Animated Solution for Chemistry - Chemical Thermodynamics: Enthalpy of sublimation of iodine is $24{\text{ cal g}}^{-1}$ at $200^{\circ }\text{C}$. If specific heat of ${\text{I}}_2(s)$ and ${\text{I}}_2(\text{vap.})$ are $0.055$ and $0.031{\text{ cal g}}^{-1}{\text{ K}}^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^{\circ }\text{C}$ in ${\text{cal g}}^{-1}$ is

Imagine you are boiling water. You know it takes a certain amount of heat to turn liquid water into steam at $100^{\circ }\text{C}$. But what if you wanted to boil it at a higher temperature under pressure? Would it take the same amount of heat?

The answer is no. The enthalpy of a phase change, or any chemical reaction, depends on the temperature at which it occurs.

This is where Kirchhoff's Equation comes to our rescue. It beautifully connects the enthalpy of a process at one temperature to its enthalpy at another temperature.

Kirchhoff's Law

Kirchhoff's law states that the difference in enthalpy of a reaction at two different temperatures is equal to the change in heat capacity multiplied by the temperature difference.

Mathematically, it is expressed as:

Here, $\Delta C_p$ is the difference in the heat capacities of the products and the reactants at constant pressure.

Why does this work? It is a direct consequence of Hess's Law and the conservation of energy. You can either heat the reactants, perform the reaction at the higher temperature, or perform the reaction at the lower temperature and then heat the products. Both paths must require the exact same total energy!

In our specific problem, we are looking at the sublimation of iodine:

We are given the enthalpy of sublimation at $T_1=200^{\circ }\text{C}$, which is $\Delta H_1=24{\text{ cal g}}^{-1}$.

We need to find the enthalpy of sublimation at $T_2=250^{\circ }\text{C}$, which we will call $\Delta H_2$.

First, we must find $\Delta C_p$. This is the specific heat of the product (gas) minus the specific heat of the reactant (solid).

Substituting the given values:

Notice that $\Delta C_p$ is negative! This means the products require less heat to raise their temperature than the reactants do.

Now, let's determine our temperature change, $\Delta T$.

Remember, a difference of $50^{\circ }\text{C}$ is exactly the same as a difference of $50\text{ K}$.

Finally, we plug everything into Kirchhoff's equation:

And there we have it! The enthalpy of sublimation at $250^{\circ }\text{C}$ is $22.8{\text{ cal g}}^{-1}$. The negative $\Delta C_p$ caused the enthalpy of sublimation to decrease as the temperature increased.