Question

$ {N_2}{O_4} $ dissociates as $ {N_2}{O_4} \to 2N{O_2} $ . At $ {55^ \circ }C $ and one atmosphere, % decomposition of $ {N_2}{O_4} $ is $ 50.3\% $ . At what P and Same temperature, the equilibrium mixture will have the ratio of $ {N_4}{O_2}:N{O_2} $ as $ 1:8 $ ?

Answer 1

Hint :This question can be solved by using the concept of equilibrium. The equilibrium constant can be defined as the ratio between the amount of reactant and the amount of product which is used to determine the chemical behaviour.Make the equation of equilibrium constants and then solve for the variable to get the required answer.

Complete Step By Step Answer:
First step will be to write the concentrations both before the equilibrium and at the equilibrium.
 $ {N_2}{O_4} \Leftrightarrow 2N{O_2} $
Moles before equilibrium $ 1{\text{ 0}} $
Moles at equilibrium $ 1 - \alpha {\text{ 2}}\alpha $
In the above expression $ \alpha $ is the degree of dissociation $ = 0.503 $
Using the definition of equilibrium constant we can write
 $ \eqalign{
  & {K_p} = \dfrac{{{{\left( {N{O_2}} \right)}^2}}}{{\left( {{N_4}{O_2}} \right)}} \times {\left( {\dfrac{P}{{\sum n }}} \right)^{\Delta n}} \cr
  & \cr} $
Substitute the values in the above equation and solve them, we get
 $ \eqalign{
  & {K_p} = \dfrac{{{{\left( {2\alpha } \right)}^2}}}{{\left( {1 - \alpha } \right)}} \times \left( {\dfrac{P}{{(1 - \alpha )}}} \right) \cr
  & \cr} $
 $ {K_p} = \dfrac{{4{\alpha ^2}P}}{{1 - {\alpha ^2}}} = \dfrac{{4 \times {{(0.503)}^2} \times 1}}{{1 - {{(0.503)}^2}}} = 1.355 $
Now, We know that
 $ {N_2}{O_4} \Leftrightarrow 2N{O_2} $ (at pressure P)
 $ 1{\text{ 1}} $
 $ (1 - x){\text{ 2x}} $
In the question itself we are given that the ratio of $ {N_4}{O_2}:N{O_2} $ is equal to $ 1:8 $ .
Using this we get,
 $ \dfrac{{{N_2}{O_4}}}{{N{O_2}}} = \dfrac{{1 - x}}{{2x}} = \dfrac{1}{8} $
After cross-multiplying we get
 $ \eqalign{
  & 8(1 - x) = 1(2x) \cr
  & 8 - 8x = 2x \cr
  & 10x = 8 \cr
  & x = \dfrac{8}{{10}} \cr
  & x = 0.8 \cr} $
Using again at pressure P
 $ \eqalign{
  & \therefore {K_p} = \dfrac{{4{x^2} \times P}}{{1 - {x^2}}} \cr
  & 1.355 = \dfrac{{4 \times {{(0.8)}^2} \times P}}{{1 - {{(0.8)}^2}}} \cr
  & \Rightarrow P = 0.19atm \cr} $

Note :
The equilibrium constant of a chemical reaction (usually denoted by the symbol K) provides the information about the relation between the reactants and the product when the chemical reaction is in the state of equilibrium. At equilibrium, the rate of the forward reaction is always equal to the rate of backward reaction. At a particular temperature the constants for a particular reaction are constant.
In these types of questions the general mistake done by the students is in the calculation part. So it is advised to pay extra attention to that part and for tough calculations use the logarithmic function.