Question

At constant temperature, the equilibrium constant $\left(K_P\right)$ for the decomposition reaction $N_2{O}_4\leftrightharpoons 2N{O}_2$ Is expressed by $K_P=\frac{4x^{2}P}{(1-x^2)}$
Where P = pressure, x = extent of decomposition. Which one of the following statements are false?

• A. $K_P$ increases with increase of P.
• B. $K_P$ increases with increase of x.
• C. $K_P$ increases with decrease of x.
• D. $K_P$ remains constant with change in P and x.

Answer 1

Answer: (A), (C), (D)

The correct options are
A $K_P$ increases with increase of P.
C $K_P$ increases with decrease of x.
D $K_P$ remains constant with change in P and x.

The number of moles of products is greater than the number of moles of reactants. Hence with increase in the pressure, the reaction shifts in the backward direction so that the value of $K_P$ decreases.
As the extent o f dissociation increases the number of moles of product increases and the number of moles of reactant decreases. hence, the value of $K_p$ increases.