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 Kp for the given equilibrium ?
$2HI (g) ⇋ H_2 (g) + I_2 (g)$

Answer 1

The initial concentration of HI is 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.
The given reaction is:
$2HI(g) ↔ H_2(g) + I_2(g)$
Initial conc. $0.2 \ atm$ $0$ $0$
At equilibrium $0.04 \ atm$ $\frac {0.16}{2}= 0.08 \ atm$ $\frac {0.16}{2}= 0.08 \ atm$
Therefore,
$K_p = \frac {P_{H_2} × P_{I_2}}{P^2_{HI}}$

$K_p= \frac {0.08 × 0.08 }{ (0.04)^2 }$

$K_p = \frac {0.0064 }{0.0016 }$

$K_p = 4.0$

Hence, the value of $K_p$ for the given equilibrium is $4.0.$