# Question f93f3

Sep 3, 2015

The answer is (d) $\Delta {H}_{3} = - \Delta {H}_{1} - \Delta {H}_{2}$

#### Explanation:

The idea here is that you can play around with your reactions to try and find a relationship between the first one and the other two.

Notice that you have

$\text{A" + "B" -> "C" + "D}$ $\text{ } \textcolor{b l u e}{\left(1\right)}$

but that the two other reactions have

$\text{C" + "X" -> "A" + "Y}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

$\text{D" + "Y" -> "B" + "X}$ $\text{ } \textcolor{b l u e}{\left(3\right)}$

If you add reactions $\textcolor{b l u e}{\left(2\right)}$ and $\textcolor{b l u e}{\left(3\right)}$ you will get

"C" + color(red)(cancel(color(black)("X"))) + "D" + color(red)(cancel(color(black)("Y"))) -> "A" + color(red)(cancel(color(black)("Y"))) + "B" + color(red)(cancel(color(black)("X")))#

$\text{C" + "D" -> "A" + "B}$ $\text{ } \left(\Delta {H}_{2} + \Delta {H}_{3}\right)$

This is none other than the reverse reaction for reaction $\textcolor{b l u e}{\left(1\right)}$.

For a general chemical reaction that has the enthalpy change of reaction $\Delta {H}_{\text{forward}}$, the enthalpy change of reaction for the reverse reaction will be

$\Delta {H}_{\text{reverse" = - DeltaH_"forward}}$

In this case, you have

$\text{A" + "B" -> "C" + "D}$, $\text{ } \Delta {H}_{1}$

and

$\text{C" + "D" -> "A" + "B}$ $\text{ } \left(\Delta {H}_{2} + \Delta {H}_{3}\right)$

This means that you have

$\Delta {H}_{1} = - \left(\Delta {H}_{2} + \Delta {H}_{3}\right)$

Rearrange this to get

$\Delta {H}_{1} = - \Delta {H}_{2} - \Delta {H}_{3} \implies \textcolor{g r e e n}{\Delta {H}_{3} = - \Delta {H}_{1} - \Delta {H}_{2}}$