Hess's Law Problem 4

Question

Consider the following reactions:

C(s) + 2H2O(g) --> CO2(g) + 2H2(g) Enthalpy = +163.0 kJ/mole

C(s) + ½O2(g) ---> CO(g) Enthalpy = -121.0 kJ/mole

H2O(g) ---> H2O(l) Enthalpy = -40.7 kJ/mole

CO(g) + ½O2(g) ---> CO2(g) Enthalpy = -283.0 kJ/mole

Calculate the Enthalpy for the reaction H2(g) + ½ O2(g) ---> H2O(l)

Answer 1

Hess' Law states that we can add up the individual enthalpies of each reaction to get the overall enthalpy of the overall reaction. In this case, we want to know the enthalpy of the overall equation. H2(g) + ½ O2(g) ---> H2O(l).

The simplest way to figure this out is to locate each molecule of our overall reaction (H2, O2, and H2O) in the reactions we are given and sum them all up:

1. Let's find H2. We see that the only reaction with H2 is in the first reaction given. However, notice how in the reaction, the H2 is a product. However, we want the H2 to be a reactant like in our overall equation. Another element to Hess' Law is that we can flip the reaction around so it becomes CO2(g) + 2H2 (g) --> 2H2O + C(s). Now our H2 is a reactant. But don't forget, we must also flip the sign of the enthalpy! So now our enthalpy for this reaction is -163 kj/mol.
2. 1/2O2 is found in two reactions listed above so let's note that down.
3. H2O is also found in two reactions, but notice how in our final equation we want H2O(l) not H2O(g). But remember our reaction that we flipped for H2? Now the H2O(g) is a product rather than a reactant. This means that if we add reaction #1 (flipped) and reaction #3, the H2O(g) almost gets canceled out. To fully cancel it out, we have to multiply reaction #3 by 2. Note that we must then multiply the enthalpy of reaction #3 by 2 as well so we get -81.4 kj/mol for reaction #3 enthalpy.

To summarize:

Reaction 1 flipped + Reaction 3 multiplied by 2 gives 2H2(g) + CO2(g) --> 2H2O(l) + C(s) (enthalpy = -244.4 kj/mol)

Now looking back at our overall equation, we need to add in 1/2O2 and cancel out the CO2 and C(s).

1. To cancel out CO2 we use reaction #4 and can add it directly to our reaction in progress:
2. 2H2(g) + CO2(g) --> 2H2O(l) + C(s) -244.4 kj/mol
3. CO(g) + 1/2O2(g) --> CO2(g) -283 kj/mol
4. Add the above two and we get: 2H2(g) + CO(g) + 1/2O2(g) --> C(s) + 2H2O(l) (enthalpy = -527.4 kj/mol)
5. Finally we can add our reaction #2
6. 2H2(g) + CO(g) + 1/2O2(g) --> C(s) + 2H2O(l) -527.4 kj/mol
7. C(s) + 1/2O2(g) --> CO(g) -121 kj/mol
8. Add the two above and we get: 2H2(g) + O2(g) --> 2H2O(l) (enthalpy = -648.4 kj/mol)

So we did it right? Almost. Look at the question, the coefficients of our molecules in the problem are half of the ones we got above. So an easy fix is to just divide the equation and enthalpy we got by 2. So we get -648.4 kj/mol / 2 = -324.2 kj/mol which is our final answer.