Question

Using the following data:

C(graphite) + O2(g) ----> CO2 (g)
ΔH °rxn = -393.5 kJ/mol

H2(g) + 1/2 O2(g) ----> H2O (l)
ΔH °rxn = -285.8 kJ/mol

2 C2H6(g) + 7 O2 (g) ----> 4 CO2 (g) + 6 H2O(l)
ΔH °rxn = -3119.6 kJ/mol

calculate the enthalpy for this reaction:
2 C(graphite) + 3 H2(g) ----> C2H6(g)

I don't really understand how these types of problems work, so if whoever answers my question could explain how they got their answer that would be great.

Answer 1

You need to use Hess' Law to find the answer to this question. Hess' Law basically says that if you have one reaction whose enthalpy you don't know, but you have several other reactions whose enthalpy you do know and, when combined, produce the unknown reaction, the enthalpy of the unknown reaction is the sum of the enthalpies of all the other reactions.

I know that sounds kind of tricky, but I'll walk you through it and hopefully it'll make some sense. We have a reaction whose enthalpy is unknown:

\[2C+3H_2 \rightarrow C_2H_6 \space \space \space \Delta H = ?\]
We need to combine the given reactions somehow to produce this reaction. The first reaction provided is the only one with C (graphite), and it's already on the reactant's side of the equation. However, our unknown equation has TWO C's on the left side...that means we need to multiply the known equation by 2 to get two C's in it as well. If we multiply the equation by 2, the enthalpy also doubles (twice as many reactants release twice as much heat). Therefore we can rewrite the equation as follows:

\[2C+2O_2 \rightarrow 2CO_2 \space \space \space \Delta H=-787 kJ\]
Does this make sense so far?

Answer 2

Yes I think I follow what you're saying

Answer 3

Ok great! So we've taken care of the C in our unknown equation, so now we need to take care of the H2 and C2H6 as well. Also, our known equation has introduced O2 and CO2, which are not in the equation we want, so we'll need to get rid of those somehow.

Let's look at the second given equation. It gives us H2, which is in our unknown, so let's focus on that. Once again it's on the correct side of the equation, but it doesn't have the correct coefficient. Just like we did for the first given equation, we need to multiply this one (including the enthalpy) by a constant, in this case 3, so it matches what we have in our unknown equation. This means we have:

\[3H_2+\frac{3}{2}O_2 \rightarrow 3H_2O \space \space \space \Delta H=-857.4kJ\]
Still following?

Answer 4

yeah that makes sense

Answer 5

Great! Now we've dealt with the C and the H2, but we still need to take care of the C2H6. We need to somehow get rid of all the compounds we don't have in our unknown equation. The third given equation will let us do both these things!

The third equation has C2H6, but it's on the wrong side of the equation - we want it to be a product, not a reactant. That means we need to flip the equation to get C2H6 where we want it. We can do that, but that means we also need to REVERSE the sign of the enthalpy, because the reaction is now going in the opposite direction (e.g. if the reaction originally released heat and therefore had a negative enthalpy, once we flip it, the reaction as written will absorb heat and therefore have a positive enthalpy).

The second thing we need to do is DIVIDE the entire reaction (including the enthalpy) by 2, in order to get the coefficient of C2H6 to match our unknown. When you flip the equation and divide by 2, here's what you get:

\[2CO_2+3H_2O \rightarrow C_6H_6+\frac{7}{2}O_2 \space \space \space \Delta H = 1559.8kJ\]
Does that step make sense?

Answer 6

So if I'm following correctly, we just have to basically flip the problem backwards and do the reverse of what we were doing for the other two steps?

Answer 7

And we want to get C2H6 by itself?

Answer 8

Pretty much! Now we've rearranged all three given equations to give us what we want in the unknown, but let's make sure everything is in order. I'd recommend you write out all the rearranged equations plus the unknown on a piece of paper - can you do that?

Answer 9

yes I just did

Answer 10

Now find everything in the rearranged equations that appears in the unknown equation, and circle it. Make sure they're on the correct side and have the same coefficient.

Answer 11

Ok got it!

Answer 12

Now everything that isn't circled we need to make sure cancels out. We're looking for "intermediates". So if something was produced in one reaction, but used up in another reaction, it cancels and we don't need to worry about it. Let's look at CO2 to start. We produce 2CO2 in the first equation, but we react 2CO2 in the third equation. All the CO2's cancel out, so you can cross out those guys in both equations.

Answer 13

Ok. I take it the H2O's cancel out too. How about the oxygens?

Answer 14

You tell me! Look at H2O and look at O2. When you add up everything on the left side of the equation, is it the same as everything you have on the right side of the equation?

Answer 15

I believe that they do add up to the same on each side

Answer 16

They do. The H2O is easy, you have 3 on one side and 3 on the other, so those cancel. The O2 is a bit trickier because it's broken up. On the right side, 7/2 O2. On the left, you have 2O2 in one equation and 3/2 O2 in another. But, 2+3/2=7/2, so really you're dealing with 7/2 on the left side as well. O2's also cancel.

Last step is to add up all the enthalpies from the rearranged equations since we know when you combine them you produce the unknown equation. What is the enthalpy of the unknown equation?

Answer 17

I got -84.6 kJ/mol. Is that correct?

Answer 18

Yup well done!

Answer 19

I know it took a while to solve this question but it's only because I was taking some time to walk you through it. Hess' Law problems are pretty easy - just look at the unknown equation and rearrange the given equations appropriately to match it, making appropriate changes to the enthalpy of each. At the end, make sure you've accounted for everything in the unknown equation, and that everything not in that equation cancels out, then add up the enthalpies and you're done!

Answer 20

Ok! Thank you so much for your help! I think these problems make much more sense now

Answer 21

No problem good luck!