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

## Given the following data: SnCl_2(s) + TiBr_2(s) → SnBr_2(s) + TiCl_2(s) ΔH = +4.2 kJ TiBr_2(s) + SnCl_4(l) → SnBr_2(s) + TiCl_4(l) ΔH = -74 kJ TiCl_4(l) → TiCl_2(s) + Cl_2(g) ΔH = +273 kJ

Apr 23, 2016

$\Delta H = - 194.8 k J$

#### Explanation:

Knowing that elemental molecules (made of only one element, like ${O}_{2} , C {l}_{2} , B {r}_{2}$) etc have $0$ enthalpy, and the equation for enthalpy

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

or overall enthalpy ($\Delta {H}_{\text{rxn}}$) is the enthalpy of products minus enthalpy of reactants, we can rearrange equation $3.$.

$T i C {l}_{4} \to T i C {l}_{2} + C {l}_{2} , \Delta H = + 273 k J$

$273 = \left(T i C {l}_{2} + C {l}_{2}\right) - T i C {l}_{4}$ (products minus reactants)

$273 = T i C {l}_{2} - T i C {l}_{4}$ (enthalpy of $C {l}_{2}$ is $0$)

$T i C {l}_{4} = T i C {l}_{2} - 273$ (rearranged algebraically)

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

$T i B {r}_{2} + S n C {l}_{4} \to S n B {r}_{2} + T i C {l}_{4} , \Delta H = - 74 k J$

$- 74 = \left(S n B {r}_{2} + T i C {l}_{4}\right) - \left(T i B {r}_{2} + S n C {l}_{4}\right)$

$- 74 = S n B {r}_{2} + T i C {l}_{2} - 273 - T i B {r}_{2} - S n C {l}_{4}$ (substituting in equation $3.$ and expanding out everything)

$199 = S n B {r}_{2} + T i C {l}_{2} - T i B {r}_{2} - S n C {l}_{4}$ (rearranging, adding $273$ to both sides)

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

$S n C {l}_{2} + T i B {r}_{2} \to S n B {r}_{2} + T i C {l}_{2} , \Delta H = + 4.2 k J$

$4.2 = \left(S n B {r}_{2} + T i C {l}_{2}\right) - \left(S n C {l}_{2} + T i B {r}_{2}\right)$ (products minus reactants)

$4.2 = S n B {r}_{2} + T i C {l}_{2} - S n C {l}_{2} - T i B {r}_{2}$ (expanding out brackets)

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

$S n B {r}_{2} + T i C {l}_{2} - S n C {l}_{2} - T i B {r}_{2} + 194.8 = S n B {r}_{2} + T i C {l}_{2} - T i B {r}_{2} - S n C {l}_{4}$

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 $S n B {r}_{2}$, so you can cancel that. They also have a $T i C {l}_{2}$ and a $- T i B {r}_{2}$, so get rid of those from each side, leaving

$- S n C {l}_{2} + 194.8 = - S n C {l}_{4}$

$S n C {l}_{4} - S n C {l}_{2} = - 194.8 k J$

This is the answer to the question in the form of products minus reactants, because you can see the product is $S n C {l}_{4}$ and the reactants are $S n C {l}_{2}$ and $C {l}_{2}$, though $C {l}_{2}$ has an enthalpy of $0$ so it can be ignored.