Hess’s Law of Constant Heat Summation: Hess’s law of constant heat summation explains how the heat change of a reaction or enthalpy change can be calculated when a reaction takes place in multiple steps. Enthalpy can be defined as the amount of heat evolved or absorbed in any chemical reaction when the complete number of moles of reactants has reacted. In any chemical reaction, the enthalpy change is a prominent factor. It is also known as the reaction heat or the enthalpy change of a chemical reaction. In this article, we will study about Hess’s Law of Constant Heat Summation in detail. Scroll down to find out.

Enthalpy Change and Phase Transitions

Enthalpy change is a very significant factor in any chemical reaction. Enthalpy can be defined as the amount of heat evolved or absorbed in any chemical reaction when the complete number of moles of reactants has reacted. It is also called the heat of reaction or enthalpy change of a chemical reaction. The enthalpy change is a common term used for all the heat exchange taking place in any reaction. The same, when applied to different reactions, assumes a different identity and is calculated depending upon the type of process that is involved in the phase change, such as fusion, sublimation, vaporization, and so on.

So, enthalpy for phase transitions is usually given as enthalpy of vaporization, enthalpy of fusion, etc. Similarly, depending upon the type of reaction involved: combustion, formation, neutralization, etc., the enthalpies are calculated as enthalpy of combustion, enthalpy of formation, enthalpy of ionization and so on. For any reaction, therefore, the enthalpy of the reactions is calculated according to the type and also according to the phase change taking place in the reaction.

However, there are aspects such as reactions happening in single or multiple steps wherein the problem of calculating enthalpy or heats of reaction can become difficult. Hess’s law of heat summation gives a comprehensive understanding of calculating the heat of reaction in such situations.

What is Hess’s Law?

A Russian Scientist, by the name of G.H.Hess, put forth a law in the year 1840 on heats of reactions based upon the experimental observations. The law is named after him as Hess’s law of constant heat summation.

The law states that:

‘The total amount of heat evolved or absorbed in a chemical reaction is the same, irrespective of whether the chemical reaction is taking place in one step or multiple steps.

Hence, according to this law, the total heat change of a reaction only depends upon the initial reactants and their nature and also the final products formed (either by a single step or multiple steps). The total heat change is independent of the path or the manner in which the final products are formed in the chemical reaction.

Hess’s law is based upon the fact that enthalpy is a state function, or in other words, the enthalpy change of a reaction depends only on the enthalpy of the initial reactants and the enthalpy of the final products formed in the reaction. The path of the reaction does not have any impact on the enthalpy change of the reaction.

Hess’s law can be easily understood with the help of some examples and theoretical explanations.

Theoretical Explanation of Hess’s Law

To explain Hess’s law theoretically, let us consider a chemical reaction:

\({\rm{W}} \to {\rm{Z}}\)

Let us assume the heat evolved in this reaction to be ‘A’ joules.

Now, if the reaction takes place in not one, but in three steps:

\({\rm{W}} \to {\rm{X}} \to {\rm{Y}} \to {\rm{Z}}\)

Then let us assume the heat evolved in the individual reactions as a₁, a₂ and a_3, respectively:

The total heat evolved would then be: \({{\rm{a}}_1}\; + {\rm{ }}{{\rm{a}}_2}\; + {\rm{ }}{{\rm{a}}_3}\; = {\rm{A’}}\)^’ joules. According to Hess’s law of constant heat summation, A = A^‘.

However, if Hess’s law is wrong, then either \({\rm{A}} > {\rm{A’}}\) or it would be \({\rm{A}} < {\rm{A’}}\).

Let us consider both possibilities:

If \({\rm{A}} < {\rm{A’}}\):

When we move from \({\rm{W}}\) to \({\rm{Z}}\) in a series of steps, the heat evolved will be more than the heat absorbed when we move in the reverse direction from \({\rm{Z}}\) to \({\rm{Z}}\) (if \({\rm{A}}’ > {\rm{A}}\)). In this case, when the cycle is completed, the \(\left( {{\rm{A}}–{\rm{A}}’} \right)\) joules of heat will be produced. By repeating the cycle multiple times, a large amount of heat can be created. However, energy can neither be created nor be destroyed according to the law of conservation of energy. Hence, the scenario of \({{\rm{A}}’}\) being greater than A cannot be simply possible.

