Question

At a certain temperature and total pressure of $ {10^5} $ Pa, iodine vapour contains $ 40 % $ by volume of I atoms.
 $ {I_2}_{(g)} \rightleftharpoons 2{I_{(g)}} $
Calculate $ {k_p} $ for the equilibrium.

Answer 1

Hint :Equilibrium constant for gaseous systems should be expressed in terms of partial pressure because it is more convenient. So, for a general reaction equilibrium constant in terms of pressure should be-
 $ aA + bB \rightleftharpoons cC + dD $
 $ {k_p} = \dfrac{{\left( {p_C^c} \right)\left( {p_D^d} \right)}}{{\left( {p_A^a} \right)\left( {p_B^b} \right)}} $
Here p is the pressure in Pa.

Complete Step By Step Answer:
 $ {k_p} $ Known as equilibrium constant calculated from the partial pressure of a reaction. Equilibrium constant for the gaseous system should be expressed in terms of partial pressure because it is more convenient. So, for a general reaction equilibrium constant in terms of pressure should be-
 $ aA + bB \rightleftharpoons cC + dD $
 $ {k_p} = \dfrac{{\left( {p_C^c} \right)\left( {p_D^d} \right)}}{{\left( {p_A^a} \right)\left( {p_B^b} \right)}} $
Here p is the pressure in Pa.
 $ {k_p} $ Is unitless.
Now let consider the above given reaction
 $ {I_2}_{(g)} \rightleftharpoons 2{I_{(g)}} $
Total pressure of mixture is given, also it is given that iodine vapours contain $ 40 % $ by volume of iodine atoms.
Now, we have
Total pressure = $ {10^5} $ Pa
We know that partial pressure is given by the formula
 $ p = x.{P_T} $
Where, $ x $ is mole fraction, $ {P_T} $ is total pressure and $ p $ is partial pressure
Partial pressure of iodine atoms = $ \dfrac{{40}}{{100}} \times {10^5} $
 $ {p_I} = 0.4 \times {10^5} $ Pa
Partial pressure of iodine molecules= $ \dfrac{{60}}{{100}} \times {10^5} $
 $ {p_{{I_2}}} = 0.6 \times {10^5} $ Pa
Now,
 $ {k_p} = \dfrac{{{p_{{I^2}}}}}{{{p_{{I_2}}}}} $
Putting all values in above equation
\[{k_p} = \dfrac{{{{\left( {0.4 \times {{10}^5}Pa} \right)}^2}}}{{\left( {0.6 \times {{10}^5}} \right)}}\]
 $ {k_p} = 2.67 \times {10^4}Pa $

Note :
It is necessary that while calculating the value of $ {k_p} $ , pressure should be expressed in bars because the standard state for pressure is $ 1 $ bar. We know that unit $ 1 $ means.
 $ 1 $ pascal, Pa= $ 1N{m^{ - 2}} $
And $ 1 $ bar = $ {10^5}Pa $
 $ {k_p} = {k_c}{\left( {RT} \right)^{\Delta n}} $
Where $ \Delta n $ = (no of moles of gaseous products) $ - $ (no of moles of gaseous reactant)
So \[{k_p}\] can be less than, greater than or equal to $ {k_c} $ . It depends only on change in $ \Delta n $ .