Hess's Law Paper

Question

Consider the formation of ethane.

a. provided the following data, calculate ΔH for the reaction:

2 C (s) + 3 H₂ (g) ---> C₂H₆ (g)

4 C₂H₆ (g) + 14 O₂ (g) --> 8 CO₂ (g) + 12 H₂O (l) ΔH = -6240 kJ

C (s) + O₂ (g) ---> CO₂ (g) ΔH = -394 kJ

2 H₂ (g) + O₂ (g) ---> 2 H₂O (l) ΔH = -572 kJ

b. Consider the enthalpy change calculated in (a). Does this represent the standard enthalpy of formation? Explain your answer.

Answer 1

2 C (s) + 3 H₂ (g) ---> C₂H₆ (g) ...TARGET EQUATION

Given:

eq.1: 4 C₂H₆ (g) + 14 O₂ (g) --> 8 CO₂ (g) + 12 H₂O (l) ... ΔH = -6240 kJ

eq.2: C (s) + O₂ (g) ---> CO₂ (g) ... ΔH = -394 kJ

eq.3: 2 H₂ (g) + O₂ (g) ---> 2 H₂O (l) ... ΔH = -572 kJ

Procedure:

eq.2 x 2: 2C (s) + 2O₂ (g) ---> 2CO₂ (g) ... ΔH = -788 kJ

reverse eq.1 and ÷4: 2CO₂(g) + 3H₂O(l) ==> C₂H₆(g) + 3.5 O₂(g) ... ∆H = +1560 kJ

eq.3 x 1.5: 3 H₂ (g) + 1.5 O₂ (g) ---> 3 H₂O (l) ... ΔH = -858 kJ

__________________________________________________ add them all together...

2C (s) + 2O₂ (g) + 2CO₂(g) + 3H₂O(l) + 3 H₂ (g) + 1.5 O₂ (g) ==> 2CO₂ (g) + C₂H₆(g) + 3.5 O₂(g) + 3 H₂O (l)

And we are left with the TARGET EQUATION:

2C(s) + 3H₂(g) ==> C₂H₆(g)

∆H = -788 kJ + 1560 kJ + (-858 kJ)

∆H = -86 kJ

Answer 2

Hess's Law is easiest when every reaction has a unique species that is in the final equation. Then it becomes a simple matter of combining the reactions in order to get those reactants in the right place and using the same multipliers on the enthalpies.

In this case we need to reverse the 1st reaction and divide all the terms by 4 in order to get 1 mole of ethane on the right. This means we have -1/4 ΔH₁

The second reaction is doubled because that is the only way to get 2C on the left: 2ΔH₂

The third reaction is multiplied by 3/2 in order to get the 3 H₂ on the left 3/2 ΔH₃

You can do the exercise of combining the equations as I stated. Everything will cancel that is not one of the species in the overall equation. (i.e. the intermediates cancel out) If the reactions are balanced correctly, you don't have to write out the reaction addition)

So the answer is to add the enthalpies as I stated above: -1/4ΔH₁ + 2ΔH₂ + 3/2ΔH₃

If the final reaction species are not uniquely determined, you can use equations to cancel intermediates created by prior reactions.

The reaction is a standard enthalpy of formation (usually C(graphite) is specified