Hence, it proves that \({{\rm{A}} = {\rm{A}}’}\)’, thereby proving that Hess’s law is correct.

Application of Hess’s Law

One significant application of Hess’s law is the calculation of heat changes for reactions where they cannot be calculated experimentally. The calculations are done using Hess’s law as follows:

The thermochemical equations, like algebraic equations, can be added, subtracted, divided or multiplied.

Using this consequence, Hess’s law can be applied to calculate the following:

i) Determination of Enthalpy of Formation

Since the enthalpies of the formation of several compounds cannot be determined experimentally, they can be calculated using Hess’s law. For example:

Enthalpy of formation of \({\rm{CO}}\) can be determined as follows:

Given:

a. \({\rm{C}}\left( {\rm{s}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);\,{\Delta _{\rm{r}}}{{\rm{H}}^0}\; = – 393.5\,{\rm{KJ}}/{\rm{mol}}\)
b. \({\rm{CO}}\left( {\rm{g}} \right) + 1/2{{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\;\left( {\rm{g}} \right);{\rm{ }}{\Delta _{\rm{r}}}{{\rm{H}}^0}\; = – 283.0\, {\rm{ KJ}}/{\rm{mol}}\)

The equation for which the enthalpy of formation needs to be found is:

c. \({\bf{C}}\left( {\bf{s}} \right) + {\bf{1}}/{\bf{2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{CO}}\left( {\bf{g}} \right)\)

This can be achieved by subtracting equation ‘b’ from equation ‘a’:

\({\rm{C}}\left( {\rm{s}} \right) + {\rm{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\rm{ }}{{\rm{O}}_2}\left( {\rm{g}} \right)–{\rm{CO}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right) – {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)\)

Or   \({\rm{C}}\left( {\rm{s}} \right) + {\rm{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\rm{ }}{{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{CO}}\left( {\rm{g}} \right)\)

\({\Delta _{\rm{r}}}{{\rm{H}}^0}\; = – 393.5 – \left( { – 283.0} \right) = – 110.5\,{\rm{KJ}}/{\rm{mol}}\)

Heat of formation of \({\rm{CO}}\), therefore, is \( = – 110{\rm{ KJ}}/{\rm{mol}}\).

ii) Determination of Enthalpy of allotropic transformation:

Some elements, such as Sulphur and carbon, exist in many allotropic forms. The change from one allotropic form to another usually involves a very limited amount of heat, and such processes are slow. Since experimental determination of heat change for such reactions is extremely difficult, Hess’s law can be used to predict them.

Example: The transformation of carbon from its allotropic form.

Graphite to Diamond is a slow process. Hess’s law of constant heat summation can be used to predict the heat change involved in the reaction.

Given:

a. \({\rm{C}}\left( {{\rm{graphite}}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right){\Delta _{\rm{c}}}{{\rm{H}}^0}\; = – 393.5{\rm{ KJ}}/{\rm{mol}}\)
b. \({\rm{C}}\left( {{\rm{diamond}}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right){\Delta _{\rm{c}}}{{\rm{H}}^0}\; = – 395.4{\rm{ KJ}}/{\rm{mol}}\)

Since we require the transformation from one allotropic form to another, the equation needed is:

c. \({\bf{C}}\left( {{\bf{graphite}}} \right) \to {\bf{C}}\left( {{\bf{diamond}}} \right){{\bf{\Delta }}_{{\bf{trans}}}}{{\bf{H}}^{\bf{0}}}\; = \)?

To obtain this, one need to subtract equation ‘b’ from equation ‘a’:

\({\rm{C}}\left( {{\rm{graphite}}} \right) – {\rm{C}}\left( {{\rm{diamond}}} \right) \to 0\)

or \({\rm{C}}\left( {{\rm{graphite}}} \right) \to {\rm{C}}\left( {{\rm{diamond}}} \right)\)

\({\Delta _{{\rm{trans}}}}{{\rm{H}}^0}\; = – 393.5–\left( { – 395.4} \right) = + {\rm{ 1}}{\rm{.9}}\,{\rm{KG}}\)

Similarly, using Hess’s law of constant heat summation, one can find enthalpies of such reactions, which are otherwise difficult to find experimentally. Enthalpies of any reaction can be predicted using the consequence given by Hess’s law wherein two or more equations can be subtracted, added, multiplied or divided until the required equation is obtained and, therefore, the required enthalpy for a particular reaction is calculated.

