Question

At 503 k the equilibrium constant kc for the dissociation of n2o4 has the value of 40.0. Calculate the fraction of n2o4 left undissociated when one mole of this gas is placed in a 1.0-l container at 503 k

Answer 1

The decomposition reaction of N2O4 will be

                      N2O4(g)  ---> 2NO2 (g)

initial conc          1                    0

Change              -x                   +2x

Equilibrium          1-x                 2x

Kc = [NO2]^2 / [N2O4]

Kc = 40 = (2x)^2 / 1-x

40 - 40x = 4x^2

x^2 + 10x - 10 = 0

On solving for x

x = 0.916

fraction of N2O4 left undissociated = 1 - 0.916 = 0.084

Answer 2

The fraction of undissociated ${{\text{N}}_2}{{\text{O}}_4}$ is $\boxed{{\text{0}}{\text{.084}}}$.

Further Explanation:

The condition at which the forward reaction rate becomes equal to the backward reaction rate is known as chemical equilibrium.

The generic equation for a general equilibrium is as follows:

$a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}$  

Here,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression for the equilibrium constant for the above reaction is as follows:

${K_{\text{c}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$

Here,

${{\text{K}}_{\text{c}}}$ is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

The dissociation reaction of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ is as follows:

${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\left( g \right) \rightleftharpoons 2{\text{N}}{{\text{O}}_2}\left( g \right)$  

The initial concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ can be calculated as follows:

$\begin{aligned}\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}} \right]&= \frac{{1{\text{ mol}}}}{{1{\text{ L}}}}\\&= 1{\text{ M}}\\\end{aligned}$  

The initial concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ is 1 M. Let x to be the change in concentration at equilibrium. Therefore, the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ becomes 1-x at equilibrium and that of ${\text{N}}{{\text{O}}_{\text{2}}}$ becomes 2x at equilibrium.

The expression of ${{\text{K}}_{\text{c}}}$ for the given reaction is as follows:

 ${{\text{K}}_{\text{c}}}=\dfrac{{{{\left[ {{\text{N}}{{\text{O}}_2}} \right]}^2}}}{{\left[ {{{\text{N}}_2}{{\text{O}}_4}} \right]}}$    …… (1)                                                                          

Substitute 2x for $\left[ {{\text{N}}{{\text{O}}_{\text{2}}}} \right]$, 1-x for $\left[ {{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}} \right]$ and 40 for ${{\text{K}}_{\text{c}}}$ in equation (1).

$40 = \dfrac{{{{\left( {{\text{2x}}} \right)}^2}}}{{\left( {1 - x} \right)}}$                                                                    …… (2)

Rearrange equation (2) to form the quadratic equation.

 ${x^2} + 10x - 10 = 0$

Solve for x,

${\text{x}} = 0.916$  

Or,

${\text{x}} =  - 10.916$

But the value of ${\text{x}} =  - 10.916$  is rejected as concentration can never be negative. So the value of x is 0.916. Therefore the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ at equilibrium is 0.916 M.

The formula to calculate the fraction of undissociated ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ is as follows:

${\text{Fraction of undissociated }}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} = \left( \begin{aligned}{\text{Initial amount of }}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} -  \\{\text{Equilibrium amount of }}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} \\\end{aligned} \right)$         …… (3)              

Substitute 1 for the initial amount of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ and 0.916 for the equilibrium amount of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ in equation (3).

 $\begin{aligned}{\text{Fraction of undissociated }}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} &= \left( {1 - 0.916} \right)\\&= 0.084\\\end{aligned}$

Learn more:

1. Calculate equilibrium constant for ammonia synthesis: https://brainly.com/question/8983893
2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior School

Chapter: Chemical Equilibrium

Subject: Chemistry

Keywords: chemical equilibrium, Kc, 0.084 M, 0.916 M, 1 M, N2O4, NO2, 1-x, 2x, undissociated, 40.