Thermochemical Equations

(See Moore , 2^nd ed., Sections 6.5 & 6.6, pp.  233-240)

The heat flow for a reaction at constant pressure, q_p, is called enthalpy, ΔH.

Exothermic reactions have negative enthalpy values (-ΔH).

Endothermic reactions have positive enthalpy values (+ΔH).

An equation which shows both mass and heat relationships between products and reactants is called a thermochemical equation.

2 H_2(g) + O_2(g) ----> 2 H₂O_(l)  ΔH = -571.6 kJ

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

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

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

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

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

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

Example 1

Based upon the thermochemical equation given, calculate the heat associated with the decomposition of 1.15 g of  NO_2(g).

N_2(g) + 2 O_2(g) ----> 2 NO_2(g)      ΔH = +67.6 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)       ΔH = -176 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)    ΔH = +1170 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)   ΔH = -3120 kJ

 Hess’s Law

(See Moore , 2^nd ed., Section 6.9, pp.  246-247)

For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps.    In other words, no matter how you go from given reactants to products (whether in one step or several), the enthalpy change for the overall chemical change is the same.

The heat transferred in a given change is the same whether the change takes place in a single step or in several steps.

ΔH_rxn for a given reaction is the same whether that reaction takes place directly through one step or via several different reactions.

    S_(s) + O_{2 (g)} ---> SO_{2 (g)}   ΔH = -296 kJ

    SO_{2 (g)} + 1/2 O_{2 (g)} ---> SO_{3 (g)}   ΔH = -98.9 kJ
    ____________________________________________
    S_(s) + 1 1/2 O_{2 (g)} ---> SO_{3 (g)}    ΔH = -394.9 kJ

In order to use Hess's Law:

1) If a rxn is reversed, the sign of ΔH is reversed.

    S_(s) + O_{2 (g)} ---> SO_{2 (g)}  ΔH = -296 kJ

    SO_{2 (g)} ---> S_(s)+ O_{2 (g)}  ΔH = +296 kJ

2) If the coefficients in a balanced equation are multiplied by some number, the value of ΔH must be multiplied by that same number.

    S_(s) + O_{2 (g)} ---> SO_{2 (g)} ΔH = -296 kJ

    2 S_(s) + 2 O_{2 (g)} ---> 2 SO_{2 (g)}   ΔH = (2)(-296 kJ) = -592 kJ

Example 5
Calculate ΔH for the vaporization of water from liquid, using the thermochemical equations given.

H₂O_(l) ---> H₂O_(g)  ΔH = ?

    a) H_{2 (g)} + 1/2 O_{2 (g)}  --->H₂O_(l) ΔH = -286 kJ

    b) H_{2 (g)} + 1/2 O_{2 (g)}  ---> H₂O_(g)  ΔH = -242 kJ
 

Example 6
Calculate ΔH for the following reaction using the thermochemical equations given.

2 C_(s) + H_{2 (g)} ---> C₂H_{2 (g)}  ΔH = ?

    a) C_(s) + O_{2 (g)}  ---> CO_{2 (g)}  ΔH = -393.5 kJ

    b) H_{2 (g)} + 1/2 O_{2 (g)}  ---> H₂O_(l)  ΔH = -285.8 k J

    c) 2 C₂H_{2 (g)} + 5 O_{2 (g)}  ---> 4 CO_{2 (g)} + 2 H₂O_(l)  ΔH = -2598.8 kJ