Hess’s law, therefore, finds use in several applications, especially to determine successfully the enthalpies of reactions for any reactions under study.

Summary

Enthalpy or heat change happening in a chemical reaction is a very significant entity and needs to be calculated experimentally or from available values. Hess’s law of heat summation provides a simple consequence by which enthalpies of all reactions, whether they take place in single or multiple steps, can be calculated easily. By using the consequence provided by the law, one can multiply, divide, subtract or add the given and relevant equations to obtain the enthalpy change of any chemical reaction. Hess’s law is applied in the calculation of the heat of reactions of those reactions, which are otherwise very hard to obtain experimentally.

FAQs

Q.1. How is the enthalpy of a reaction calculated according to Hess’s law?
Ans: Hess’s law of constant heat summation uses the method of obtaining a particular equation by multiplying, subtracting, adding or dividing two or more given equations. The heats of reactions of all the equations are also evaluated to calculate the enthalpy of the reactions.

Q.2. What is Hess’s law of constant heat summation? Explain with an example?
Ans: Hess’s law of constant heat summation states that the total amount of heat evolved or absorbed in a chemical reaction is the same, irrespective of whether the chemical reaction is taking place in one step or in multiple steps.

Q.3. What is the importance of Hess’s law of constant heat summation?
Ans: Hess’s law of heat summation and its consequence can be applied to any reaction to calculate the enthalpies of the reaction easily.

Q.4. What is the most important application of Hess’s law?
Ans: The most significant application of Hess’s law is to calculate the enthalpies of those reactions, such as the change from one allotropic form to another, hydration reactions, etc., for which experimental calculation may be very difficult. Using Hess’s law of constant heat of summation, one can easily calculate the enthalpies of such reactions.

Q.5. Give an example of an enthalpy calculation using Hess’s law?
Ans: Hess’s law can be used to calculate the enthalpy of the formation of methane.

For given equations:

a. \({\rm{C}}\left( {\rm{s}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)\),                                          \({\Delta _{\rm{r}}}{{\rm{H}}^0} = – 393.5{\rm{ KJ}}/{\rm{mol}}\)

b. \({{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{ + 1/2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{ }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)\),                                    \({\Delta _{\rm{r}}}{{\rm{H}}^0} = – 285.8{\rm{ KJ}}/{\rm{mol}}\)

c. \({\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ + 2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{\;}}\left( {\rm{g}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)\),                    \({\Delta _{\rm{r}}}{{\rm{H}}^0} = – 890.2{\rm{ KJ}}/{\rm{mol}}\)

To calculate the enthalpy of the formation, the reaction is:

\({\rm{C}}\left( {\rm{s}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right)\),                    \({\Delta _{\rm{f}}}{{\rm{H}}^0}\; = \) ?

We can multiply equation: \({\rm{‘b’}}\) with 2, and add it to equation \({\rm{‘a’}}\). Equation \({\rm{‘c’}}\) is then subtracted from the sum \(\left( {{\rm{2 }} \times {\rm{b + a}}} \right)\) to get:

\({\rm{C}}\left( {\rm{s}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right)\)

\({{\rm{\Delta }}_{\rm{f}}}{{\rm{H}}^{\rm{0}}}{\rm{ = – 393}}{\rm{.5 + 2}}\left( {{\rm{ – 285}}{\rm{.8}}} \right){\rm{–}}\left( {{\rm{ – 890}}{\rm{.3}}} \right){\rm{ = – 74}}{\rm{.8}}\,{\rm{KJ/mol}}\)

Enthalpy of formation of methane is \({\rm{ – 74}}{\rm{.8}}\,{\rm{KJ/mol}}\).

Study About Types Of Chemical Reactions Here

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