Hess's Law Worksheet (2) – Chemistry Name: _______________________________________ (1) Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl 5 (g) → PCl 3 (g) + Cl 2 (g) P 4 (s) + 6Cl 2 (g) → 4PCl 3 (g) ΔH = -2439 kJ 4PCl 5 (g) → P 4 (s) + 10Cl 2 (g) ΔH = 3438 kJ answer = 249.8 kJ (2) Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: 2CO 2 (g) + H 2 O(g) → C 2 H 2 (g) + 5/2O 2 (g) C 2 H 2 (g) + 2H 2 (g) → C 2 H 6 (g) ΔH =-94.5 kJ H 2 O(g) → H 2 (g) + 1/2O 2 (g) ΔH =71.2 kJ C 2 H 6 (g) + 7/2O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ΔH =-283 kJ answer = 235 kJ (3) Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: N 2 H 4 (l) + H 2 (g) → 2NH 3 (g) N 2 H 4 (l) + CH 4 O(l) → CH 2 O(g) + N 2 (g) + 3H 2 (g) ΔH = -37 kJ N 2 (g) + 3H 2 (g) → 2NH 3 (g) ΔH = -46 kJ CH 4 O(l) → CH 2 O(g) + H 2 (g) ΔH = -65 kJ answer = -18 kJ (4) Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: H 2 SO 4 (l) → SO 3 (g) + H 2 O(g) H 2 S(g) + 2O 2 (g) → H 2 SO 4 (l) ΔH = -235.5 kJ H 2 S(g) + 2O 2 (g) → SO 3 (g) + H 2 O(l) ΔH = -207 kJ H 2 O(l) → H 2 O(g) ΔH = 44 kJ answer = 72 kJ
First you have to design your cycle.
Write down the enthalpy change you want to find as a simple horizontal equation, and write ΔH over the top of the arrow. (In diagrams of this sort, we often miss off the standard symbol just to avoid clutter.)
Then fit the other information you have onto the same diagram to make a Hess's Law cycle, writing the known enthalpy changes over the arrows for each of the other changes.
Finally, find two routes around the diagram, always going with the flow of the various arrows. You must never have one of your route arrows going in the opposite direction to one of the equation arrows underneath it.
In this case, what we are trying to find is the standard enthalpy change of formation of benzene, so that equation goes horizontally.
You will notice that I haven't bothered to include the oxygen that the various things are burning in. The amount of oxygen isn't critical because you just use an excess anyway, and including it really confuses the diagram.
Why have I drawn a box around the carbon dioxide and water at the bottom of the cycle? I tend to do this if I can't get all the arrows to point to exactly the right things. In this case, there is no obvious way of getting the arrow from the benzene to point at both the carbon dioxide and the water. Drawing the box isn't essential - I just find that it helps me to see what is going on more easily.
Notice that you may have to multiply the figures you are using. For example, standard enthalpy changes of combustion start with 1 mole of the substance you are burning. In this case, the equations need you to burn 6 moles of carbon, and 3 moles of hydrogen molecules. Forgetting to do this is probably the most common mistake you are likely to make.
How were the two routes chosen? Remember that you have to go with the flow of the arrows. Choose your starting point as the corner that only has arrows leaving from it. Choose your end point as the corner which only has arrows arriving.
Now do the calculation:
Hess's Law says that the enthalpy changes on the two routes are the same. That means that:
ΔH - 3267 = 6(-394) + 3(-286)
Rearranging and solving:
ΔH = 3267 + 6(-394) + 3(-286)
ΔH = +45 kJ mol-1