Question

1 mol of Hl converts into H2 and I2 at 400 K. Value
of Kp will be, if 2/3rd of HI left at gaseous
equilibrium 2HI <=> H2 + I2
(1)1/4
(2) 1/2
(3) 1/16
(4) 1​

Answer 1

Answer:

Explanation:

Refer to the attachment

Answer 2

Answer: The value of $K_p$ for the given equation is $\frac{1}{16}$

Explanation:

We are given:

Initial moles of HI = 1 mole

Moles of HI left unreacted = $\frac{2}{3}\times 1=\frac{2}{3}mol$

For the given chemical equation:

                  $2HI\rightarrow H_2+I_2$

Initial:           1        -    -

At eqllm:     1-2x       x    x

Calculating the value of 'x' from unreacted HI:

Moles of HI unreacted = 1 - 2x

Equating the values, we get:

$1-2x=\frac{2}{3}\\\\1-\frac{2}{3}=2x\\\\x=\frac{1}{6}$

The expression of $K_p$ for the above equation follows:

$K_p=\frac{p_{H_2}\times p_{I_2}}{p_{HI}^2}$

We are given:

$p_{H_2}=\frac{1}{6}\\\\p_{I_2}=\frac{1}{6}\\\\p_{HI}=1-[2\times (\frac{1}{6})]=\frac{2}{3}$

Putting values in above expression, we get:

$K_p=\frac{\frac{1}{6}\times \frac{1}{6}}{(\frac{2}{3})^2}\\\\K_p=\frac{1}{36}\times \frac{9}{4}=\frac{1}{16}$

Hence, the value of $K_p$ for the given equation is $\frac{1}{16}$