Question

State Hess’s law. Give its mathematical form.

Answer 1

Hint: If the reaction is occurring in several steps then the overall enthalpy of that reaction is always equal to the sum of the enthalpies of the intermediate reactions and there is no enthalpy change whether the reaction takes place in one step or several steps. Now answer it.

Complete step by step solution:
Hess’s law states that when the reaction takes place in several steps then its standard enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
In mathematical form;
Consider , the enthalpy of an overall reaction i.e. A to B along one path is $\Delta H $and $\Delta {{H}_{1}}$, $\Delta {{H}_{2}}$, $\Delta {{H}_{3}}$ and $\Delta {{H}_{4}}$ representing enthalpies of reactions leading to the same product B, along another path, then;
$\Delta H=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}+\Delta {{H}_{4}}$
Now, let’s discuss it with the help of an example as;
Consider the enthalpy change for the following :
$C(graphite, s)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$---------(1)
In this, carbon monoxide is the major product but some carbon dioxide gas is also produced in this reaction and we cannot measure the enthalpy change for the above reaction directly. But by finding the other reactions involving related species , we can find its enthalpy.
The rest two equations are as;
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(2)
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=-z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(3)
We can combine these two reactions i.e. equation (1) and (2) , in such a way so as to get the equation(1). To get 1 mole of carbon monoxide on the right side, reverse the equation (3) and heat is absorbed instead of being released, so we change the sign of ${{\Delta }_{r}}{{H}^{\circ }}$ as ${{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{e}^{-1}}$.
So, the equation becomes:
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(4)
On adding, equations (2) and (4) , we get equation(1) as;
$C(graphite, s)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}(y)=x+ z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$
So, this example clarifies and explains Hess's law of constant heat summation.
Note: The enthalpy of formation of elements which are present in their molecular forms like oxygen gas, or in any solid form etc. their standard enthalpy of formation is always taken as zero as they undergo no change in their formation.