Question

${K_p}$ for the reaction $N{H_4}HS(s) \rightleftarrows N{H_3}(g) + {H_2}S(g)$ at certain temperature is $4 ba{r^2}$.Calculate the equilibrium pressure.
A. 2
B. 4
C. 3.2
D. 5.4

Answer 1

According to the question, with the help of given Equilibrium Constant, ${K_p}$ of the reaction, we will conclude the Equilibrium Pressure of the given reaction.

Complete solution:
Given reaction is:-
$N{H_4}HS(s) \rightleftarrows N{H_3}(g) + {H_2}S(g)$
$\because {K_p} = 4$
here, ${K_p}$ is an Equilibrium constant for the given reaction. The equilibrium constant is the value of the reaction quotient that is calculated from the expression for chemical equilibrium. It depends on the ionic strength and temperature and is independent of the concentrations of reactants and products in a solution. Equilibrium constant being the ratio of the concentrations raised to the stoichiometric coefficients.
$\Rightarrow {K_p} = {2^2}$
Therefore, $P = 2$
If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change. If the pressure is increased the equilibrium will shift to favour a decrease in pressure. If the pressure is decreased the equilibrium will shift to favour an increase in pressure.
Equilibrium pressure= $2 p$
So, Equilibrium pressure =$4 bar$

So the correct option is (B).

Note: Equilibrium constants aren't changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium may be changed if you change the pressure.