Question

Variation of heat of reaction with temperature is given by Kirchhoff’s equation. Which of the following is not the equation for Kirchhoff’s?
(A) $\Delta {H_2} = \Delta {H_1} + \Delta {C_V}\left( {{T_2} - {T_1}} \right)$
(B) $dH = TdS + VdP$
(C) \[\dfrac{{\Delta {H_2} - \Delta {H_1}}}{{\Delta T}} = \Delta {C_P}\]
(D) $\dfrac{{d\left( {\Delta H} \right)}}{{dT}} = \Delta {C_P}$

Answer 1

Hint:The heat change involved in a process depends on the temperature at which the process is taking place. Mathematically, this dependency is given by Kirchhoff’s equation. The thermal quantities linked with the process depend on temperature and are expressed in terms of the difference between the heat capacities of product and reactants. Consider a process in which the state of a reactant is changed from initial state $i$ to final state $f$, assume the pressure to be constant throughout the process. Also use Hess’s Law when required. Derive the equation.

Complete step by step solution:
The Kirchhoff’s equation in general is given as $\dfrac{{dq}}{{dt}} = {C_f} - {C_i}$, where ${C_f}\&
{C_i}$ are the specific heats of products and reactants.

Consider a process in which a reactant with specific heat ${C_i}$ at temperature ${T_1}$ is
converted into a product with specific heat ${C_f}$ at temperature${T_2}$. This process can be performed in two ways but the total heat change must be equal in both the processes according to Hess’s Law.

First way: The reactant at temperature ${T_1}$ is converted into product keeping temperature constant. The heat change will be ${\left( {{H_f} - {H_i}} \right)_1} = \Delta {H_1}$.
 Now, the temperature of the product is changed from ${T_1}$ to ${T_2}$. The heat absorbed will be
${C_{{P_f}}}\Delta T$. The total heat change will be $\Delta {H_1} + {C_{{P_f}}}\Delta T$

Second way: The reactant at temperature ${T_1}$ is heated and raised to temperature ${T_2}$. The heat absorbed will be ${C_{{P_i}}}\left( {{T_2} - {T_1}} \right)$. Now, the temperature is kept constant and the reactant converts into a product. The change in heat will be ${\left( {{H_f} - {H_i}}
\right)_2} = \Delta {H_2}$. The total heat change will be \[\Delta {H_2} + {C_{{P_i}}}\left( {\Delta T}
\right)\]

From Hess’s Law, the total heat change must be equal, we get,
\[
\Delta {H_2} + {C_{{P_i}}}\left( {\Delta T} \right) = \Delta {H_1} + {C_{{P_f}}}\left( {\Delta T}
\right) \\
\Delta {H_2} - \Delta {H_1} = \left( {{C_{{P_f}}} - {C_{{P_i}}}} \right)\Delta T \\
\Delta {H_2} - \Delta {H_1} = \left( {\Delta {C_P}} \right)\left( {\Delta T} \right) \\
\]

Therefore, we have $\Delta {H_2} = \Delta {H_1} + \Delta {C_P}\left( {{T_2} - {T_1}} \right)$,
\[\dfrac{{\Delta {H_2} - \Delta {H_1}}}{{\Delta T}} = \Delta {C_P}\] and for infinitesimally small
change, we also have \[\dfrac{{d\left( {\Delta H} \right)}}{{dT}} = \Delta {C_P}\].
As you can see that the equation $\Delta {H_2} = \Delta {H_1} + \Delta {C_V}\left( {{T_2} - {T_1}}
\right)$ is not Kirchhoff's equation.

Hence ,Option (A) is correct.

Note:Remember the method we used to derive the Kirchhoff’s equation. We looked at the processes of change in state in two different ways and then using Hess’s Law. We equated the total heat change. Also remember the rest three equations in the options given above.

Note that throughout the process, the pressure is kept constant and the specific heat we considered is the one at constant pressure.