Solving Hess' Law Problems

Here, again, is our problem:

We know that just adding these reactions won't work.

So, in order to make this work, we will need to make some changes to the three working reactions.

The changes we can make are:

• we can multiply (or divide) them by any number
• we can reverse them (run them backwards). 

Of course if we change a reaction, we will also change the amount of heat that it takes in or gives away. However those changes are very predictable.

When we reverse a reaction, we change the sign on the heat. In other words, if a reaction is exothermic, its reverse will be endothermic to the exact same extent.

If we multiply a reaction by a factor, we multiply the heat by the same factor. In other words, if we react twice as much of the reactant we will get out (or put in) exactly twice as much heat. The same goes for dividing.

To apply this to our problem, there are a few simple rules that you should follow:

1. Look for elements or compound that appear in the reaction you want (the one at the top in our problem) which show up in one (and ONLY one) of the other reactions. When you find it, focus ONLY on that element or compound 
2. For each of the things you find (in step 1) ask yourself two questions: A. is it on the correct side of the equation, and B. do you have the correct amount? If the answer to either (or both) of those questions is no, fix it. 
3. Once you have made something (from step 1) match the main reaction, do NOT touch that reaction again. 
4. If there is a reaction that you have not used, look to see if there are things that need to be cancelled out and do whatever math is needed to make that work.  

Solving the problem

Step 1 says to "Look for elements or compound that appear in the reaction you want (the one at the top in our problem) which show up in one (and ONLY one) of the other reactions. When you find it, focus ONLY on that element or compound"

In this case, that would be N₂ (which appears only in the 3rd reaction) and N₂O₅ (which appears only in the middle reaction). Everything in the first reaction appears in more that one reaction. The H₂ appears in the third reaction, as does the O₂ and the H₂O is also in the middle reaction.

Let's look at the N₂ in the last reaction. According to step 2 we need to ask ourselves two questions:

Is the N₂ on the correct side of the equation and do we have the correct amount?
The answer to the first is YES. The N₂ is on the left in the third reaction and it is on the left in the main reaction.

The answer to the second question, however, is NO. The main reaction contains 1 N₂, while the third reaction contains only 1/2 N₂. To fix this, we need to double the third reaction.

2 x { 1/2 N_{2 (g)} + 3/2 O_{2 (g)} + 1/2 H_{2 (g)} → HNO_{3 (l)} }

Which results in

N_{2 (g)} + 3 O_{2 (g)} + H_{2 (g)} → 2 HNO_{3 (l)}

Remember that when we double a reaction, we also double the heat, so the heat of this reaction is now -83.2 kcal.

Now, if we look at the N₂O₅ we see that we have the correct amount, but it is on the wrong side of the reaction. Reversing the reaction gives us

2 HNO_{3 (l)} → N₂O_{5 (g)} + H₂O _(l)

Reversing the reaction makes an exothermic reaction into an endothermic one, so the heat for this reaction is now +18.3 kcal

We've now dealt with 2 of the three reactions. Let's look at the third...

It is not apparently obvious what to do with the first of the bottom reactions.
This is because everything in this reaction shows up in at least one other reaction and two of the three ingredients in this reaction don't even show up in the main reaction.

That does not mean that we can't figure out what to do.

There are a number of ways to determine what to do with the remaining reaction (the first of the three). What is crucial to remember is that we are trying to create the main reaction and that we CANNOT change the reactions we have already dealt with.

Here are several ways you could figure out what to do with the first reaction.

First way
The third reaction gives us 3 oxygen molecules on the left side, but we only need 5/2 (that's 2.5 for those of you who hate fractions). Therefore, we need to cancel out 1/2 O2. This could be done if we had 1/2 O2 on the right side of the first reaction. That can be achieved by flipping the reaction and dividing it by 2.

Second way
There is no water in the main reaction, but the middle reaction has a water on the right side. To cancel it out, we need a water on the left side. This can be accomplished by flipping the first reaction and dividing it by two.

Third way
The bottom reaction gives us an H2 on the left side, but we shouldn't have any (according to the main reaction). To get rid of it, we need an H2 on the right side. This can be accomplished by flipping the first reaction and dividing it by 2.

Notice that each way of thinking about it gives you the same answer. We need to flip the first reaction and divide it by 2. So here's the reaction:

2 H_{2 (g)} + O_{2 (g)} → 2 H₂O _(l)

If we flip the reaction, we get

2 H₂O _(l) → 2 H_{2 (g)} + O_{2 (g)}

and the heat will become +136.6 kcal. Then, if we divide it by two we have 

H₂O _(l) → H_{2 (g)} + 1/2 O_{2 (g)}

and the heat will become + 68.3 kcal.

Here is our problem with the reactions transformed as described above: Determine the ΔH_rxn for the reaction N_{2 (g)} + 5/2 O_{2 (g)} → N₂O_{5 (g)} from the following:

If we add up these three reactions we get

H₂O _(l) + 2 HNO_{3 (l)} + N_{2 (g)} + 3 O_{2 (g)} + H_{2 (g)} → 2 HNO_{3 (l)} + N₂O_{5 (g)} + H₂O _(l) + H_{2 (g)} + 1/2 O_{2 (g)}

We can now cancel out the water (one on each side of the equation) and the hydrogen (same reason). In addition, we can subtract 1/2 O₂ from each side. In addition, without even paying attention to it, the HNO₃ will also cancel out. This leaves us with:

N_{2 (g)} + 5/2 O_{2 (g)} → N₂O_{5 (g)}

which is the reaction we wanted. According to Hess' Law, since adding the three reactions now gives us the main reaction, we should be able to add the heats to get the heat of the main reaction. That means that the ΔH_rxn for our reaction is 

68.3 kcal + 18.3 kcal -83.2 kcal = 3.4 kcal

If you want to try some on your own, take a look at this worksheet. The answers are here.