# 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

Mar 29, 2015

To calculate the enthalpy of combustion of acetylene, ${C}_{2} {H}_{2}$, all you have to do is use the standard enthalpies of formation of the reactants, ${C}_{2} {H}_{2}$ and ${O}_{2}$, and of the products, $C {O}_{2}$ and ${H}_{2} O$.

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

$\textcolor{red}{2} {C}_{2} {H}_{2 \left(g\right)} + 5 {O}_{2 \left(g\right)} \to \textcolor{g r e e n}{4} C {O}_{2 \left(g\right)} + \textcolor{b l u e}{2} {H}_{2} {O}_{\left(g\right)}$

You were given the enthalpies of formation for 1 mole of ${C}_{2} {H}_{2}$, $C {O}_{2}$, and ${H}_{2} O$ - 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 ${C}_{2} {H}_{2}$ reacting to produce 2 moles of ${H}_{2} O$ and 4 moles of $C {O}_{2}$. To get the enthalpy of combustion for 1 mole of acetylene, divide the balanced equation by 2

${C}_{2} {H}_{2 \left(g\right)} + \frac{5}{2} {O}_{2 \left(g\right)} \to \textcolor{g r e e n}{2} C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(g\right)}$

Now the expression for the enthalpy of combustion will be

$\Delta {H}_{\text{comb}} = \left(\textcolor{g r e e n}{2} \cdot \Delta {H}_{C {O}_{2}}^{0} + \Delta {H}_{{H}_{2} O}\right) - \left(\Delta {H}_{{C}_{2} {H}_{2}}\right)$

$\Delta {H}_{\text{comb}} = \left[2 \cdot \left(- 393.5\right) + \left(- 241.6\right)\right] - \left(226.7\right)$

DeltaH_("comb") = -"1255.3 kJ"

Therefore, the enthalpy of combustion for acetylene is

DeltaH_("comb") = -"1253.3 kJ/mol"

SIDE NOTE I'm not sure what you mean by $\Delta H$ 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.