Hess’s law

Hess’s law is named after Germain Hess, a Russian chemist, who published it in 1840. It is based on the principle of conservation of energy, and states that:

The total enthalpy change in a chemical reaction is sum of enthalpy changes of all steps from the reactants to the products regardless of the route taken.

For example, there are two possible chemical routes (indicated by red arrows in the diagram below) for the combustion of methane to form carbon dioxide and water:

On closer scrutiny, ΔH1o is composed of the following standard enthalpy changes of formation:

C\left ( graphite \right )+2H_2\left ( g \right )\rightarrow CH_4\left ( g \right )\; \; \; \; \; \; \left ( \Delta H_{f}^{o} \right )_{R=1}

2O_2\left ( g \right )\rightarrow 2O_2\left ( g \right )\; \; \; \; \; \; 2\times \left ( \Delta H_{f}^{o} \right )_{R=2}

Similarly, ΔH2o is composed of the following standard enthalpy changes of formation:

C\left ( graphite \right )+O_2\left ( g \right )\rightarrow CO_2\left ( g \right )\; \; \; \; \; \; \left ( \Delta H_{f}^{o} \right )_{P=1}

2H_2\left ( g \right )+O_2\left ( g \right )\rightarrow 2H_2O\left ( g \right )\; \; \; \; \; \; 2\times \left ( \Delta H_{f}^{o} \right )_{P=2}

The choice of the indices R and P will be apparent shortly. Hence,

\Delta H_1^{\: o}=\sum _Rv_R\left ( \Delta H_f^{\: o} \right )_R

\Delta H_2^{\: o}=\sum _Pv_P\left ( \Delta H_f^{\: o} \right )_P

where vR and vP are the stoichiometric coefficients of the products of the respective standard enthalpy change of formation reactions.

In general, for a reaction:

where R denotes the reactants and P denotes the products. Hess’ law states that:

\Delta H_r^{\: o}=-\Delta H_1^{\: o}+\Delta H_2^{\: o}

\Delta H_r^{\: o}=\sum_Pv_P\left ( \Delta H_f^{\: o} \right )_P-\sum_Rv_R\left ( \Delta H_f^{\: o} \right )_R\; \; \; \; \; \; \; \; 6

Eq 6 is a very useful formula for calculating standard enthalpy changes.



a) Calculate the standard enthalpy change of formation of CO2(g) given:

\Delta H_f^{\: o}\left [ C_2H_2(g) \right ]=226\: kJmol^{-1}

\Delta H_c^{\; o}\left [ C_2H_2(g) \right ]=-1300\; kJmol^{\: -1}

\Delta H_c^{\: o}\left [ H_2(g) \right ]=-286\: kJmol^{-1}

b) Deduce the relationship between ΔHsolo, ΔHhydo and ΔHlatto using MgCl2 as an example.



2C(graphite)+H_2(g)\rightarrow C_2H_2(g)\; \; \; \; \; \; \Delta H_f^{\: o}=226\: kJmol^{-1}

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)\; \; \; \; \; \; \Delta H_c^{\; o}=-1300\: kJmol^{\: -1}

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)\; \; \; \; \; \; \Delta H_f^{\: o}=-286\: kJmol^{-1}

Using eq6, the standard enthalpy change of C(graphite) + O2(g) → CO2(g) is:

\Delta H_f^{\: o}=\frac{226+(-1300)-(-286)}{2}=-394\: kJmol^{-1}


\Delta H_{sol}^{\: o}=\Delta H_{latt}^{\: o}+\sum _iv_i\left ( \Delta H_{hyd}^{\: o} \right )_i

where vi is the stoichiometric coefficient of the reactant of the respective standard enthalpy change of hydration reaction. If ΔHlatto is defined as the change in enthalpy when one mole of an ionic solid is formed from its gaseous ions that are initially infinitely apart under standard conditions, the relation becomes:

\Delta H_{sol}^{\: o}=-\Delta H_{latt}^{\: o}+\sum _iv_i\left ( \Delta H_{hyd}^{\: o} \right )_i



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