Question

Calculate the heat of formation of $KOH$ from the following data in $K.Cal$
 $$K_{(s)}+H_{2}O+aq\to KO{H}_{(aq)}+{\displaystyle \frac{1}{2}}{H}_2;\mathrm{\Delta }H=-48.4K.Cal$$
 $$H_{2(g)}+{\displaystyle \frac{1}{2}}{O}_{2(g)}\to H_2{O}_{(l)};\mathrm{\Delta }H=-68.44KCal$$
 $$KO{H}_{(s)}+aq\to KO{H}_{(aq)};\mathrm{\Delta }H=-14.0K.Cal$$
A. $+102.83$
B. $+130.85$
C. $-102.83$
D. $-130.85$

Answer 1

Hint: Standard enthalpy of formation is the change in enthalpy during the formation of $1$ mole of substance. Hess’ law gives an expression of the principle of conservation energy, which states that energy can neither be created nor be destroyed but can only be transferred from one form to another.

Complete step by step answer:
Let us see the reaction as follows:
$K_{(s)}+{\displaystyle \frac{1}{2}}{H}_{2(g)}+{\displaystyle \frac{1}{2}}{O}_{2(g)}\to KO{H}_{(s)}$
It is given that,
$$K_{(s)}+H_{2}O+aq\to KO{H}_{(aq)}+{\displaystyle \frac{1}{2}}{H}_2;\mathrm{\Delta }H=-48.4K.Cal\left(1\right)$$
$$H_{2(g)}+{\displaystyle \frac{1}{2}}{O}_{2(g)}\to H_2{O}_{(l)};\mathrm{\Delta }H=-68.44K.Cal\left(2\right)$$
$$KO{H}_{(s)}+aq\to KO{H}_{(aq)};\mathrm{\Delta }H=-14.0K.Cal\left(3\right)$$
If we add equation $$\left(1\right)$$ and $$\left(2\right)$$ , and subtract the equation $$\left(3\right)$$ , we get
$K_{(s)}+{\displaystyle \frac{1}{2}}{H}_{2(g)}+{\displaystyle \frac{1}{2}}{O}_{2(g)}\to KO{H}_{(s)}$
so the corresponding enthalpy can be calculated as follow:
$$\mathrm{\Delta }{H}_1=\mathrm{\Delta }{H}_1+\mathrm{\Delta }{H}_2-\mathrm{\Delta }{H}_3$$
on substituting the values ,we get
$$\Rightarrow \mathrm{\Delta }{H}_1=(-48.4-68.44-(-14))$$
$$\Rightarrow \mathrm{\Delta }{H}_1=-102.83K.Cal$$
Therefore, the heat of formation of $KOH$ is $-102.83K.Cal$ .
So, the correct option is (C), that is, $-102.83K.Cal$

Additional information:
-The standard enthalpy of formation is defined as the enthalpy change during the formation of $1$ mole of substance. The standard enthalpy change of formation is calculated in kilo Joule per mole $(kJmo{l}^{-1})$ and also in kilocalorie per mole $(K.Calmo{l}^{-1})$ .
-Standard enthalpy is zero for the elements in the standard states.
-There are factors that affect the standard enthalpy of formation and they are as follows:
-The temperature of the system.
-The partial pressure of gas.
-The concentration of reactant and product
-Hess’ law states that the change in enthalpy is independent of the path taken from initial state to final state.

Note: Enthalpy of formation is a state function that does not depend on paths.
-If change in enthalpy $(\mathrm{\Delta }H)$ is less than zero, then it is an exothermic reaction, otherwise, it is an endothermic reaction.
-If $(\mathrm{\Delta }H)$value is negative, then it is a spontaneous reaction.
-If $(\mathrm{\Delta }H)$value is positive, then it is a non – spontaneous reaction.
-Change in enthalpy is additive property.