Question

At ${25^ \circ }C$ and one atmospheric pressure, the partial pressures in an equilibrium mixture of ${N_2}{O_4}$ and $N{O_2}$ are 0.7 and 0.3 atmosphere, respectively. Calculate the partial pressures of these gases when they are in equilibrium at ${25^ \circ }C$ and at a total pressure of 10 atmospheres.

Answer 1

Hint: To solve this question we must know how to calculate partial pressure. We should also be able to relate equilibrium constant to partial pressure using the pressure and concentration equilibrium constant.

Complete step by step answer:
${K_p}$ 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 unitless number, although it relates the pressures.
We know that ${N_2}{O_4}$ dissociates into $N{O_2}$
The reaction takes the form:
${N_2}{O_4} \to 2N{O_2}$
The equilibrium pressures give here are:
For ${N_2}{O_4}$ = 0.7
For $N{O_2}$= 0.3
We can now calculate the equilibrium constant from the formula:
${K_p} = \dfrac{{{{\left( {{\rm{partial}}\,{\rm{pressure}}\,{\rm{of}}\,{\rm{product}}} \right)}^{coeff}}}}{{{{\left( {{\rm{partial}}\,{\rm{pressure}}\,{\rm{of}}\,{\rm{reactant}}} \right)}^{coeff}}}}$
On substituting the given values in the formula above we get,
${K_p} = \dfrac{{{{\left( {pN{O_2}} \right)}^2}}}{{{{\left( {p{N_2}{O_4}} \right)}^1}}}$
Or, ${K_p} = \dfrac{{0.3 \times 0.3}}{{0.7}} = 0.1285\,atm$
Let us assume the degree of dissociation of ${N_2}{O_4}$ to be x when the total pressure is 10 atmosphere.
Therefore, the equilibrium concentration is:
For ${N_2}{O_4}$ = 1-x
For $N{O_2}$ = 2x
Since, they will dissociate according to their stoichiometric coefficient.
The total number of moles at equilibrium is = 1 - x + 2x = 1 + x
We may now obtain the partial pressures:
${p_{{N_2}{O_4}}} = \dfrac{{1 - x}}{{1 + x}} \times 10$ and ${p_{N{O_2}}} = \dfrac{{2x}}{{1 + x}} \times 10$
The equilibrium constant now takes the form:
${K_p} = 0.1285 = \dfrac{{{{\left( {\dfrac{{2x}}{{1 + x}}} \right)}^2} \times 100}}{{\left( {\dfrac{{1 - x}}{{1 + x}}} \right) \times 10}} = \dfrac{{40{x^2}}}{{1 - {x^2}}}$
Since x is negligibly small, we can consider $\left( {1 - {x^2}} \right) \to 1$
So, ${x^2} = \dfrac{{0.1285}}{{40}}$
Or, $x = 0.0566$

Substituting the values of x, we will get the partial pressure of each component.
${p_{{N_2}{O_4}}} = \dfrac{{1 - x}}{{1 + x}} \times 10 = \dfrac{{1 - 0.0566}}{{1 + 0.0566}} \times 10 = \dfrac{{0.9436 \times 10}}{{1.0566}} = 8.93\,atm$
And
${p_{N{O_2}}} = \dfrac{{2x}}{{1 + x}} \times 10 = \dfrac{{2 \times 0.0566}}{{1 + 0.0566}} \times 10 = 1.07\,atm$

Hence, the correct answer is 8.93 atm and 1.07 atm.

Note: While calculating pressure equilibrium constant, the partial pressures of gases are used. The partial pressures of pure solids and liquids are not included. It can also be obtained from concentration equilibrium constant.