THE Hess's law it was proposed in 1840 by the Swiss physician and chemist Germain Henri Hess. During his work on energy in the form of heat in neutralization reactions in acids and bases, he concluded that the sum of the energies in this type of reaction was always constant.
The Swiss scientist's studies led to the proposition of the following law:
“The variation of enthalpy involved in a chemical reaction, under certain experimental conditions, depends exclusively on the enthalpy initial and final products, whether the reaction is carried out directly in a single step or is carried out indirectly in a series of phases."
In general, the calculation of the ?H of a reaction is independent of the number of steps and the type of reaction and is done by the following expression:
?H = Hp-Hr
When we are unable to calculate the ?H of a specific chemical reaction, we can determine it by the sum of the ?Hs of the steps that make up this reaction:
?H = ?H1 + ?H2 + ?H3 + ...
An example is the determination of the energy involved in the transformation of graphite carbon into diamond carbon (C
(g) → C(d)). To determine the ?H of this process, we have the following steps at our disposal:Ç(g) + O2(g) → CO2(g) ?H = -94 Kcal
CO2(g) → C(d) + O2(g) ?H = +94.5 Kcal
As there are compounds that repeat themselves (CO2 it's the2) in both equations, but in different areas (reagents or products), they are eliminated. So, just add the ?Hs provided, since both the O2 how much the CO2 are on opposite sides of the equation:
?H = ?H1 + ?H2
?H = -94 + 94.5
?H = 0.5 Kcal
Fundamentals of Hess' Law
when do we have to calculate the enthalpy change of a reaction from its steps and its enthalpy variations, we have to keep in mind that the final reaction is who will dictate this calculation.
All steps provided are worked out in such a way that they are fully in agreement with the final reaction. For example, if we have a final reaction:
Overall reaction: X + Y → Z
And the exercise provides the following steps:
Step 1: X + D → W + E
Step 2: Z + D → F + E
Step 3: F → Y + W
It is clear that steps 2 and 3 are not obeying the final reaction, since in 2, A is in the reactant, and in 3, Y is in the product. In this case, these steps need “handling” to comply with the final or global reaction. Understand what this "treatment" is:
Possibilities of working with the steps of a reaction in Hess' Law
a) Invert the entire equation
An equation can be inverted (reagents become products, and products become reactants) in order to equalize the position of the participants. In this case, the value of ?H will have its sign inverted.
In the example below, it is evident that steps 2 and 3 must be reversed:
Overall reaction: X + Y → Z
Step 1: X + D → W + E
Step 2: Z + D → F + E
Step 3: F → Y + W
b) Multiply the equation
An equation can be multiplied by any numerical value in order to equalize the number of participants. In that case, the value of ?H must be multiplied.
In the example below, it is evident that step 2 must be multiplied by 2 to equal the number of participants B and C relative to the global equation.
Overall reaction: A + 2B → 2C
Step 1: A + 2D → 2Z
Step 2: Z + B → C + D
c) Split the entire equation
An equation can be divided by any numerical value in order to equalize the number of participants. In this case, the value of ?H must also be divided.
In the example below, it is evident that step 2 must be divided by 2 to equal the number of participants F and C in relation to the global equation.
Overall reaction: W + F → 2C
Step 1: W + 2D → 2Z
Step 2: 4Z + 2F → 4C + 4D
Hess' Law Application Example
Example: The complete combustion reaction (formation of carbon dioxide and water) of butane gas is given by the following equation:
Ç4H10(g) + 13/2O2(g) → 4CO2(g) + 5 hours2O(g)
Knowing that butane, C4H10, is the gas present in the greatest quantity in cooking gas (LPG), determine the value of its enthalpy, with reference to the following data for standard enthalpies of formation of each of its components:
Ç(s) + 5h2(g) → 1C4H10(g) ?H = -125 Kcal
Ç(s) + O2(g) → CO2(g) ?H = -394 Kcal
H2(g) + 1/2O2(g) → H2O(g) ?H = -242 Kcal
Resolution:
1O Step: Step 1 must be reversed because, according to the global equation, the substance must be reactant, not product. With this, the sign of the value of ?H is also inverted:
1C4H10(g) → 4C(s) + 5h2(g) ?H = + 125 Kcal
2O Step: Step 2 must be kept, but it will have to be multiplied by four, because, according to the global equation, it must have 4 mol of CO2. Thus, the value of ?H must be multiplied by 4 as well:
(4x) Ç(s) + O2(g) → CO2(g) ?H = -394 Kcal
soon:
4C(s) + 4 O2(g) → 4 CO2(g) ?H = -1576 Kcal
3O Step: Step 3 must be kept, but it will have to be multiplied by five, because, according to the global equation, it must have 5 mol of H2O. Thus, the value of ?H must be multiplied by 5 as well:
(5x) H2(g) + 1/2O2(g) → H2O(g) ?H = -242 Kcal
soon:
5 hours2(g) + 5/2O2(g) → 5h2O(g ?H = -1210 Kcal
4O Step: Perform deletions:
Step 1: 1C4H10(g) → 4C(s) + 5h2(g) ?H = + 125 Kcal
Step 2: 4C(s) + 4 O2(g) → 4 CO2(g) ?H = -1576 Kcal
Step 3: 5 hours2(g) + 5/2O2(g) → 5h2O(g ?H = -1210 Kcal
5 hours ago2 in the step 1 product and in the step 3 reagent, therefore, they are eliminated;
There is 4 C in the product from step 1 and the reagent from step 2, so they are eliminated.
Thus, the steps remain as follows:
Step 1: 1C4H10(g) → ?H = + 125 Kcal
Step 2: + 4 O2(g) → 4 CO2(g) ?H = -1576 Kcal
Step 3: + 5/2O2(g) → 5h2O(g ?H = -1210 Kcal
By adding the steps after the eliminations, we find that they are in line with the overall reaction.
Ç4H10(g) + 13/2O2(g) → 4CO2(g) + 5 hours2O(g)
5O Step: Add the values of ?hrs of the steps to determine the ?H of the global reaction.
?H = ?H1 + ?H2 + ?H3
?H = 125 + (-1576) + (-1210)
?H = 125 – 1576 – 1210
?H = 125 - 2786
?H = - 661 Kcal
By Me. Diogo Lopes Dias
Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/quimica/o-que-e-lei-hess.htm
