Hess's Law Of Constant Heat Summation

Hess’s Law- The law for constant heat summation was derived in the year 1840, a Swiss-born Russian chemist and physician, Germain Hess, derived a relationship in thermochemistry for calculating the standard reaction enthalpy for multistep reactions. It basically exploits the properties of state functions, that is the value of state functions do not depend on the path taken for formation or dissociation rather it depends only on state at the moment (temperature, pressure, volume, etc). As we know enthalpy is a state function, hence, it is independent of the path taken to reach the final state starting from the initial state. According to Hess’s law, “for a multistep reaction the standard reaction enthalpy is independent of the pathway or number of steps taken, rather it is the sum of standard enthalpies of intermediate reactions involved at the same temperature”.

Hess's law

Application of Hess Law:

Let us look at some practical areas where Hess’s law is applied, for example, the formation of Sulphur Trioxide gas from Sulphur is a multistep reaction involving the formation of Sulphur Dioxide gas. Let us find out the standard reaction enthalpy for the formation of Sulphur Trioxide gas from Sulphur.

Step 1: Formation of Sulphur Dioxide gas

\( S + O_2 \rightarrow SO_2  ~~~~~~~~~~~~~~~~~~~\triangle H_1 = -70.96KCal/mol\)

Step 2: Conversion of Sulphur Dioxide gas to Sulphur Trioxide gas

\(SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ~~~~~~~~~~~~ \triangle H_2 \)= \(- 23.49 KCal/mol\)

Standard reaction enthalpy as per Hess’s Law:

\(\triangle H_R \)=\( \triangle H_2 + \triangle H_1\) = \((-70.96) + (-23.49)\) = \(-94.95 KCal/mol\)

Net reaction:

\(S + \frac{3}{2} O_2 \rightarrow SO_3 ~~~~~~~~~~~~~~~ \triangle H_R \)=\( -94.95 KCal/mol\)

Hence, in simple words we can state:

\(\triangle H_R \)= \(\triangle H_2 + \triangle H_1 + \triangle H_3 + \triangle H_4 +…………….\)

Determination of free energy and entropy:

As with enthalpy, Hess’s Law can be used to determine other state functions such as free energy and entropy, for example: Bordwell thermodynamic cycle which takes advantage of easily measured equilibriums and Redox potentials to determine experimentally inaccessible Gibbs free energy values.

\(\triangle G_{(reaction)}\) = \(\sum \triangle G_{(product)} – \sum \triangle G_{(reactants)}\)

As entropy can be measured as an absolute value, hence in case of entropy there is no need to use entropy of formation.

\(\triangle S_{(reaction)}\) = \(\sum S_{(products)} – \sum S_{(reactants)}\)

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Practise This Question

Equal volumes of 1M HCl& 1M H2SO4 are neutralized by 1M NaOH solution and x & y KJ/equivalent of heat are liberated respectively. Which of the following relations is correct?