Question

State Hess's law of constant heat summation and explain it.

Answer 1

Hess Law

1. The enthalpy change in a chemical or physical process is the same whether the process is carried out in several steps.
2. Hess's law is a direct consequence of the fact that enthalpy is a state function and so enthalpy change depends only upon the initial and final states of the system and not on the path by which the reaction takes place.
3. Suppose the conversion of A to C can take place directly $\mathrm{A}\to \mathrm{C}$ then according to Hess's law $∆{\mathrm{H}}^{°}=∆{\mathrm{H}}_1$
4. The reaction can take place in two steps for which $∆{\mathrm{H}}^{°}$values are known. Step 1: $\mathrm{A}\to \mathrm{B}∆{\mathrm{H}}^{°}=∆{\mathrm{H}}_2,\mathrm{B}\to \mathrm{C}∆{\mathrm{H}}^{°}=∆{\mathrm{H}}_3$
5. According to Hess's law: $∆{\mathrm{H}}^{°}=∆{\mathrm{H}}_2+∆{\mathrm{H}}_3$
6. Example: There are two ways in which Carbon dioxide (${\mathrm{CO}}_2$) can be formed.
7. In the first one burning the carbon in excess of oxygen. $\underset{\mathrm{Carbon}}{\mathrm{C}\left(\mathrm{s}\right)}+\underset{\mathrm{Dioxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}$given ${∆}_{\mathrm{r}}\mathrm{H}=-393.5\mathrm{kJ}$
8. In second one by burning carbon in a limited supply of oxygen to form $\mathrm{CO}$ and then $\mathrm{CO}$ is converted into ${\mathrm{CO}}_2$.
9. $\underset{\mathrm{Carbon}}{\mathrm{C}\left(\mathrm{s}\right)}+\underset{\mathrm{dioxygen}}{12O_2\left(g\right)}\to \underset{\mathrm{Carbon}\mathrm{Monoxide}}{\mathrm{CO}\left(\mathrm{g}\right)}$given ${∆}_{\mathrm{r}}\mathrm{H}=-110.54\mathrm{KJ}$
10. $\underset{\mathrm{Carbon}\mathrm{Monoxide}}{\mathrm{CO}\left(\mathrm{g}\right)}+\underset{\mathrm{Dioxygen}}{12O_2\left(g\right)}\to \underset{\mathrm{Carbon}\mathrm{Dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}$ given ${∆}_{\mathrm{r}}\mathrm{H}=-283.5\mathrm{KJ}$
11. On Adding the above two equations we get $\underset{\mathrm{Carbon}}{\mathrm{C}\left(\mathrm{s}\right)}+\underset{\mathrm{Dioxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}$ ${∆}_{\mathrm{r}}\mathrm{H}=-393.5\mathrm{kJ}$
12. Thus in both the cases $∆{\mathrm{H}}^{°}$ is same, This proves the law.

Therefore, Hess's law states that enthalpy is constant for several steps.