O). To do this we simply add up the enthalpy changes:
-426.5 + -286 + -394 = -1106.5KJmol-1
H1 + H3 + H4
So we know that the total enthalpy change for the left hand side of the diagram will also have to equal -1106.5KJmol-1, but we just want to find the enthalpy change for the formation of calcium carbonate.
We know the enthalpy change for the bottom half of the left hand side is the enthalpy change of reaction for the reaction of calcium carbonate with hydrochloric acid, which we are also give in data (H2) as -24.1KJmol-1.
So to find the enthalpy of formation of calcium carbonate all we have to do is subtract H2 (the bottom half of left hand side) from our total for the left hand side (which is equal to the total for the right hand side).
-1106.5 - -24.1 = -1082.4KJmol-1.
So we can see the enthalpy change of formation of calcium carbonate to be -1082.4KJmol-1.
Hope this helps.