Answer

Verified

53.7k+ views

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.

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.

Recently Updated Pages

What is the common property of the oxides CONO and class 10 chemistry JEE_Main

What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main

Ferrous sulphate when heated decomposes with the evolution class 10 chemistry JEE_Main

Aluminium is not used A In silvery paints B For making class 10 chemistry JEE_Main

What is DDT among the following A An organic fertiliser class 10 chemistry JEE_Main

What is the chemical name of quicklime A Calcium oxide class 10 chemistry JEE_Main

Other Pages

A given ray of light suffers minimum deviation in an class 12 physics JEE_Main

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Oxidation of succinate ion produces ethylene and carbon class 12 chemistry JEE_Main

In the ground state an element has 13 electrons in class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main