


Hess’s Law
The total enthalpy of a reaction is independent of the
reaction pathway. This means that if a reaction is carried out in
a series of steps, the enthalpy change (DH) for
the overall reaction will be equal to the sum of the enthalpy changes
for the individual steps. This idea is also known as Hess’s
law. Here are some rules for using Hess’s law in solving
problems:
 Make sure to rearrange the given equations so that reactants and products are on the appropriate sides of the arrows.
 If you reverse equations, you must also reverse the sign of DH.
 If you multiply equations to obtain a correct coefficient, you must also multiply the DH by this coefficient.
Finally, in doing Hess’s law problems, it’s often helpful
to begin by working backward from the answer that you want. In other
words—write the final equation first. Try it out.
Example
Given the following equations
H_{3}BO_{3(aq)}HBO_{2(aq)} + H_{2}O_{(l)} DH_{rxn} =
0.02 kJ
H_{2}B_{4}O_{7(aq)} +
H_{ 2}O_{(l)}4HBO_{2(aq) }DH_{rxn} =
11.3 kJ
H_{2}B_{4}O_{7(aq)}2B_{2}O_{3(s)} +
H_{2}O_{(l)} DH_{rxn} =
17.5 kJ
find the DH for this overall reaction:
2H_{3}BO_{3(aq)}B_{2}O_{3(s)} +
3H_{2}O_{(l)}
Explanation
Multiply the first equation by 4:
4H_{3}BO_{3(aq)}4HBO_{2(aq)} + 4H_{2}O_{(l)} DH_{rxn} =
4(0.02 kJ) = 0.08
Reverse the second equation:
4HBO_{2(aq)}H_{2}B_{4}O_{7(aq)} +
H_{2}O_{(l)} DH_{rxn} =
+11.3 kJ
Leave the last equation as is:
H_{2}B_{4}O_{7(aq)}2B_{2}O_{3(s)} +
H_{2}O_{(l)} DH_{rxn} =
17.5 kJ
Cross out common terms and you are left with:
4H_{3}BO_{3(aq)}2B_{2}O_{3(s)} +
6H_{2}O_{(l)} DH_{rxn} =
28.8 kJ
Divide the above equation and the enthalpy by 2 and you
see that the answer is 14.4 kJ (the reaction is endothermic).
Bond Energies
As we mentioned earlier, another way of calculating the
enthalpy change in a chemical reaction is by using bond energies.
You are probably aware that energy must be added or absorbed to
break bonds and that energy is released when bonds are formed. Therefore, you
can calculate the total enthalpy of the reaction using the following
formula:
DH =
bonds broken  bonds formed
Now try this on a problem.
Example
Using bond energies, calculate the change in energy that
accompanies the following reaction:
H_{2(g)} + F_{2(g)}2HF_{(g)}
Bond type  Bond energy 

H—H  432 kJ/mol 
F—F  154 kJ/mol 
H—F  565 kJ/mol 
Explanation
[1(432) + 1(154)]  [2(565)] = 544 kJ
The answer is 544 kJ.
If the problem you’re trying to solve involves more complex
molecules, be sure to draw out their structures to determine how
many bonds of each type you have.
