Ncert Exercises & Solutions

7.11 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) H2 (g) + I2 (g)

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:

$2 {HI}_{(g)} \leftrightarrow H_{2 (g)} + I_{2 (g)}$
$Initial conc . 0.2 atm 0 0$
$At equilibrium 0.04 atm \frac{0.16}{2} \frac{0.16}{2}$
$= 0.08 atm = 0.08 atm$
$Therefore,$
$K_{p} = \frac{p_{H_{2}} \times p_{I_{2}}}{{p^{2}}_{HI}}$
$= \frac{0.08 \times 0.08}{{(0.04)}^{2}}$
$= \frac{0.0064}{0.0016}$
$= 4.0$

Hence, the value of $K_{p}$ for the given equilibrium is 4.0.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions