Question

Given,(1) $CaO(s) + {H_2}O(l) = Ca{(OH)_2}(s),\Delta H_{180}\circ C = - 15.26kcal$
(2) ${H_2}O(l) = {H_2}(g) + \dfrac{1}{2}{O_2}(g),\Delta H_{180}\circ C = 68.37kcal$
(3) $Ca(s) + \dfrac{1}{2}{O_2}(g) = CaO(s),\Delta H_{180}\circ C = - 151.80kcal$
From the following data, heat of formation of $Ca{(OH)_2}$ at $18^\circ C$ is:
A. -98.68 Kcal
B. -235.43 Kcal
C. 194.91Kcal
D. 98.69Kcal

Answer 1

Hint:The heat of formation is also known as standard enthalpy of formation which is the change in enthalpy in the formation of compound from its pure elements. It is calculated by subtracting heat of formation of product by heat of formation of reactant.

Complete step by step answer:Given,
$CaO(s) + {H_2}O(l) = Ca{(OH)_2}(s),\Delta {H_{180^\circ C}} = - 15.26kcal$……(i)
In this reaction, one mole of calcium oxide reacts with one mole of water to form one mole of calcium hydroxide.
${H_2}O(l) = {H_2}(g) + \dfrac{1}{2}{O_2}(g),\Delta {H_{180^\circ C}} = 68.37kcal$……(ii)
In this reaction one mole of water gives one mole of hydrogen and half mole of oxygen.
$Ca(s) + \dfrac{1}{2}{O_2}(g) = CaO(s),\Delta {H_{180^\circ C}} = - 151.80kcal$……(iii)
In this reaction one mole of calcium reacts with half mole of oxygen to give one mole of calcium oxide.
The formation of $Ca{(OH)_2}$ is shown below.
\[Ca + {H_2} + {O_2} \to Ca{(OH)_2}\]……..(iv)
In this reaction, one mole of calcium reacts with one mole of hydrogen and one mole of oxygen to form one mole of calcium hydroxide.
The standard heat of formation is defined as the enthalpy change (consumed or released) during the formation of one mole of compound from the elements at standard condition. The standard heat of formation is denoted by
\[\Delta {H_f}\]
Where,
\[\Delta\] is the change in enthalpy
H is the enthalpy
f is the substance formed from the elements
The formula to calculate the standard enthalpy of formation is shown below.
\[\Delta H_{reaction}^0 = \sum {\Delta H_f^0} (product) - \sum {\Delta H_f^0} (reac\tan t)\]
The heat of formation of $Ca{(OH)_2}$ is shown below.
\[(iv) = (iii) - (ii) - (i)\]
Substitute the values in the equation.
\[\Rightarrow \Delta H\;of\;(iv) = ( - 151.8 - 68.37 - 15.26)Kcal\]
\[\Rightarrow \Delta H\;of\;(iv) = - 235.48Kcal\]
Thus, heat of formation of $Ca{(OH)_2}$ at $18^\circ C$ is -235.48 Kcal.
Therefore,the correct option is B.

Note:
You can see that the value for heat of formation of calcium hydroxide is negative; the heat is released therefore it is an exothermic reaction.