Question

State and explain Hess’s law of constant heat summation? Give its application?

Answer 1

Hint: Energy nor be created or destroyed but it can transfer from one form to another is the law of conservation of energy also known as the first law of thermodynamics. Hess's law is also expressed in the first law of thermodynamics.

Complete answer:
Hess law of constant heat summation states that the total enthalpy change during a reaction is the same whether the reaction takes place in one step or in several steps.
The enthalpy change in a chemical or physical process is the same whether the process is carried out in one step or in several steps.
For example:
Carbon in a solid-state is heated with oxygen give carbon-dioxide here the enthalpy change is -94Kcal

$C(s) + {O_2}(g) \to C{O_2}(g):\Delta H = - 94Kcal$

If the same reaction takes place in several steps, in the first step carbon is converted to carbon monoxide and further, it is further oxidized to carbon dioxide. The enthalpy change takes place in the first step is -26.4k cal and in the second step it is -67.6k cal.
$
C(s) + \dfrac{1}{2}{O_2}(g) \to CO(g);\Delta {H_1} = - 26.4Kcal \\
CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g);\Delta {H_2} = - 67.6Kcal \\
\Delta H = \Delta {H_1} + \Delta {H_2} \\
- 94 = - 26.4 - 67.6 \\
- 94 = - 94 \\
$
Now according to Hess law the enthalpy change is the same whether the reaction takes place in one step or several steps.
Application: Hess law is useful to calculate heats of many reactions which do not take place directly. It is useful to find out heat from extremely slow reactions. It is useful to find out the Heat of reaction, Heat of transition, Lattice energy of the crystal, Heat of formation of intermediate compounds which are unstable.

Note: Heat of formation of intermediate compounds can be calculate, if the enthalpy change for direct step and the final step are known then the intermediate step enthalpy change can be calculated by;
$
C(s) + {O_2}(g) \to C{O_2}(g):\Delta H = - 94Kcal \\
C(s) + \dfrac{1}{2}{O_2}(g) \to CO(g);\Delta {H_1} = ? \\
CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g);\Delta {H_2} = - 67.6Kcal \\
\Delta H = \Delta {H_1} + \Delta {H_2} \\
- 94 = \Delta {H_1} + ( - 67.6) \\
\Delta {H_1} = ( - 67.6) - 94 = - 26.4Kcal \\
$