Chem 122
Thermochemical Equations
Name _________________________
In chemistry, we write many chemical equations to show reactants forming products. Often, energy or heat
is involved.
If a change is endothermic, energy/heat would be a reactant since the energy/heat would be required for the
change to occur. If heat/energy is to be written as a term in the equation, it will appear on the left side of the
equation. The system is absorbing energy/heat from the surroundings, therefore, the system has a “ + “ change in
energy/heat.
If a change is exothermic, energy/heat would be a product because energy/heat would be produced along
with the chemical products. If heat/energy is to be written as a term in the equation, it will appear on the right side
of the equation. The system is releasing energy/heat to the surroundings, therefore, the system has a “ - “ change in
energy/heat.
There are two common methods of writing thermochemical equations:
1. Write the energy/heat numerical value, with unit, within the equation.
2. Write the energy/heat numerical value, with unit, as an enthalpy change (∆H) showing the value as either
positive or negative after the equation.
EX: A
1.
2.
EX: B
1.
2.
EX: C
1.
2.
129 KJ is involved in the endothermic reaction of sodium hydrogen carbonate producing
sodium carbonate, water and carbon dioxide.
NaHCO3 + 129 KJ  Na2CO3 + H2O + CO2
NaHCO3  Na2CO3 + H2O + CO2
∆H = + 129 KJ
65.2 KJ of heat is produced when calcium hydroxide is formed from calcium oxide and water.
CaO + H2O  Ca(OH)2 + 65.2 KJ
CaO + H2O  Ca(OH)2
∆H = - 65.2 KJ
Calcium chloride has an enthalpy change value of – 14 KJ when it ionizes into its ions in a solution.
CaCl2  Ca +2 + Cl - + 14 KJ
CaCl2  Ca +2 + Cl -
∆H = - 14 KJ
PRACTICE – write thermochemical equations (method 1 and 2) for each of the following….
1.
An exothermic change occurs when 23KJ of heat is released when lithium chloride is ionized into
lithium ions and chloride ions.
LiCl  Li+ + Cl- + 23KJ
LiCl  Li+ + Cl∆H = -23KJ
2.
56 KJ of heat is involved in an endothermic reaction between acetic acid and sodium carbonate to
produce sodium acetate and carbonic acid.
CH3COOH + Na2CO3 + 56KJ  NaCH3COO + H2CO3
CH3COOH + Na2CO3  NaCH3COO + H2CO3
∆H = +56KJ
3.
The ∆H = + 49 KJ when C6H6 is formed from its elements.
49KJ + C + H2  C6H6
C + H2  C6H6
4.
∆H = +49KJ
The combustion of CH4 has an exothermic change of 890 KJ of heat.
CH4 + O2  CO2 + H2O + 890KJ
CH4 + O2  CO2 + H2O
∆H = -890KJ
5.
The ionization of ammonium nitrate into its ions has a + 25.7 KJ energy change.
NH4NO3 + 25.7KJ  NH4+ + NO3NH4NO3  NH4+ + NO3∆H = +25.7KJ
6.
C6H6 is broken down into its elements with 49 KJ of heat released.
C6H6  49KJ + C + H2
C6H6  C + H2
∆H = -49KJ