Question

A sample of $H{I_{\left( g \right)}}$ is placed in a flask at a pressure of $0.2$ atm . At equilibrium the partial pressure of $H{I_{\left( g \right)}}$ is $0.04$ atm . What is ${K_p}$ for the given equilibrium ?
$2H{I_{\left( g \right)}} \rightleftharpoons {H_2}_{\left( g \right)} + {I_{2\left( g \right)}}$ :
A. 6
B. 16
C. 4
D. 2

Answer 1

Hint: ${K_p}$ is the equilibrium constant calculated from the partial pressure of a reaction equation . It is used to express the relationship between product pressure and reactant pressure.

Complete step by step answer: At the initial concentration , $HI$ has a pressure of $0.2$ atm. At equilibrium , it has a partial pressure of $0.04$ atm. Therefore , a decrease in the pressure of $HI$ is $0.2 - 0.04 = 0.16$ atm.
The given reaction is :-
 Initial Conc. $2H{I_{\left( g \right)}} \rightleftharpoons {H_{2\left( g \right)}} + {I_{2\left( g \right)}}$
 $0.2$atm- $0$ $0$ (for both the species in the product)
At equilibrium $0.04$ atm- $\dfrac{{0.16}}{2}$ $\dfrac{{0.16}}{2}$ (for both the species in the product)
 Acc to the formula of ${K_p}$
 ${K_p} = \dfrac{{{P_{{H_2}}} \times {P_{{I_2}}}}}{{{p_{HI}}}}$
Here ${P_{{H_2}}}$ - partial pressure of ${H_2}$ - $0.08$ atm.
  ${P_{{I_2}}}$ - partial pressure of ${I_2}$ - $0.08$atm.
  ${P_{HI}}$ - partial pressure of $HI$ - $0.04$ atm.
On substituting the values in the formula , we get
$\Rightarrow$ ${K_p} = \dfrac{{0.08 \times 0.08}}{{{{\left( {0.04} \right)}^2}}} = \dfrac{{0.0064}}{{0.0016}} = 4$
So , the correct answer is (C).

Additional Information:
There are various factors which affect the equilibrium , like
-Change in concentration of reactant and product - If In a reaction in eclipse Priyam the concentration of reactant is increased then the equilibrium shifts in the forward direction where as if the concentration of the product is increased the equilibrium ships in the backward direction.
-Change of temperature of the system - If the reaction is endothermic then increase of temperature will favour the forward direction and if the reaction is exothermic then the decrease in temperature will favour the forward direction.
-Change of pressure on the system - If action is accompanied by an increase in total number of moles then low pressure favours the reaction whereas if a reaction is accompanied by a decrease in total number of holes then vice versa happens .
-When catalyst is added - When catalyst is added it does not change the equilibrium instead it helps in attainment of equilibrium quickly.
-The addition of some inert gas - When inert gas is added at constant pressure the equilibrium shifts towards the larger number of moles.

Note: When a reaction is at equilibrium , the forward reaction and reverse reaction have the same rate.
An increase in equilibrium constant with increase in temperature shows that the reaction is endothermic whereas the decrease shows that the reaction is exothermic.