Kirchoff’s Equation

According to Kirchhoff's Law, the enthalpy of a reaction is temperature-dependent. Generally, with an increase in temperature, the enthalpy of any substance also increases, and as such, the enthalpy of both the products and the reactants is increased. If the change in enthalpy of products and reactants is not the same, it will alter the overall enthalpy of the reaction. The heat capacity at constant pressure is equal to the change in enthalpy with respect to the change in temperature. Now, if the heat capacities are independent of temperature, then the change in enthalpy will be some function of this difference in temperature and heat capacities.

Kirchhoff's Law

The variation of enthalpy of a reaction with changes in temperature. In general, the enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.

At constant pressure, the heat capacity is equal to the change in enthalpy divided by the change in temperature.

$c_p=\frac{\Delta H}{\Delta T}$...............1

Therefore, if the heat capacities do not vary with temperature then the change in enthalpy is a function of the difference in temperature and heat capacities. The amount that the enthalpy changes by is proportional to the product of temperature change and change in heat capacities of products and reactants. A weighted sum is used to calculate the change in heat capacity to incorporate the ratio of the molecules involved since all molecules have different heat capacities at different states.

$\Delta H_{T_f}=\Delta H_{T_i}+\int_{T_i}^{T_f} \Delta C_p d T$...........2

If the heat capacity is temperature-independent over the temperature range, then Equation 1 can be approximated as

$\Delta H_{T_f}=\Delta H_{T_i}+\Delta C_p\left(T_f-T_i\right)$................3

where

• $\Delta C_p$ is the change in heat capacity of the reaction and is defined as

$\Delta C_P=\sum_{f(\mathrm{P})} \mathrm{i}_{\mathrm{P}} \mathrm{C}_{\mathrm{p}}-\sum_{\mathrm{f}(\mathrm{R})} \mathrm{i}_{\mathrm{R}} \mathrm{C}_{\mathrm{P}}$

It has exact same treatment for Cp as we have for the calculation of change in enthalpy

• H_Ti and H_Tf are the enthalpy at the respective temperatures.

Equation 3 can only be applied to small temperature changes, (<100 K) because over a larger temperature change, the heat capacity is not constant. There are many biochemical applications because it allows us to predict enthalpy changes at other temperatures by using standard enthalpy data.

In case the variation of C_p with temperature is given then we can use equation 2 and integrate the function of Cp as f(T) with respect to T and obtain the result.

Recommended topic video on ( Kirchoff's Equation)

Some Solved Examples

Example 1: Calculate the change in enthalpy (in kJ/mol) at 358K for the given reaction

$\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{H}_2 \rightarrow 2 \mathrm{Fe}+3 \mathrm{H}_2 \mathrm{O}$

Given $\Delta H_{298}^{\circ}=-35$ KJ/mol

Substance $\mathrm{Fe}_2 \mathrm{O}_3 \mathrm{Fe} \mathrm{H}_2 \mathrm{OH}_2$ Cp (J/K.mol) 100 25 75 30

1)29.7

2)-29.9

3)40.6

4)-40.6

Solution

According to the equation,

$\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{H}_2 \rightarrow 2 \mathrm{Fe}+3 \mathrm{H}_2 \mathrm{O}$

So Change in specific heat capacity,

$\begin{aligned} & \Delta \mathrm{Cp}=\sum \mathrm{C}_{\text {pProducts }}-\sum \mathrm{C}_{\text {pereactants }} \\ & \Delta \mathrm{Cp}=(2 \times 25+3 \times 75)-(100+3 \times 30)=85 \mathrm{~J} / \mathrm{K} . \mathrm{mol}\end{aligned}$

Also, according to Kirchoff’s equation,

$\Delta \mathrm{Cp}=\frac{\Delta \mathrm{H}_{\mathrm{T}_2}-\Delta \mathrm{H}_{\mathrm{T}_1}}{\mathrm{~T}_2-\mathrm{T}_1}$

$85 \times 10^{-3}=\frac{\Delta \mathrm{H}_{358}-(-35)}{358-298}$

$\Delta \mathrm{H}_{358 \mathrm{~K}}=-29.9 \mathrm{~kJ} / \mathrm{mol}$

Hence, the answer is (-29.9kJ/mol).

