THERMOCHEMICAL EQUATIONS

Many chemical reactions are accompanied by heat.  Some reactions produce heat, while others require heat for the reaction to occur.  The heat flow associated with a reaction at constant pressure, qp, is called enthalpy.  The symbol for enthalpy is H.  For reasons not covered here, only the difference in enthalpy for the reaction is important.  The symbol for this enthalpy change is DH. (DH is read as "delta H".  For some reason the Greek capital letter delta is not working.)

A reaction that produces heat is referred to as exothermic.  For exothermic reactions, enthalpy values are assigned a negative value (-DH).

A reaction that requirs heat is referred to as endothermic.  For endothermic reactions, enthalpy values are assigned a positive value (+DH).

An equation which shows both mass and heat relationships between products and reactants is called a thermochemical equation.  The following four reactions are examples of thermochemical equations.  The first two are exothermic and the last two are endothermic reactions.

                2 H2(g) + O2(g) ----> 2 H2O(l)     DH = -571.6 kJ

                CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(g)     DH = -890 kJ

                CH3OH(l) ---> CO(g) + 2H2(g)     DH = +90.7 kJ

                NH4Cl(s)  ----> NH3(g) + HCl(g)     DH = +176 kJ

The magnitude of DH is directly proportional to the amount of reactants or products.

                 A + 2 B ----> C     DH = -100 kJ

                1/2 A + B ----> 1/2 C     DH = -50 kJ

DH for a reaction is equal in magnitude but opposite in sign for the reverse reaction.

                A + 2 B ----> C     DH = -100 kJ

                C ----> A + 2 B     DH = +100 kJ

The value of DH depends upon the phase of the reactants and the products.

                2H2(g) + O2(g) ---> 2H2O(g) DH = -483.7 kJ

                2H2(g) + O2(g) ----> 2H2O(l) DH = -571.7 kJ

Stoichometric Calculations Involving Thermochemical Equations

1.  When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by multiplying the value of DH in the original equation by that same factor.

2.  When a chemical equation if reversed, the value of DH is reversed in sign.

Example 1
Based upon the thermochemical equation given, calculate the heat associated with the decomposition of 1.15 g of NO2(g).
    N2(g) + 2 O2(g) ----> 2 NO2(g)    DH = +67.6 kJ

    A. Calculate moles of NO2(g).
                [1.15 g NO2(g)][1mol NO2(g) / 46.01 g NO2(g)] = 0.0250 mol NO2(g)

    B. Reverse the thermochemical equation and change sign of enthalpy.
                2 NO2(g) ----> N2(g) + 2 O2(g)    DH = -67.6 kJ

    C. Calculate heat.
                [0.0250 mol NO2][-67.6 kJ / 2 mol NO2] = - 0.845 kJ

Example 2
Using the following thermochemical equation, calculate how much heat is associated with the decomposition of 4.00 moles of NH4Cl.
NH3(g) + HCl(g) ---> NH4Cl(s)       DH = -176 kJ

(4.00 mol NH4Cl)(+176 kJ/1 mol NH4Cl) = +704 kJ

Example 3
What quantity of heat is associated with the synthesis of 45.0 g of NH3 according to the following equation?
4 NO(g) + 6 H2O(l) ---> 4 NH3(g) + 5 O2(g)    DH = +1170 kJ

(45.0 g NH3)(1 mol/17.04 g NH3) = 2.641 mol NH3

(2.641 mol NH3)(+1170kJ/4 mol NH3) = +772 kJ

Example 4
Calculate the mass of ethane, C2H6, which must be burned to produce 100 kJ of heat.

2 C2H6(g) + 7 O2(g) ---> 4 CO2(g) + 6 H2O(l)   DH = -3120 kJ

(-100 kJ)(2 mol C2H6 / -3120 kJ) = 0.0641 mol C2H6

(0.0641 mol C2H6)(30.08 g C2H6/1 mol C2H6) = 1.93 g C2H6