Hess's Law is one of the topic that we 12-STEM encountered for the last semester of our Chemistry Class. Discussing it was a bit confusing since it requires a lot of practices in order to solve each problem and one must really understood the concept for him to arrive at the right answer. Similarly, our class used up a lot of time just for the whole to fully understand and be able to solve the problem solving without the help of our teacher. We would admit that at first we really found it very difficult but with the help of our classmates, subject teacher and even from other sources like Khan academy; soon we were able to overcome those struggles and be familiar to the process of solution.
This blog will focus on helping the upcoming Grade 12 or for those who wanted to understand the Concept and solution of Hess's Law since this will give you some overview of the solutions and guidelines on how to easily solved a problem solution of Hess's Law.
But first, there are a couple notes to keep straight before beginning. 1. If a reaction is reversed, the sign of the change in enthalpy (ΔHf) changes. For example: the reaction C(s) + O2(g) → CO2(g) has an ΔHf of -393.5 kJ/mol. The reverse reaction CO2(g) → C(s) + O2(g) has a ΔHf of +393.5 kJ/mol.
2. If a reaction is multiplied by a constant, the change in enthalpy is changed by the same constant. Example, for the previous reaction, if three times the reactants are allowed to react, ΔHf is changed by three times.
3. If ΔHf is positive, the reaction is endothermic. If ΔHf is negative, the reaction is exothermic.
EXAMPLE:
Find the enthalpy change for the reaction
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
when:
C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol
SOLUTION:
Hess’s Law problems can take a little trial and error to get started. One of the best places to begin is with a reaction with only one mole of reactant or product in the reaction. Our reaction needs one CO2 in the product and the first reaction also has one CO2 product. C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ/mol This reaction gives us the CO2 needed on the product side and one of the O2 needed on the reactant side. The other two O2 can be found in the second reaction. S(s) + O2(g) → SO2(g) ΔHf = -296.8 kJ/mol Since only one O2 is in the reaction, multiply the reaction by two to get the second O2. This doubles the ΔHf value. 2 S(s) + 2 O2(g) → 2 SO2(g) ΔHf = -593.6 kJ/mol Combining these equations gives 2 S(s) + C(s) + 3 O2(g) → CO2(g) + SO2(g) The enthalpy change is the sum of the two reactions: ΔHf = -393.5 kJ/mol + -593.6 kJ/mol = -987.1 kJ/mol This equation has the product side needed in the problem but contains an extra two S and one C atom on the reactant side. Fortunately, the third equation has the same atoms. If the reaction is reversed, these atoms are on the product side. When the reaction is reversed, the sign of the change in enthalpy is reversed. CS2(l) → C(s) + 2 S(s); ΔHf = -87.9 kJ/mol Add these two reactions together and the extra S and C atoms cancel out. The remaining reaction is the reaction needed in the question. Since the reactions were added together, their ΔHf values are added together. ΔHf = -987.1 kJ/mol + -87.9 kJ/mol ΔHf = -1075 kJ/mol Answer: The change in enthalpy for the reaction CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) is ΔHf = -1075 kJ/mol. Hess’s law problems require reassembling the component reactions until the needed reaction is achieved. While Hess’s law applies to changes in enthalpy, this law can be used for other thermodynamic state equations such as Gibbs energy and entropy.
Hoping that through this blog you are able to understand the problem solving in relation to Hess’s Law. This will be a very big contribution of us If you can share your experience upon reading this tips and guidelines by commenting on this blog and by giving this a heart.

















