Answer to Question #330291 in General Chemistry for Lwsom

Question

Given the following thermochemical equations and the corresponding enthalpy change for each reaction, calculate the heat of combustion of acetylene gas, C2H2

1. 2C (graphite) + H2 (g)               C2H2(g)      ∆H= + 226.7 kJ
2. C(graphite) + O2 (g)             CO2(g)         ∆H= - 393.5 kJ
3.  H2(g) + ½ O2 (g)            H2O (l)                ∆H= - 285.8 kJ

Answer 1

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.

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) + 5/2O₂(g) = 2CO₂(g) + H₂O(g)

Now the expression for the enthalpy of combustion will be

∆H_comb = (2×∆H⁰_CO2 + ∆H_H2O) - (∆H_C2H2) = (2×(-393.5) + (-285.8)) - (226.7) = -1299.5 kJ