Calculate the delta H for 2C + H2 = C2H2

Question

Given the following data:

C2H2 + 5/2 O2 = 2CO2 + H2O delta H= -1300kJ

C + O2 = CO2 delta H=-394 kJ

H2 + 1/2 O2 = H2O delta H=-286 kJ

Answer 1

Short answer: delta H = +226 kJ. The reaction is endothermic, and that positive sign is not an arithmetic slip -- acetylene is one of the very few hydrocarbons that sits uphill from its own elements.

Write the target first, then ask what each given equation has to do to land on it. That is the whole method; everything else is bookkeeping.

Target:  2 C(s) + H2(g) -> C2H2(g)      delta H = ?

Given (1):  C2H2 + 5/2 O2 -> 2 CO2 + H2O    -1300 kJ
Given (2):  C + O2 -> CO2                    -394 kJ
Given (3):  H2 + 1/2 O2 -> H2O               -286 kJ

Step 1 -- equation (2), doubled. The target needs 2 C on the left. Equation (2) supplies carbon on the left one atom at a time, so multiply the entire equation by 2 -- and multiply its delta H by 2 as well.

2 C + 2 O2 -> 2 CO2        delta H = 2(-394) = -788 kJ

Step 2 -- equation (3), used exactly as written. The target needs one H2 on the left, which is where (3) already has it. Nothing to change.

H2 + 1/2 O2 -> H2O         delta H = -286 kJ

Step 3 -- equation (1), reversed. The target needs C2H2 as a product, but (1) burns it as a reactant. Flip the equation, and flip the sign of delta H with it.

2 CO2 + H2O -> C2H2 + 5/2 O2   delta H = +1300 kJ

Add all three and cancel.

left:   2 C + 2 O2 + H2 + 1/2 O2 + 2 CO2 + H2O
right:  2 CO2 + H2O + C2H2 + 5/2 O2

2 CO2 cancels. H2O cancels. And the oxygen cancels too: 2 + 1/2 = 5/2 on the left against 5/2 on the right. What survives is exactly the target reaction, so the enthalpies simply add:

delta H = (-788) + (-286) + (+1300) = +226 kJ

The oxygen cancellation is a free check on your work. You never chose those O2 coefficients -- they fell out of doubling (2) and keeping (3) as written. The fact that they add to exactly the 5/2 that the reversed equation (1) demands means your multipliers are right. If the O2 had not cancelled, you would know a coefficient was wrong before you ever touched the arithmetic. Get in the habit of checking the species you are not asked about; those are the ones that catch errors.

A second route that gets the same number in one line. Notice that the target reaction builds C2H2 from carbon and hydrogen in their standard states -- that is the definition of a formation reaction. So apply the standard rule to equation (1):

delta H(1) = [2 dHf(CO2) + dHf(H2O)] - [dHf(C2H2) + 0]
-1300 = [2(-394) + (-286)] - dHf(C2H2)
-1300 = -1074 - dHf(C2H2)
dHf(C2H2) = -1074 + 1300 = +226 kJ/mol

O2 drops out because the enthalpy of formation of an element in its standard state is zero by definition. Two independent methods landing on +226 kJ is much stronger evidence than one method giving it once.

Cross-check against a table. Look up the standard enthalpy of formation of acetylene and you will find +227 kJ/mol. Your answer of +226 differs only because the data you were handed are rounded to three figures. That agreement is the sign you did it right.

The two rules doing all the work:

Reverse an equation  ->  reverse the sign of delta H
Multiply by n        ->  multiply delta H by n

Both follow from enthalpy being a state function: the change depends only on where you start and where you finish, never on the path taken. That is exactly why you are allowed to route through CO2 and H2O even though nobody actually makes acetylene that way.

Why the positive sign is the most interesting thing here. Almost every hydrocarbon has a negative enthalpy of formation, meaning it sits lower in energy than the elements it came from. Acetylene does not -- the carbon-carbon triple bond leaves it about 226 kJ/mol above graphite and hydrogen gas. That stored energy gets released on top of the normal combustion energy, which is precisely why an oxyacetylene torch reaches roughly 3500 degrees C and can cut steel while a propane torch tops out far lower. It is also why acetylene is shipped dissolved in acetone inside its cylinder rather than simply compressed; pure compressed acetylene can decompose back to its elements on its own.

One thing to watch in your data. Equation (3) at -286 kJ is the value for liquid water, so the H2O in equation (1) has to be liquid too. That happens to be safe here, because H2O appears once on each side of the sum and cancels -- but if a problem ever hands you one equation with H2O(l) and another with H2O(g), they do not cancel, and you have to add the vaporization step (about +44 kJ/mol) to reconcile them. Check the physical states before you start cancelling, every time.