calculate the heat of combustion for the following reactions from the standard enthalpies of formation
2H2(g) + O2(g) = 2H2O(l)
2C2H2(g) + 502(g) = 4CO2(g) + 2 H2O(l)
To calculate the heat of combustion for a given reaction using standard enthalpies of formation, you need to follow these steps:
1. Write down the balanced chemical equation for the reaction.
For the first reaction:
2H2(g) + O2(g) → 2H2O(l)
For the second reaction:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
2. Determine the standard enthalpies of formation (ΔHf°) for each compound involved in the reaction. The standard enthalpy of formation is the heat absorbed or released when one mole of a compound is formed from its elements, with all substances in their standard states. These values can be found in reference tables.
The standard enthalpies of formation for the compounds in the first reaction are:
H2(g) = 0 kJ/mol
O2(g) = 0 kJ/mol
H2O(l) = -285.8 kJ/mol
The standard enthalpies of formation for the compounds in the second reaction are:
C2H2(g) = 226.7 kJ/mol
O2(g) = 0 kJ/mol
CO2(g) = -393.5 kJ/mol
H2O(l) = -285.8 kJ/mol
3. Calculate the overall change in enthalpy (ΔH) for the reaction using the standard enthalpies of formation.
For the first reaction:
ΔH = 2(ΔHf°[H2O(l)]) - 2(ΔHf°[H2(g)]) - ΔHf°[O2(g)]
= 2(-285.8 kJ/mol) - 2(0 kJ/mol) - 0 kJ/mol
= -571.6 kJ/mol
For the second reaction:
ΔH = 4(ΔHf°[CO2(g)]) + 2(ΔHf°[H2O(l)]) - 2(ΔHf°[C2H2(g)]) - 5(ΔHf°[O2(g)])
= 4(-393.5 kJ/mol) + 2(-285.8 kJ/mol) - 2(226.7 kJ/mol) - 5(0 kJ/mol)
= -2597.6 kJ/mol
Therefore, the heat of combustion for the first reaction is -571.6 kJ/mol, and for the second reaction is -2597.6 kJ/mol.
DHfrxn = (n*DHfproducts) - (n*DHfreactants)
Look up DHf for the material, multiply by n etc.