How do you find the `ΔH` of the following reaction: `SnCl_2(s) + Cl_2(g) → SnCl_4(l)`?

Knowing that elemental molecules (made of only one element, like `O_2, Cl_2, Br_2`) etc have `0` enthalpy, and the equation for enthalpy

`DeltaH_("rxn")=Sigman*"products"-Sigman*"reactants"`,

or overall enthalpy (`DeltaH_("rxn")`) is the enthalpy of products minus enthalpy of reactants, we can rearrange equation `3.`.

`TiCl_4->TiCl_2+Cl_2, DeltaH=+273kJ`

`273=(TiCl_2+Cl_2)-TiCl_4` (products minus reactants)

`273=TiCl_2-TiCl_4` (enthalpy of `Cl_2` is `0`)

`TiCl_4=TiCl_2-273` (rearranged algebraically)

Now you can substitute this into equation `2.`, again using the products minus reactants equation

`TiBr_2+SnCl_4->SnBr_2+TiCl_4, DeltaH=-74kJ`

`-74=(SnBr_2+TiCl_4)-(TiBr_2+SnCl_4)`

`-74=SnBr_2+TiCl_2-273-TiBr_2-SnCl_4` (substituting in equation `3.` and expanding out everything)

`199=SnBr_2+TiCl_2-TiBr_2-SnCl_4` (rearranging, adding `273` to both sides)

You also have equation `1.`, which you can use products minus reactants to find

`SnCl_2+TiBr_2->SnBr_2+TiCl_2, DeltaH=+4.2kJ`

`4.2=(SnBr_2+TiCl_2)-(SnCl_2+TiBr_2)` (products minus reactants)

`4.2=SnBr_2+TiCl_2-SnCl_2-TiBr_2` (expanding out brackets)

Now you can equate equations `1.` and `2.` in their new form, as so

`SnBr_2+TiCl_2-SnCl_2-TiBr_2+194.8=SnBr_2+TiCl_2-TiBr_2-SnCl_4`

because if you add `194.8` to equation `1` it equals `199` in total, the same as equation `2.` With these two equal to each other, you can cancel out the similarities on both sides and find what's left.

Both sides have a `SnBr_2`, so you can cancel that. They also have a `TiCl_2` and a `-TiBr_2`, so get rid of those from each side, leaving

`-SnCl_2+194.8=-SnCl_4`

`SnCl_4-SnCl_2=-194.8kJ`

This is the answer to the question in the form of products minus reactants, because you can see the product is `SnCl_4` and the reactants are `SnCl_2` and `Cl_2`, though `Cl_2` has an enthalpy of `0` so it can be ignored.