Hess Law – Aqdix

by Snir Mac
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This law was enunciated and formulated a Russian-Swiss chemist Germain Henri Hess. This law is commonly known as Hess Law of constant heat summation which was a basic principle in early thermodynamics during the 19th century. It states that “The amount of energy evolved or absorbed in a chemical reaction is the same whether the reaction takes place in a single or several steps”

This statement means that the net heat of chemical reaction depends only upon the intial and final state and not on intermediate steps. The total heat absorbed or evolved in two or more chemical reactions may be used to calculate net heat of intermediate chemical reaction. You will fully understand its background and applications starting below.

The law is diagrammatically illustrated below:

Hess’s Law illustration

Let there be a substance A which changes to B in two separate ways:

A changes directly to B where ‘Q’ is the net amount of head absorbed in this process.
A changes to B indirectly in four steps are represented diagrammatically in the figure given below:

According to Hess Law, the total heat absorbed ‘Q’ is the sum of individual heats ‘qn involved in four steps for this process.

Hence, Q = q1 + q2 + q3 + q4 —> (1)

Let’s take a quick comprehension;

Q1) Sodium Hydroxide reacts directly and indirectly with Carbon Dioxide forming Sodium Carbonate along with the release of Water. Verify the net heat absorbed or evolved from the two process is same.

Ans: Single-step process
2NaOH (aq) + CO2 (g) —> Na2CO3 (aq) + H2O (l) ; ΔH = -89.08 kJ
Two-step process
NaOH(aq) + CO2 (g) —> NaHCO3 (aq) ; ΔH = -48.06 kJ
NaHCO3 (s) + NaOH (aq) —> Na2CO3 + H2O (l) ; ΔH = -41.02 kJ

Solution: We are asked to verify the net heat sum of intermediate steps with the net heat sum of initial and final steps. These two steps net heat sum must be equal.

According to Hess Law; ΔH = ΔH1 + ΔH2 —> (1)

Putting the given values in (1), we get:
-89.08 = -48.06 + (-41.02)
-89.08 = -89.08
L.H.S = R.H.S
Hence Verified

Applications

Every chemical reaction heat cannot be calculated directly. For which, we will be using Hess Law to calculate the net heat of every step or the individual heat enthalpy associated with each step that are involved in a chemical reaction. This can be well understood by the following example.

Q2) Find out the net heat when Carbon is reacted with half mole of Oxygen forming Carbon Monoxide with the help of two chemical equations given below.
C (s) + O2 (g) —> CO2 (g) ; ΔH1 = -394.5 kJ mol-1 —> (1)
CO (s) + ½O2 (g) —> CO2 (g) ; ΔH2 = -285.7 kJ mol-1 —> (2)

Solution: We need to determine the net heat or enthalpy change of the following chemical reaction;
C (s) + ½O2 (g) —> CO (g) ΔH = ? kJ mol-1
Subtracting equation (2) from equation (1) and similarly their respective net heat energies, we get:
C (s) + ½O2 (g) —> CO (g) ΔH = -108.8 kJ mol-1

Therefore, the net heat energy evolved for the required chemical reaction is 108.8 kJ mol-1.

Born-Haber Cycle

The Born-Haber Cycle is based on the principle that the “Sum of energy change which occurs in a closed cycle, from the intial to final states and vice versa, is equal to zero”. The principle is involves ‘Law of Conservation of Energy’ in accordance with the ‘First Law of Thermodynamics’. A chemical reaction begins and ends eventually releasing the net ‘Heat of Formation’ (ΔH0f). The ΔH0f develops the following formula; ΔH0f = ΔH0sub + ΔH0IE + ΔH0diss + ΔH0latt + ΔH0EA

Applications

Using this cycle, we can easily calculate intermediate stages of heat absorbed or evolved of a particular chemical reaction. For instance, we need to calculate the ‘Heat of Sublimation’ (ΔH0sub), ‘Electron Affinity’ (ΔH0EA), ‘Lattice Energy’ (ΔH0latt), etc. For this, you will be provided with relevant necessary information in order to answer a question.


Q1) Calculate the ‘Lattice Energy’ of KBr for K (s) + 1/2 Br2 (g) —> KBr (s). The ‘Heat of Sublimation’ of (K) is 88 kJ mol-1, the ‘Heat of Dissociation’ of (Br) gas is 192.5 kJ mol-1, the ‘Ionization Energy’ of (K) is 414 kJ mol-1, the ‘Electron Affinity’ of (Br) is -334.7 kJ mol-1 and the ‘Heat of Formation’ of KBr is -405.8 kJ mol-1. (KBr – Potassium Bromide, K – Potassium, Br – Bromine).

Diagrammatic illustration of various enthalpies

Solution: Given: ΔH0sub = 88 kJ mol-1
ΔH0IE = 414 kJ mol-1
ΔH0EA = -334.7 kJ mol-1
ΔH0f = -405.8 kJ mol-1
ΔH0diss = 192.5/2 kJ mol-1 =
ΔH0latt = ?

Using the Born-Haber Cycle formula to calculate ‘Lattice Energy’, the following formula must be solved for ‘Lattice Energy’ (ΔH0latt):

ΔH0f = ΔH0sub + ΔH0IE + ΔH0diss + ΔH0latt + ΔH0EA
ΔH0latt = -(ΔH0sub + ΔH0IE + ΔH0diss + ΔH0EA) + ΔH0f —> (1)

Plugging in the given values into the equation (1), we get:

ΔH0latt = – (88 + 414 + 96.25 – 334.7 ) – 405.8
ΔH0latt = – (88 + 414 + 192.5 – 334.7 ) – 405.8
ΔH0latt = – (359.8) – 405.8
ΔH0latt = -359.8 – 405.8
ΔH0latt = -669.35 kJ mol-1

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