Question

If ${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O;\Delta H=-68.39kCal$
$K+{{H}_{2}}O\to KO{{H}_{aq}}+\dfrac{1}{2}{{H}_{2}};\Delta H=-48.0kCal$
$KOH+{{H}_{2}}O\to KO{{H}_{aq}};\Delta H=-14.0kCal$; the heat formation of $KOH$ is?

Answer 1

We know that standard enthalpy of formation is the change in enthalpy during the formation of a 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 answer: Here we also know that $KOH$ has a molecular weight of $56.10$. It is a corrosive solution. $KOH$ is a clear aqueous solution. Heat formation is the process when heat is evolved or absorbed when one mole of compound is formed during a chemical reaction. Here we need to find the heat formation of $KOH$ while heat formed by other elements are given. To make $KOH$ we need to add the equation.
$K+\dfrac{1}{2}{{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to KOH$
As per the given equations we must add equations
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$ and $K+{{H}_{2}}O\to KO{{H}_{aq}}+\dfrac{1}{2}{{H}_{2}}$
And later subtract equation
$KOH+{{H}_{2}}O\to KO{{H}_{aq}}$
Therefore we get,
$-68.39+(-48.0)-(-14)$
Thus, we get; $\Rightarrow -68.39-48.0+14$
$=-102.39kCal$
Therefore heat of $KOH$ is $-102.39kCal$

Note:
Remember that the $KOH$ potassium hydroxide is also known as lye. It is used as a stabilizer, thickening agent, etc. 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.