Hess's Law problem, please help.

Question

6. Considering the following data:

ΔH KJ

Ca (s) + 2 Cgraphite → CaC2 (s) -62.8

Ca (s) + ½ O2 → CaO (s) -635.5

C2H2 (g) + 5/2 O2 (g) → 2 CO2 (g) + H2O (l) -1300

Cgraphite + O2 → CO2 (g) -393.5

CaO (s) + H2O (l) → Ca(OH)2 (aq) -653.1

Use Hess’s law to find the change in enthalpy at 25oC for the following reaction:

CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g)

Answer 1

The easiest way to begin is to switch the equations (and change the enthalpy sign!) to match the reaction.

 

So, we can tell that the first reaction: Ca_(s)+ 2 C_(graphite) → CaC_{2 (s)} ΔH= -62.8  needs to flipped to this.

 

CaC_{2 (s)} → Ca_(s) +2 C_(graphite) ΔH= 62.8

 

Looking down the list we can see that third reaction has a water on the product side, when we need it on the reactant side, so the third equation needs to be flipped (as well as the ethyne).

 

2 CO_{2 (g)}+ H₂O_(l) → C₂H₂ + 5/2 O₂ ΔH= 1300

 

Your full reaction synthesis will now look like this in:

Ca₂C₂ → Ca + 2 C_(graphite)

Ca + 1/2 O₂ → CaO

2 CO₂ + H₂O → C₂H₂ + 5/2 O₂

C_(graphite) + O₂ → CO₂

CaO + H₂O → Ca(OH)₂

 

Now comes hard part. My preference is to tally how much of each species is on the reactant and product side.

 

Reactant side: 1 CaC₂, 1 Ca, 3/2 O₂, 1 C_(graphite), 1 CaO, 2 H₂O, 2 CO₂

Product side: 1 Ca, 5/2 O₂, 2 C_(graphite), 1 CaO, 1 CO₂, 1 Ca(OH)₂

 

So it looks like to cancel out the O₂, graphite, and CO₂, we need to times one of the reactions by two. Looking back at the all the reactions, C_(graphite) + O₂ → CO2, fits the bill.

 

The overall reaction will look like this when put together:

 

CaC₂ + Ca + 5/2 O₂ + 2 CO₂ + 2 H₂O + 2 C_(graphite) → Ca + 2 C_(graphite) + CaO + C₂H₂ + 5/2 O₂ + 2 CO₂ + Ca(OH)₂

 

However, when simplified, it becomes:

 

CaC₂ + 2 H₂O → Ca(OH)₂ + C₂H₂

 

Which is the overall net reaction.

 

Now, we can calculate the change in enthalpy (finally!), which is just adding up the ΔH values (keeping in mind the switching of signs and the multiplication):

 

ΔH= 62.8 - 635.5 +1300 - 2(393.5) - 653.1= -712.8 kJ

 

So the final answer would ΔH= -712.8 kJ