Question

For the reaction, $S{{O}_{2\left( g \right)}}+\frac{1}{2}{{O}_{2\left( g \right)}}{\leftrightarrows}S{{O}_{3\left( g \right)}}$, if ${{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{x}}$ where the symbols have usual meaning then the value of x is (assuming ideality):
(A) $\dfrac{1}{2}$
(B) 1
(C) −1
(D) $-\dfrac{1}{2}$

Answer 1

Hint: ${{K}_{C}}$ is the Equilibrium Constant of Concentration and ${{K}_{P}}$ is the Equilibrium Constant of Pressure. They are related through the equation ${{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{x}}$ where x is the number of moles of reactants subtracted from number of moles of product of gas molecules. Thus by subtracting the number of moles of reactants from the number of moles of products of given reaction we will get the value of x.

Complete step by step answer:
  - Let's start with the concept of equilibrium constants. In equilibrium constants the ratio of the product concentrations to reactant concentrations is calculated. Since the concentrations are measured at equilibrium, the equilibrium constant will remain the same for a given reaction independent of initial concentrations of reactants.
- There are mainly two types of equilibrium constants ${{K}_{P}}$ and ${{K}_{C}}$ .The difference between the two equilibrium constants is that is ${{K}_{C}}$ defined by molar concentrations whereas ${{K}_{P}}$ is defined by the partial pressures of the gasses inside a closed system.
  - The equilibrium constants do not comprise the concentrations of single constituents such as solids and liquids and also they do not have any units. For converting the Equilibrium Constant of Concentration (${{K}_{C}}$) and Equilibrium Constant of Pressure (${{K}_{P}}$) the following equation is used
\[{{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta {{n}_{gas}}}}\]
Where $\Delta {{n}_{gas}}$ is the number of moles of reactants subtracted from the number of moles of product of gas molecules. So in the given question the term x represents $\Delta {{n}_{gas}}$.Hence we can write as follows
\[\Delta {{n}_{g}}=moles\text{ }of\text{ }product-moles\text{ }of\text{ }reactants\]
 In the given reaction there is one mole of $S{{O}_{2}}$ and $\dfrac{1}{2}$ moles of oxygen. These are the reactants and the number of moles of the product is one. Hence we can write
\[\Delta {{n}_{g}}=1-\left( \dfrac{1}{2}+1 \right)\]
\[x=1-\dfrac{3}{2}=-\dfrac{1}{2}\]
So, the correct answer is “Option D”.

Note: Keep in mind that ${{K}_{P}}$ is the equilibrium constant written using pressures to represent the activity of the different chemicals and it is only useful for gas phase reactions where the pressure is proportional to the amount of each kind of gas. Also, ${{K}_{C}}$ is that same equilibrium constant written using concentrations to signify the activity of the different chemicals and it is useful in gas phase reactions and in liquid solution phase reactions where the concentration is proportional to the activity or amount of each component but only one or the other.