Acetylene torches utilize the following reaction: 2 C2H2 (g) + 5 O2 (g) -----> 4 CO2 (g) + 2 H20 (g) How to calculate Delta H in Celsius for this reaction ? Species Delta H C2H2 (g )= +226.7 CO2 (g )= -393.5 H20 (g) = -241.6

1 Answer
Mar 29, 2015

To calculate the enthalpy of combustion of acetylene, C2H2, all you have to do is use the standard enthalpies of formation of the reactants, C2H2 and O2, and of the products, CO2 and H2O.

More specifically, you need to subtract from the sum of enthalpies of formation of the products the sum of the enthalpies of formation of the reactants.

Keep in mind that each of these enthalpies must be multiplied or divided in accordance to their stoichiometric coefficients from the balanced chemical reaction.

So, the balanced chemical reaction is

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

You were given the enthalpies of formation for 1 mole of C2H2, CO2, and H2O - the enthalpy of formation for oxygen is zero (read more on that here: http://socratic.org/questions/why-is-the-enthalpy-of-formation-of-oxygen-zero).

But notice that the balanced chemical reaction has 2 moles of C2H2 reacting to produce 2 moles of H2O and 4 moles of CO2. To get the enthalpy of combustion for 1 mole of acetylene, divide the balanced equation by 2

C2H2(g)+52O2(g)→2CO2(g)+H2O(g)

Now the expression for the enthalpy of combustion will be

ΔHcomb=(2⋅ΔH0CO2+ΔHH2O)−(ΔHC2H2)

ΔHcomb=[2⋅(−393.5)+(−241.6)]−(226.7)

ΔHcomb=−1255.3 kJ

Therefore, the enthalpy of combustion for acetylene is

ΔHcomb=−1253.3 kJ/mol

Here's a link to a similar answer on this topic:

http://socratic.org/questions/how-do-you-calculate-enthalpy-change-of-combustion

SIDE NOTE I'm not sure what you mean by ΔH in Celsius, since the heat of combustion is expressed as energy per mole; in this case, burning 1 mole of acetylene will release 1253.3 kJ of energy.