Question

At $500K$ temperature and a total pressure of $100KPa$ , iodine vapour contains $40%$ of iodine atoms by volume, calculate the ${{K}_{P}}$ and ${{K}_{C}}$ for the reaction
$I_2 \leftrightharpoons 2I$

Answer 1

Hint:In order to determine the equilibrium constants of gaseous mixture, firstly, we must calculate the partial pressure of iodine and the partial pressure of iodine atom. In general, partial pressure is defined as the force exerted by the gas. Mathematically, partial pressure is equal to the product of number of moles and total pressure.
Formula used:
There is a relation defined in equilibrium between these constants is given as:
 ${{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta ng}}$
Where, ${{K}_{p}}$ and ${{K}_{C}}$ are equilibrium constants and $\Delta ng$ is the change in number of moles
$\Delta ng=$ (Moles of product $-$ moles of reactant)

Complete step by step answer:
Here it is given that temperature $\left( T \right)=500K$
Total pressure is $100KPa$
Percentage of iodine vapour $=40%$
As we know, partial pressure $=n\times {{P}_{T}}$
Where, ${{P}_{T}}$ is the total pressure
Therefore, partial pressure of iodine atom can be calculated by simple mathematical approach that is:
$\left( {{P}_{I}} \right)=\dfrac{40}{100}\times {{10}^{5}}=40\times {{10}^{3}}Pa$
Similarly, the partial pressure of iodine gas can also be calculated by same way:
 $\left( {{P}_{{{I}_{2}}}} \right)=\dfrac{60}{100}\times {{10}^{5}}=60\times {{10}^{3}}Pa$
Now,$I_2 \leftrightharpoons 2I$
so, the change in number of moles between the number of moles of reactant and number of moles of product will be:
$\Delta ng=2-1=1$
${{K}_{P}}$ is defined as the ratio of the pressure of a product to the pressure of the reactants.
Now, one the basis of the above reaction${{K}_{P}}$ can be calculated by the formula:
${{K}_{P}}=\dfrac{{{P}_{I}}^{2}}{{{P}_{{{I}_{2}}}}}$
Therefore, on substituting the values in the relation we will get the value of gas constant:
${{K}_{P}}=\dfrac{{{\left( 40\times {{10}^{3}} \right)}^{2}}}{\left( 60\times {{10}^{3}} \right)}$
${{K}_{P}}=2.66\times {{10}^{4}}$
As, we also have a relation between these two gas constants given as:
 ${{K}_{C}}=\dfrac{{{K}_{P}}}{{{\left( RT \right)}^{\Delta ng}}}$
By substituting the values we will get the second one gas constant:
${{K}_{C}}=\dfrac{\left( 2.66\times {{10}^{4}} \right)}{\left( 8.314\times 500 \right)}$
${{K}_{C}}=6.414\times {{10}^{-3}}$
So, the answer for ${{K}_{P}}$ and ${{K}_{C}}$ is $2.66\times {{10}^{4}}$ and $6.414\times {{10}^{-3}}$

Additional information:
-${{K}_{P}}$ is the equilibrium constant of pressure and ${{K}_{C}}$ is the equilibrium constant of concentration.
-${{K}_{P}}$ is the equilibrium constant which is calculated from the partial pressure of the reaction equation. It helps to show the relationship between reactant pressure and product pressure.
-Algebra of equilibrium constant
-If a reversible reaction is reversed then the equilibrium constant for the new reaction is reciprocal of the previous one.
-When a reaction is multiplied with the constant value $n$ then the equilibrium constant ${{n}^{th}}$ power of the previous value.
-When a reaction is divided with constant value $m$ , then the equilibrium constant becomes the ${{m}^{th}}$ route of the previous one.
-When more number of reactions are added then the final reaction becomes the product of the individual equilibrium constant.

Note:
-Don’t forget to convert $KPa$ to $Pa$
-To calculate $\Delta ng$ , compounds that are in gaseous form should be taken.
-The temperature should be taken in Kelvin
-The unit ${{K}_{P}}$ and ${{K}_{C}}$ are dimensionless.