Hess's Law alternative method

We start with the reaction that we are trying to find the enthalpy change for:

2KHCO₃(s) ⇒ K₂CO₃(s) + H₂O(l) + CO₂(g)

We now find a practical reaction for the reactants and one for the products which give the same end products. These may be given to you in an assessment. In this case the reaction with dilute acid works well:

KHCO₃(s) + HCl(aq) ⇒ KCl(aq) + H₂O(l) + CO₂(g)	DH₁
K₂CO₃(s) + 2HCl(aq) ⇒ 2KCl(aq) + H₂O(l) + CO₂(g)	DH₂

These reactions can be carried out in a plastic cup and the enthalpy changes measured. What we now have to do is to construct the original equation from the two practical ones. This is easier than it looks. In the original equation we can see that there are two moles of KHCO₃, so we double up our first equation:

2KHCO₃(s) + 2HCl(aq) ⇒ 2KCl(aq) + 2H₂O(l) + 2CO₂(g)

2DH₁

Also in the original equation, the K₂CO₃ is a product, but in our second practical reaction it is a reactant. This is solved by reversing the second equation (but remember this will also change the sign of the enthalpy change):

2KCl(aq) + H₂O(l) + CO₂(g) ⇒ K₂CO₃(s) + 2HCl(aq)

- DH₂

We now add together these two adjusted equations (everything on the left of the arrows turns into everything on the right of the arrows):

2KHCO₃(s) + 2HCl(aq) + 2KCl(aq) + H₂O(l) + CO₂(g) ⇒ 2KCl(aq) + 2H₂O(l) + 2CO₂(g) + K₂CO₃(s) + 2HCl(aq)

2DH₁ - DH₂

This looks complicated but a lot of substances are common to both sides (shown in blue) and can be cancelled to give:

2KHCO₃(s) ⇒ K₂CO₃(s) + H₂O(l) + CO₂(g)

This is the original equation we were looking for. The enthalpy change is 2DH₁ - DH₂. Note that in the first method shown (with the triangle) DH1 represented the change for two moles of KHCO₃. In this second method it represents one mole of KHCO₃ but we've doubled it. This amounts to the same thing, but the second method is probably clearer as we usually calculate an enthalpy change per mole of reactant.