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2NaHCO3(aq) + CaCl2(aq) ⇒ CaCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g) |
We have carried out the following two reactions and obtained enthalpy change values for them:
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NaHCO3(aq) + HCl(aq) ⇒ NaCl(aq) + H2O(l) + CO2(g) |
DH1 |
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CaCO3(s) + 2HCl(aq) ⇒ CaCl2(aq) + H2O(l) + CO2(g) |
DH2 |
I'll start by looking at the problem in a different way without the Hess's Law triangle. If we look at the reactants, we have two moles of NaHCO3 so we could start by doubling up reaction 1 to give 2DH1:
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2NaHCO3(aq) + 2HCl(aq) ⇒ 2NaCl(aq) + 2H2O(l) + 2CO2(g) |
2DH1 |
We can also see that there is a mole of CaCl2 as a reactant. This is present in reaction 2, but as a product. We can reverse this reaction and expect a change in the sign of DH2:
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CaCl2(aq) + H2O(l) + CO2(g) ⇒ CaCO3(s) + 2HCl(aq) |
- DH2 |
We now add these two processes together:
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2NaHCO3(aq) + 2HCl(aq) + CaCl2(aq) + H2O(l) + CO2(g) ⇒ 2NaCl(aq) + 2H2O(l) + 2CO2(g) + CaCO3(s) + 2HCl(aq) |
Certain substances are found on both sides of the equation (indicated in blue above) and can be cancelled to give:
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2NaHCO3(aq) + CaCl2(aq) ⇒ 2NaCl(aq) + H2O(l) + CO2(g) + CaCO3(s) |
2DH1 - DH2 |
What we find is that this is the reaction that we were originally looking for. So, DH = 2DH1 - DH2.
We can still draw a Hess's Law triangle:
