Lesson: Chapter - 9
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
H3BO3(aq) ? HBO2(aq) + H2O(l) DHrxn = -0.02 kJ
H2B4O7(aq) + H2O(l) ? 4HBO2(aq)
DHrxn = -11.3 kJ
H2B4O7(aq)?2B2O3(s) + H2O(l) DHrxn = 17.5 kJ
find the DH for this overall reaction:
2H3BO3(aq) ? B2O3(s) + 3H2O(l)
Explanation
Multiply the first equation by 4:
4H3BO3(aq) ? 4HBO2(aq) + 4H2O(l) DHrxn = 4(-0.02 kJ) = -0.08
Reverse the second equation:
4HBO2(aq) ? H2B4O7(aq)
+ H2O(l) DHrxn = +11.3 kJ
Leave the last equation as is:
H2B4O7(aq) ? 2B2O3(s)
+ H2O(l) DHrxn = 17.5 kJ
Cross out common terms and you are left with:
4H3BO3(aq) ? 2B2O3(s) + 6H2O(l) DHrxn = 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.
Next to display next topic in the chapter.
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