Question

The enthalpy of formation of CO and steam are -110.5 and -243.0 KJ respectively. Calculate the heat of reaction when steam is passed over the coke as
\[C(s) + {\text{ }}{H_2}O(g) \to CO(g) + {H_2}(g)\]

Answer 1

Hint: Combustion and formation both are the opposite processes. A combustion reaction is always exothermic i.e. heat of that reaction is negative then the other one can be exothermic or endothermic i.e. heat of reaction can be negative or positive.

Complete step by step answer:
-The heat of combustion and heat of formation are two kinds of heat of reaction. The heat of reaction is an energy that is released or absorbed during the reaction. This can be calculated from the change of energy between all reactants and all products.
-The definition of heat of formation is the amount of energy change due to the formation of 1mole of a compound from its constituent elements.
-The definition of heat of combustion is the amount of heat or energy developed due to the total combustion of 1mole of a compound.
-The enthalpy of formation of an element at its elementary state is zero. i.e. \[\sum {\Delta {H_f}(C(s))} = 0\] , \[\Delta {H_f}({H_2}) = 0\] , \[\Delta {H_f}({H_2}O(l)) = 0\]
Now the heat of the combustion reaction of glucose can be calculated based on Hess’ law. Using formula,
 \[\Delta H = \sum {\Delta {H_f}(product)} - \sum {\Delta {H_f}(reac\tan t)} \]
The equation of formation of CO is,
 \[C(s) + \dfrac{1}{2}{O_2}(g)\xrightarrow{\Delta }CO(g),\Delta H = - 0110.5KJ/mole\] ……(1)
The equation of combustion of H2O (steam) is,
 \[{H_2} + \dfrac{1}{2}{O_2}\xrightarrow{\Delta }{H_2}O,\Delta H = - 243.0KJ/mole\] ……(2)

 Now according to the Hess’ law, the heat of reaction is,
  \[
  \Delta H = \sum {\Delta {H_f}(product)} - \sum {\Delta {H_f}(reac\tan t)} \\
   = \sum {\Delta {H_f}(CO) + } \Delta {H_f}({H_2}) - \sum {\Delta {H_f}(C(s))} + \Delta {H_f}({H_2}O(g)) \\
   = \sum { - 110.5 + } \Delta {H_f}({H_2}) - \sum {\Delta {H_f}(C(s))} - 243.0 \\
   = \sum { - 110.5 + } 0 - \sum 0 - 243.0 \\
   = 132.5KJ/mole \\
 \]

Note:
The Can't Hoff equation relates the equilibrium constant of a reaction to change in temperature by assuming that the change in enthalpy of a reaction is constant as a function of temperature.
Using the integrated form of Can't Hoff equation is shown below, we can determine the change in enthalpy.
 \[\ln \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{R}[\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}]\]