Question

The equilibrium constant ${{{K}}_{{p}}}$ for the reaction is 9. The equilibrium constant ${{Kp}}\left( {{{in atm}}} \right)$ for the reaction is 9 at ${{7 atm}}$ and ${{300 K}}$ ${{{A}}_{{2}}}\left( {{g}} \right) \rightleftharpoons {{{B}}_{{2}}}\left( {{g}} \right) + {{{C}}_{{2}}}\left( {{g}} \right)$ Calculate the average molar mass of an equilibrium mixture.
Given: Molar mass of ${{{A}}_{{2}}}{{,}}{{{B}}_{{2}}}{{,\;}}{{{C}}_{{2}}}{{\;:}}70,49\& 21{{ gm/mol}}$ respectively:
(A) 50
(B) 45
(C) 40
(D) $37.5$

Answer 1

To answer this question, you should recall the concept of the equilibrium constant. We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. We shall substitute the values given in the question to calculate ${K_c}$ and then use it to calculate the concentration of the reactants and products.
${{{K}}_{{p}}} = {{{K}}_{{c}}}{{R}}{{{T}}^{{{\Delta ng}}}}$
where ${K_p}$ is pressure constant, ${K_c}$ is equilibrium constant, $R$ is the universal gas constant, $T$ is temperature and $\Delta ng$ is the change in the number of moles.

Complete Step by step solution
The equilibrium constant value of a reaction is independent of the initial values of the concentration of reactants. It is defined as the product of molar concentrations of the products divided by the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical equation at equilibrium. It is only dependent on the temperature at which the reaction is taking place.
The relation between ${{{K}}_{{{p}}}}{{\;and\;}}{{{K}}_{{{c}}}}{{\;is}}\;$ ${K_p} = {K_c}R{T^{\Delta ng}}$ .
Change in the number of moles $= 1$ .
Substituting these values in the equation:
$9 = {K_c} \times 0.0821 \times 300$ .
Rearranging and solving:
${K_c} = 0.3654$ .
Let the equilibrium number of moles of $A,B$ and $C$ be 1, $x$ respectively:
${K_c} = \frac{{[A][B]}}{{[C]}} = {X^2} = 0.3654$ .
Solving:
$x = 0.6044$ .
Thus equilibrium moles of $A,B$ and $C$ are $1,0.6{\text{ and }}0.6$ respectively.
The mole fractions of $A,B$ and $C$ are $0.455,{{ }}0.27{\text{ and }}0.27$ respectively.
$\therefore$ Average molecular weight $= 70 \times 0.455 + 49 \times 0.27 + 21 \times 0.27 = 50$
Hence, the correct option is A.

Note
You should know the change in values of concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. Le Chatelier's principle states that the temperature, concentration, pressure, catalyst, inert gas addition to a chemical reaction can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction. Collisions of particles lead to reactions. Only particles that collide sufficiently can react. Now comes the important point. From the kinetic and collision model it can be said that the rate of a reaction increases with increase in temperature due to more energy and more collisions. However, this increase in the rate of reaction depends on the value of the energy of activation of the reaction which is different for both the forward and the backward reaction. So, a given increase in temperature leads to an increase not only in the rate of forward reaction but also the backward reactions to a different extent. So, this proves that the value of the equilibrium constant changes with the temperature only.