Question

Hydrogen iodide can decompose into hydrogen and iodine gases. 2HI(g) H2(g) I2(g) K for the reaction is 0.016. Of 0.148 atm of HI(g) is sealed in a flask, what is the pressure of each gas when equilibrium is established

Answer 1

Solution :

Given :

Hydrogen iodide decomposes to hydrogen and iodine gas

    $2 HI \ \ \ \Leftrightarrow \ \ \ \ H_2 \ \ \\ + \ \ I_2$

I    0.148            0          0

C     -2a           +a          +a

E  0.148-2a      a            a

We know

$k_p=\frac{P(H_2)P(I_2)}{P(HI)^2}$

$0.016=\frac{a^2}{(0.148-2a)^2}$

$0.016^{1/2}=\frac{a}{0.148-2a}$

$0.12649=\frac{a}{0.148-2a}$

0.0187 = 1.2529 a

a = 0.0149

Therefore

P(HI) = 0.148 - 2a

        = 0.148 - 2(0.0149)

        = 0.1182 atm

P($H_2$) =  a

        = 0.0149 atm

P($I_2$) =  a

        = 0.0149 atm