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Question: Given,(1) \(CaO(s) + {H_2}O(l) = Ca{(OH)_2}(s),\Delta H_{180}\circ C = - 15.26kcal\) (2) \({H_2}O(...

Given,(1) CaO(s)+H2O(l)=Ca(OH)2(s),ΔH180∘C=−15.26kcalCaO(s) + {H_2}O(l) = Ca{(OH)_2}(s),\Delta H_{180}\circ C = - 15.26kcal
(2) H2O(l)=H2(g)+12O2(g),ΔH180∘C=68.37kcal{H_2}O(l) = {H_2}(g) + \dfrac{1}{2}{O_2}(g),\Delta H_{180}\circ C = 68.37kcal
(3) Ca(s)+12O2(g)=CaO(s),ΔH180∘C=−151.80kcalCa(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)2Ca{(OH)_2} at 18∘C18^\circ C is:
A. -98.68 Kcal
B. -235.43 Kcal
C. 194.91Kcal
D. 98.69Kcal

Explanation

Solution

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)+H2O(l)=Ca(OH)2(s),ΔH180∘C=−15.26kcalCaO(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.
H2O(l)=H2(g)+12O2(g),ΔH180∘C=68.37kcal{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)+12O2(g)=CaO(s),ΔH180∘C=−151.80kcalCa(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)2Ca{(OH)_2} is shown below.
Ca+H2+O2→Ca(OH)2Ca + {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
ΔHf\Delta {H_f}
Where,
Δ\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.
ΔHreaction0=∑ΔHf0(product)−∑ΔHf0(reactan⁡t)\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)2Ca{(OH)_2} is shown below.
(iv)=(iii)−(ii)−(i)(iv) = (iii) - (ii) - (i)
Substitute the values in the equation.
⇒ΔH  of  (iv)=(−151.8−68.37−15.26)Kcal\Rightarrow \Delta H\;of\;(iv) = ( - 151.8 - 68.37 - 15.26)Kcal
⇒ΔH  of  (iv)=−235.48Kcal\Rightarrow \Delta H\;of\;(iv) = - 235.48Kcal
Thus, heat of formation of Ca(OH)2Ca{(OH)_2} at 18∘C18^\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.