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 $\Delta n$ .