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 9.1 Enthalpy 9.2 Spontaneous Reactions 9.3 Heat Capacity and Specific Heat 9.4 Enthalpies of Reactions 9.5 Hess’s Law

 9.6 More About Entropy 9.7 Gibb’s Free Energy 9.8 Practice Questions 9.9 Explanations
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:
1. Make sure to rearrange the given equations so that reactants and products are on the appropriate sides of the arrows.
2. If you reverse equations, you must also reverse the sign of DH.
3. 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) + H 2O(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:
H2(g) + F2(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