Welders often use acetylene torches heat up the metal. The reaction is:
C2H2(g) + 2.5O2(g) = 2CO2(g) + H2O(g)
Calculate the ΔH of reaction for this reaction.
dHrxn = (n*dHf products) - (n*dHf reactants)
To calculate the ΔH (enthalpy change) of the reaction, you need to use the standard enthalpy of formation values for the reactants and products involved in the reaction. The standard enthalpy of formation (ΔH°f) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
Here are the standard enthalpy of formation values for the compounds involved in the reaction you provided:
ΔH°f of C2H2(g) = 226.7 kJ/mol
ΔH°f of O2(g) = 0 kJ/mol (by definition)
ΔH°f of CO2(g) = -393.5 kJ/mol
ΔH°f of H2O(g) = -241.8 kJ/mol
Now, let's calculate the ΔH of reaction:
Reactants:
C2H2(g): 1 mole
O2(g): 2.5 moles
Products:
CO2(g): 2 moles
H2O(g): 1 mole
Using the stoichiometric coefficients, we can calculate the ΔH of reaction:
ΔH of reaction = (ΣΔH°f of products) - (ΣΔH°f of reactants)
ΔH of reaction = (2 mol CO2 * ΔH°f of CO2) + (1 mol H2O * ΔH°f of H2O) - (1 mol C2H2 * ΔH°f of C2H2) - (2.5 mol O2 * ΔH°f of O2)
ΔH of reaction = (2 mol * -393.5 kJ/mol) + (1 mol * -241.8 kJ/mol) - (1 mol * 226.7 kJ/mol) - (2.5 mol * 0 kJ/mol)
ΔH of reaction = -787 kJ/mol
Therefore, the ΔH of reaction for the given reaction is -787 kJ/mol.