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Wednesday, July 17, 2013

How to Calculate Enthalpies of Reaction

The enthalpy (abbreviated as dH) of a reaction indicates amount of heat absorbed (dH >0) or released (dH<0) in this chemical reaction. Enthalpies of reactions are usually computed from enthalpies of formation of reagents at standard conditions (pressure of 1 bar and temperature 25 degrees Celsius). For the purposes of explaining this principle of thermodynamics, we will calculate the enthalpy of the reaction for methane (CH4) combustion in the following steps.

Instructions

    1

    Write down the chemical reaction equation (if it is not given initially) paying attention to the substance phase i.e. gas (g), liquid (l) or solid (s). The combustion of methane follows the equation: CH4(g) + 2O2(g)= CO2(g) + 2H2O(l).

    2

    Navigate to the enthalpy of formation table or consult the reference and find enthalpies of formation for all components of the reaction. Note that enthalpies of formation of elements at standard conditions are equal to zero.
    In our example,
    dH(CH4(g))= -74.87 kJ/mol
    dH(O2(g)=0
    dH(CO2(g))= -393.509 kJ/mol
    dH(H2O(l))= -285.83 kJ/mol

    3

    Add up enthalpies of formation of the initial reagents, multiplying each enthalpy value on the corresponding reaction coefficient. In our example, the initial reaction components are CH4 and O2 and
    dH(initial)= -74.87 kJ/mol + 2 x 0=-74.87 kJ/mol.

    4

    Add up enthalpies of formation of the final reagents, multiplying each enthalpy value on the corresponding reaction coefficient. In our example, the final reaction components are CO2 and H2O and
    dH(final)=-393.509 kJ/mol + 2 x (-285.83 kJ/mol) =-965.169 kJ/mol.
    Note that the coefficient at H2O in the reaction is 2 hence the H2O enthalpy of formation is multiplied by 2.

    5

    Subtract the value from Step 4 from the value obtained in Step 3 to calculate the enthalpy of the reaction: dH(reaction) = dH(final)-dH(initial). In our example, dH(reaction) = -965.169 kJ/mol- (-74.87 kJ/mol)= -890.299 kJ/mol.

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