JAMB OJCA

➖CHEMISTRY
➖TOPIC: HESS’s LAW & ENTHALPY CHANGES
➖DEPARTMENT OF SCIENCE

UNDERSTANDING TO HESS’S LAW: Hess’s Law, also known as “Hess’s Law of Constant Heat Summation,” states that the total enthalpy of a chemical reaction is the sum of the enthalpy changes for the steps of the reaction.

Therefore, you can find enthalpy change by breaking a reaction into component steps that have known enthalpy values. This example problem demonstrates strategies for how to use Hess’s Law to find the enthalpy change of a reaction using enthalpy data from similar reactions.

Hess’s Law Enthalpy Change Problem
What is the value of ΔH for the following reaction?

▪️CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
Given:

▪️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 says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps.

To solve this type of problem, organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that you must follow when manipulating a reaction.

  1. The reaction can be reversed. This will change the sign of ΔHf.
  2. The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the same constant.
  3. Any combination of the first two rules may be used.

-Finding a correct path is different for each Hess’s Law problem and may require some trial and error. A good place to start is to find one of the reactants or products where there is only one mole in the reaction. You need one CO2, and the first reaction has one CO2 on the product side.

C(s) + O2(g) → CO2(g), ΔHf = -393.5 kJ/mol

This gives you the CO2 you need on the product side and one of the O2 moles you need on the reactant side. To get two more O2 moles, use the second equation and multiply it by two. Remember to multiply the ΔHf by two as well.

2 S(s) + 2 O2(g) → 2 SO2(g), ΔHf = 2(-326.8 kJ/mol)
-Now you have two extra S’s and one extra C molecule on the reactant side that you don’t need. The third reaction also has two S’s and one C on the reactant side. Reverse this reaction to bring the molecules to the product side. Remember to change the sign on ΔHf.

CS2(l) → C(s) + 2 S(s), ΔHf = -87.9 kJ/mol
When all three reactions are added, the extra two sulfur and one extra carbon atoms are canceled out, leaving the target reaction. All that remains is adding up the values of ΔHf.

ΔH = -393.5 kJ/mol + 2(-296.8 kJ/mol) + (-87.9 kJ/mol)
ΔH = -393.5 kJ/mol – 593.6 kJ/mol – 87.9 kJ/mol
ΔH = -1075.0 kJ/mol

I.e the Answer is: The change in enthalpy for the reaction is -1075.0 kJ/mol.

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PART II: https://youtu.be/7iQ9hCX3hzU


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