Question

Calculate Ho298 for the process
Sb + 5/2Cl2 --> SbCl5
from the following information:
Sb + 3/2Cl2 --> SbCl3 Ho298 = -314kJ
SbCl3 + Cl2 --> SbCl5 Ho298 = -80KJ

Answer 1

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$    $\Delta H^0_1 =  -314 kJ$  ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$    $\Delta H^0_2 = -80kJ$   ..............(2)

The final reaction is as follows:  

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$  $\Delta H^0_3 = ?$  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

             $\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.