# For the reaction 4"XY"_3 + 7"Z"_2 -> 6"Y"_2"Z" + 4"XZ"_2, what is the enthalpy change? The steps are shown below.

## ${\text{X"_2 + 3"Y"_2 -> 2"XY}}_{3}$, $\Delta {H}_{1} = - \text{320 kJ}$ ${\text{X"_2 + 2"Z"_2 -> 2"XZ}}_{2}$, $\Delta {H}_{2} = - \text{140 kJ}$ $2 \text{Y"_2 + "Z"_2 -> 2"Y"_2"Z}$, $\Delta {H}_{3} = - \text{230 kJ}$

May 6, 2017

DeltaH_(rxn) = DeltaH_1' + DeltaH_2' + DeltaH_3' = ???

($\Delta {H}_{i} ' \ne \Delta {H}_{i}$.)

This is asking you to apply Hess's law in a VERY general setting. The basic ideas are:

• Multiply reaction step $i$ by a constant $\to$ multiply $\Delta {H}_{i}$ by that constant.
• Reverse reaction step $i$ $\to$ switch the sign of $\Delta {H}_{i}$.
• Anything that matches on the reactants and products side of any reaction steps in the mechanism cancels out.
• You must achieve the overall reaction when adding up the individual steps.

In general, this will guarantee you get the correct answer:

1. Look for all intermediates and catalysts, and plan your actions out so that they will cancel out.
2. Use the overall reaction as a guide, as you must get the right number of products and reactants on each side.
3. Apply the above methods to flip reactions, scale them, etc., so that all intermediates and catalysts cancel out.

So, for this reaction, I will demonstrate the above by blindly looking to produce the correct product and reactant coefficients:

1. Multiply step $1$ by $2$ and reverse it. That brings $4 X {Y}_{3}$ to the left side.
2. Multiply step $2$ by $2$. That allows $4 X {Z}_{2}$ to be produced.
3. Multiply step $3$ by $3$. That allows $6 {Y}_{2} Z$ to be produced.

$- 2 \left({X}_{2} + 3 {Y}_{2} \to 2 X {Y}_{3}\right)$, $- 2 \left(\Delta {H}_{1} = - \text{320 kJ}\right)$

$2 \left({X}_{2} + 2 {Z}_{2} \to 2 X {Z}_{2}\right)$, $2 \left(\Delta {H}_{2} = - \text{140 kJ}\right)$

$3 \left(2 {Y}_{2} + {Z}_{2} \to 2 {Y}_{2} Z\right)$, $3 \left(\Delta {H}_{3} = - \text{230 kJ}\right)$

Now apply the coefficients. A negative coefficient takes into account reversing the reaction.

Note that I foolishly did not look for what were intermediates or catalysts... and yet... magic. If this doesn't work, then there is a typo in the reaction steps!

$4 X {Y}_{3} \to \cancel{2 {X}_{2}} + \cancel{6 {Y}_{2}}$, $\Delta {H}_{1} ' = + \text{640 kJ}$

$\cancel{2 {X}_{2}} + 4 {Z}_{2} \to 4 X {Z}_{2}$, $\Delta {H}_{2} ' = - \text{280 kJ}$

$\cancel{6 {Y}_{2}} + 3 {Z}_{2} \to 6 {Y}_{2} Z$, $\Delta {H}_{3} ' = - \text{690 kJ}$

$\text{----------------------------------------------------}$

$4 X {Y}_{3} + 7 {Z}_{2} \to 6 {Y}_{2} Z + 4 X {Z}_{2}$,

color(blue)(DeltaH_(rxn) = DeltaH_1' + DeltaH_2' + DeltaH_3' = ???)

And just to see if you're paying attention, I'll let you add this up. Rather than give you the straight-up answer, I want you to go through the process, because this is best learned by doing.