Example 2: Enthalpy of sublimation of iodine is $24 \mathrm{calg}^{-1}$ at $200^{\circ} \mathrm{C}$. If the specific heat of $I_2(s)$ and $I_2(vap)$ are 0.055 and $0.031 \mathrm{calg}^{-1} \mathrm{~K}^{-1}$ respectively, then enthalpy of sublimation of iodine at $250^{\circ} \mathrm{C}$ in $\operatorname{cal~}^{-1}$ is :
1)2.85

2) 22.8

3)5.7

4)11.4

Solution

Enthalpy of Sublimation -

Amount of enthalpy change to sublimise 1 mole solid into 1-mole vapor at a temperature below its melting point

$I_2(\mathrm{~s}) \rightarrow I_2(\mathrm{~g}), \Delta H_{200^{\circ}}=24 \mathrm{cal} / \mathrm{g}$

Specific heat of $I_2(s)$ & $I_2(vap)$ are respectively, $0.055 \mathrm{calg}^{-1} k^{-1}$ & $0.031 \mathrm{calg}^{-1} k^{-1}$

$\therefore \triangle S=$ Heat capacity of Product - Reactant

$\therefore \Delta S=0.031-0.055$

$=-0.024$

Applying Kirchoff's equation, we get:

$\begin{aligned} \Delta H_2 & =\Delta H_1+\Delta S\left(T_2-T_1\right) \\ \Delta H_2 & =24+(-0.024) \times 50 \\ & =24-1.2=22.8\end{aligned}$

$\therefore$ Enthalpy of sublimation of iodine at

$250^{\circ} \mathrm{C}=22.8 \mathrm{cal} / \mathrm{g}$

Hence, the answer is (22.8 cal/g).

Example 3: The standard heat of formation for gaseous $\mathrm{NH}_3$ is -11 kcal mol^-1 at 298 K. Given that at 298 K, the constant pressure heat capacities of gaseous $N_2$, $H_2$ and $\mathrm{NH}_3$ are 7 , 6 and 8 cal/mol respectively. Find change in enthalpy (in Kcal/mol) at 558K:

1)12.17

2)-9.83

3) -12.17

4)9.83

Solution

The heat of the formation of NH₃ is represented according to the equation:

$\left.\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g})\right) \rightarrow \mathrm{NH}_3(\mathrm{~g})$

$\Delta \mathrm{H}_{298 \mathrm{~K}}^{\circ}=-11.0 \mathrm{kcal} \mathrm{mol}^{-}$

Now, we know from Kirchoff's Equation

$\frac{\Delta H_2-\Delta H_1}{T_2-T_1}=\Delta C_P$

where $\Delta C_P=\sum_{f(P)} i_P C_p-\sum_{f(R)} i_R C_p$

Putting the given values in the equation, we have

$\frac{\Delta \mathrm{H}_2-(-11.1)}{558-298}=\left(8.0-\frac{1}{2} \times 7.0-\frac{3}{2} \times 6.0\right) \times 10^{-3}$

$\Delta \mathrm{H}_2=-12.17 \mathrm{kcal} \mathrm{mol}^{-1}$

Hence, the answer is the option(3).

Example 4: The heat capacity at constant pressure of a gas varies with temperature as Cp = 9+10T J/(mole-K). What will be the enthalpy of formation (in kJ/mol) when the temperature rises to 350 K from 300 K? Given that the enthalpy of formation at 300K is 45 kJ/mol?

1) 208

2)20

3)220

4)45

Solution

According to Kirchoff’s equation,

$\Delta H_{T_2}=\Delta H_{T_1}+\int_{T_1}^{T_2} C p d T$

We have given,

$\begin{aligned} & \Delta \mathrm{H}_{\mathrm{T}_1}=45 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{T}_1=300 \mathrm{~K} \\ & \mathrm{~T}_2=350 \mathrm{~K}\end{aligned}$

$\mathrm{C}_{\mathrm{P}}=(9+10 \mathrm{~T}) \frac{\mathrm{J}}{\mathrm{mol}-\mathrm{K}}$

Putting all these values in the equation we get,

$\begin{aligned} & \Delta \mathrm{H}=45 \times 10^3+\int_{300}^{350}(9+10 \mathrm{~T}) \mathrm{dT} \\ & \Rightarrow \Delta \mathrm{H}=45 \times 10^3+\left[9 \mathrm{~T}+5 \mathrm{~T}^2\right]_{300}^{350} \\ & \Rightarrow \Delta \mathrm{H}=45 \times 10^3+\left[9(350-300)+5 \times\left(350^2-300^2\right)\right] \\ & \Rightarrow \Delta \mathrm{H}=45 \times 10^3+[450+162500] \\ & \Rightarrow \Delta \mathrm{H}=208 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Hence, the answer is ($\Delta H=208 \mathrm{~kJ} / \mathrm{mol}$).

Summary

It finds extensive application in thermodynamics and physical chemistry for understanding temperature effects on reactions, optimizing reaction conditions, and performing energy balance calculations. By providing insights into the temperature sensitivity of chemical reactions, Kirchhoff's equation helps in designing efficient and safe chemical processes, making it an indispensable tool in both academic and industrial settings. By knowing the heat capacities, one can calculate the enthalpy change at different temperatures, aiding in the analysis and design of chemical reactions and processes.