Question

A sample of HI (g) is placed in flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI (g) is 0.04 atm. What is $K_{p}$ for the given equilibrium ?

Answer 1

${ K }_{ p }$  is the equilibrium constant calculated with the partial pressures of the gaseous reactants and products in an equilibrium reaction.  

The formula to calculate Kp: ${ K }_{ p }\quad =\quad \frac { p{ H }_{ 2 }\quad \times \quad p{ I }_{ 2 } }{ p{ HI }_{ 2 } }$

Reaction equation: ${ 2HI }_{ (g) }\quad \rightleftharpoons \quad { H }_{ 2(g) }\quad +\quad { I }_{ 2(g) }$

The equilibrium pressure of ${ H }_{ 2(g) }$ and ${ I }_{ 2(g) }$ is calculated by calculating the decrease in the pressures of HI (0.2 - 0.04 = 0.16 atm)  

As per the equation, 2 moles of HI dissociates to give 1 mole of ${ H }_{ 2(g) }$ and 1 mole of ${ I }_{ 2(g) }$.

Hence, the equilibrium pressure of ${ H }_{ 2(g) }$ is $\frac { 0.16 }{ 2 } \quad =\quad 0.08 atm$ and of ${ I }_{ 2(g) }$ is $\frac { 0.16 }{ 2 } \quad =\quad 0.08 atm.$.

Initial Pressure of HI: 0.2 atm

Equilibrium Pressure of HI: 0.04 atm

Initial Pressure of ${ H }_{ 2(g) }: 0$

Equilibrium Pressure of ${ H }_{ 2(g) }: 0.08 atm$

Initial Pressure of ${ I }_{ 2(g) }: 0$

Equilibrium Pressure of ${ I }_{ 2(g) }: 0.08 atm$

Substituting the values in the formula:

${ K }_{ p }\quad =\quad \frac { p{ H }_{ 2 }\quad \times \quad p{ I }_{ 2 } }{ p{ HI }_{ 2 } }{ K }_{ p }\quad =\quad \frac { 0.8\quad \times \quad 0.8 }{ \left( 0.04 \right) ^{ 2 } }$

So, ${ K }_{ p } = 4.0$