In 1840, Chemist G.H. Hess put forward an important law governing the enthalpies of reaction. According to Hess' law:
The enthalpy of a reaction depends upon nature and state reactants and products. It is independent of the path followed by the chemical reaction. The law is, in fact, merely a special case of the first law of thermodynamics (law of conservation of energy).
Proof: A theoretical proof of Hess law can be obtained from the following considerations.
Let a substance A be converted into B by a process involving a single step reaction which is accompanied by the evolution of Q kJ of heat. In the second process, let A be converted into B in two steps, i.e., A to C and then C to B. Let Q1and Q2 be the heat involved in the two-step respectively, so that Q1+Q2=Q'where Q' is the total heat produced in going from A to B. Suppose Q is greater than Q'.
Now if we go from A to B, in a single step, Q kJ of heat will be evolved while returning from B to A through C, Q'kJ of heat will be absorbed. In this way, Q-Q'kJ heat will be produced by transforming A to B in a single step and then B to A via C.
In other words, a large amount of heat can be produced without doing any external work by merely repeating the above cycle several times. However, this is against the law of conservation of energy. Hence, Q' must be equal to Q as required by Hess’ law.
Illustration 1) As a further illustration of the law, consider the formation of carbon dioxide. There are two ways in which CO2 can be formed.
(i) In the first step, by burning carbon in excess of oxygen.
C(s)+O2CO2(g); ΔrH= -393.5 k J
(ii) In the second step, by burning carbon in a limited supply of oxygen to form CO and then CO is converted to CO2.
C(s)+12O2(g)CO(g); ΔrH = -110.54 kJ
CO(g)+12O2(g)CO2(g); ΔrH = -393.5 kJ
On adding, we get
C(s)+O2(g)CO2(g); ΔrH = -393.5 kJ
Thus, in both cases, ΔrH is the same. This proves the law.
Illustration 2) Now consider the formation of an aqueous solution of ammonium chloride from NH3HCL and water. There are two ways in which the reaction can be brought about.
(i) NH3(g) + HCL(g) → NH4CL(s); ΔrH =175.73 kJ
NH4CL(s)+aq → NH4CL(aq); ΔrH = +16.32 kJ
NH3(g)+HCL(g)+aq → NH4CL(aq); ΔrH = -159.41kJ
(ii) NH3(g) +aq → NH3(aq); ΔrH = -35.15 kJ
HCL (g)+aq →HCL(aq); ΔrH = -72.38 kJ
NH3(aq)+HCL(aq) → NH4CL(aq); ΔrH = -51.36 kJ
On adding NH3(g) + HCL(g)+aq → NH4CL(aq) ΔrH = -158.99 kJ
Thus, in both cases, the net enthalpy change is nearly the same. The difference in 0.42kJ may be due to experimental error.
Example 1) Calculate the enthalpy of formation of methane from the following data:
i) C(s)+O2(g)CO2(g); ΔrHⒽ = -394 kJ
ii) H2(g)+12O2(g)H2O(l); ΔrHⒽ = -286 kJ
iii) CH4(g) +2O2(g)CO2(g) +2H2O(l); ΔrHⒽ = -890.0 kJ
Solution 1) The required Hess law equation is:
C(s)+2H2(g)CH4(g); ΔrHⒽ =?
In order to get the required Hess law equation from equation (i), (ii), and (iii), multiply equation (ii) by two and add to equation (i), i.e., 2 x equation (ii) + equation (i)
2H2(g)+O2(g)2H2O(l); ΔrHⒽ = -572 kJ
C(s)+O2(g)CO2(g); ΔrHⒽ = -394
(iv) C(s) + 2H2(g)+2O2(g)CO2(g) +2H2O(l) ΔrHⒽ = -966 kJ
Subtracting equation (iii) from (iv) we get,
C(s)+2H2(g)CH4(g) ΔrHⒽ = -76.0 kJ
Thus, enthalpy of formation of methane is -76.0 kJ
Practice more Hess law examples to have a clear underlying of the concept.
1) What is the Significance of Hess Law?
Answer: Each and every substance such as atoms or molecules possesses energy inside them. The internal energy that these substances possess depends on the nature of the force that exists in the substance and the temperature. If these substances have to undergo chemical reactions, some of the bonds that are connected to some atoms are broken and some bonds are made new. This break and make of the bonds involve energy.
So, in reactions, the product substances might have either less or more or the same energy than the reacting substances. Accordingly, the reactions may release heat to become exothermic or absorb heat to become endothermic. Having the knowledge of the energy changes in any reaction is useful for manipulating the reactants and the products in a chemical process to our necessity.
Heat energy changes of reactions that are measured at a constant volume which is known as internal energy change ΔE and the energy measured at constant pressure is known as enthalpy change ΔH.
2) What are the Hess law Examples or the Application of Hess Law?
Answer: One of the most useful applications of Hess law is the calculation of the value of rH for a reaction whose rH is unknown or cannot be measured. According to Hess’s law, we can add thermochemical equations to obtain some desired thermochemical equation and its rH.
Applying this concept, we can calculate the heat of formation, the heat of transition of allotropic modifications, the heat of hydration and heat of various reactions. There is yet much other application of Hess law.