Question

The enthalpy of a reaction at $ 273K $ is $ - 3.57kJ $ . What will be the enthalpy of a reaction at $ 373K $ if $ \Delta {C_p} = zero? $
(A) $ - 3.57 $
(B) Zero
(C) $ - 3.57 \times \dfrac{{373}}{{273}} $
(D) $ - 375 $

Answer 1

Hint: The enthalpy of a reaction can be given by the difference between the enthalpy of the products and enthalpy of reactants. Thus, enthalpy of other reaction at a different temperature can be calculated from Kirchhoff’s law. The specific heat at constant pressure and the temperatures were needed to calculate the enthalpy of a reaction.
 $ \Delta {C_p} = \dfrac{{\Delta {H_2} - \Delta {H_1}}}{{{T_2} - {T_1}}} $
 $ \Delta {C_p} $ is specific heat at constant pressure
 $ \Delta {H_2} $ is final enthalpy having to be determined
 $ \Delta {H_1} $ is initial enthalpy
 $ {T_2} $ is final temperature
 $ {T_1} $ is the initial temperature.

Complete answer:
Given that a chemical reaction at $ 273K $ has the enthalpy $ - 3.57kJ $ .
The specific heat at constant pressure is zero
The enthalpy at $ 373K $ has to be determined
Kirchhoff’s law is the law that gives the relation between the enthalpy or heat of a reaction, temperature and specific heat at constant pressure. Generally, enthalpy increases with increase in temperature. But in this case the specific heat at constant pressure given is zero.
In the above Kirchhoff’s law, substitute all the values in that equation
 $ 0 = \dfrac{{\Delta {H_2} + 3.57}}{{373 - 273}} $
By simplification, we will get
 $ \Delta {H_2} = - 3.57 $
Thus, the enthalpy of a reaction at $ 373K $ is $ - 3.57 $ .
Thus, option A is the correct one.

Note:
Generally, according to Kirchhoff’s law, the enthalpy of a reaction increases with the increase in temperature. But the specific heat capacity at constant pressure is zero in the given reaction. The enthalpy is constant at the two given temperatures.