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A final challenge! In this exercise we have three measured enthalpy changes from which we can calculate the DH value of the unmeasured change by applying Hess's Law. Try using the technique in exercise 2 to evaluate the enthalpy change for:
|
Ca(s) + ½O2(g) ⇒ CaO(s) |
We have carried out the following three reactions and obtained enthalpy change values (all in kJ mol-1) for them:
|
Ca(s) + 2H+(aq) ⇒ Ca2+(aq) + H2(g) |
DH1 = - 1926 |
|
2H2(g) + O2(g) ⇒ 2 H2O(l) |
DH2 = - 572 |
|
CaO(s) + 2H+(aq) ⇒ Ca2+(aq) + H2O(l) |
DH3 = - 2275 |
If we look at the original equation we need to start with one mole of calcium (from equation 1) and half a mole of oxygen (from equation 2 x ½). We need one mole of calcium oxide as a product. We could get this by reversing equation 3 (this will change the sign of the enthalpy change):
|
Ca(s) + 2H+(aq) ⇒ Ca2+(aq) + H2(g) |
DH1 = - 543 |
|
H2(g) + ½O2(g) ⇒ H2O(l) |
DH2 = - 572 x ½ |
|
Ca2+(aq) + H2O(l) ⇒ CaO(s) + 2H+(aq) |
DH3 = + 194 |
If we add these three equations together:
|
Ca(s) + 2H+(aq) + H2(g)+ ½O2(g) + Ca2+(aq) + H2O(l) ⇒ Ca2+(aq) + H2(g) + H2O(l) + CaO(s) + 2H+(aq) |
and cancel anything common to both sides (shown in blue) we get:
|
Ca(s) + ½O2(g) ⇒ CaO(s) |
The total enthalpy change is calculated by adding the three enthalpy changes together:
DH = - 543 + - 572 × ½ + + 194 = - 635 kJ mol-1