Question

The degree of dissociation of ${I_2}$ molecules at ${1000^0}C$ and under atmospheric pressure is 40% by volume. The total pressure on the gas at equilibrium so that dissociation is reduced to 20% at the same temperature, will be.
A. 4.57 atm
B. 2.83 atm
C. 5.33 atm
D. 7.57 atm

Answer 1

Hint- In order to solve the problem use the concept of equilibrium. Consider the degree of dissociation as some unknown value at the given temperature. Then find the total number of moles of both the dissociated particles in terms of the same unknown variable and use the values given in question and take atmospheric pressure as 1 atm.

Complete answer:
As we know that
​${I_2}$ dissociates as
${I_2} \rightleftharpoons 2I$
If x is the degree of dissociation at ${1000^0}C$ under atmospheric pressure, then
Initial concentration: $\left[ {{I_2}} \right] = 1,I = 0$
At equilibrium ${I_{2}} = 1 - x,I = 2x$
The total number of moles at equilibrium $= 1 - x + 2x = 1 + x$
Partial pressure of ${I_2}$​ and $I$ will be

{{P_{{I_2}}} = \dfrac{{\left( {1 - x} \right)p}}{{\left( {1 + x} \right)}}} \\\ {{P_I} = \dfrac{{\left( {2x} \right)p}}{{\left( {1 + x} \right)}}} \end{array}$$ We can find out the pressure constant by these two equations in terms of x. Now let us find out ${K_p}$ in terms of x.

{K_p} = \dfrac{{{{\left( {{P_I}} \right)}^2}}}{{{P_{{I_2}}}}} = \dfrac{{{{\left( {\dfrac{{\left( {2x} \right)p}}{{\left( {1 + x} \right)}}} \right)}^2}}}{{\dfrac{{\left( {1 - x} \right)p}}{{\left( {1 + x} \right)}}}} \\
{K_p} = \dfrac{{{{\left( {2x} \right)}^2}{p^2}\left( {1 + x} \right)}}{{\left( {1 - x} \right)p{{\left( {1 + x} \right)}^2}}} \\
{K_p} = \dfrac{{{{\left( {2x} \right)}^2}p}}{{\left( {1 - {x^2}} \right)}}.............(1) \\

Given in the problem at atmospheric pressure, when $p = 1atm$ and $x = 40\% = 0.4$ Let us substitute the value to find the value of ${K_p}$

\because {K_p} = \dfrac{{{{\left( {2x} \right)}^2}p}}{{\left( {1 - {x^2}} \right)}} \\
\Rightarrow {K_p} = \dfrac{{{{\left( {2 \times 0.4} \right)}^2} \times 1}}{{\left( {1 - {{\left( {0.4} \right)}^2}} \right)}} \\
\Rightarrow {K_p} = \dfrac{{{2^2} \times {{0.4}^2} \times 1}}{{\left( {1 + 0.4} \right)\left( {1 - 0.4} \right)}} \\
\Rightarrow {K_p} = \dfrac{{4 \times 0.16 \times 1}}{{1.4 \times 0.6}} \\
\Rightarrow {K_p} = 0.76 \\

We have to find pressure when dissociation is 20 percent. At x=0.2 let pressure P Let us substitute the value along with the value of ${K_p}$ in equation (1)

\because {K_p} = \dfrac{{{{\left( {2x} \right)}^2}p}}{{\left( {1 - {x^2}} \right)}} \\
\Rightarrow 0.76 = \dfrac{{{{\left( {2 \times 0.2} \right)}^2}P}}{{\left( {1 - {{\left( {0.2} \right)}^2}} \right)}} \\
\Rightarrow 0.76 = \dfrac{{{2^2} \times {{0.2}^2} \times P}}{{\left( {1 + 0.2} \right)\left( {1 - 0.2} \right)}} \\
\Rightarrow 0.76 = \dfrac{{4 \times 0.04 \times P}}{{1.2 \times 0.8}} \\
\Rightarrow P = \dfrac{{0.76 \times 1.2 \times 0.8}}{{4 \times 0.04}} \\
\Rightarrow P = 4.57atm \\

Hence, the total pressure on the gas at equilibrium so that dissociation is reduced to 20% at the same temperature, will be 4.57 atm. _So, option A is the correct option._ Note- The constant used above ${K_p}$ is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unit less number, although it relates the pressures.