If $33 %$ HI decompose at a particular temperature, $K_{c}$ for the reaction is equal to?
$2 H I ⇌ H_{2} + I_{2}$ .

1 / 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 / 16

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 / 8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1 / 16$
33% means one third of HI decompose as $\frac{33}{100} = \frac{1}{3}$
$2 H I ⇌ H_{2} + I_{2}$
I: 1 0 0
C: $- \frac{1}{3}$ $+ \frac{1}{6}$ $+ \frac{1}{6}$
E: $\frac{2}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

$K_{c} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}}$

$K_{c} = \frac{1 / 6 \times 1 / 6}{2 / 3} = \frac{1}{16}$

Suggest Corrections

Similar questions

If one-third of HI decomposes at a particular temperature $K_{c}$ for the reaction : $2 H I ⇌ H_{2} + I_{2}$ is:

If one third of HI decomposed at a particular temperature. What is the value of K_cfor 2HI<======> H₂ + I₂ is

A) 1/16

B)1/4

C)1/6

D)1/2

33 %

H I

K_{r}

2 H I ⇌ H_{2} + I_{2}

H I ⇌ 1 / 2 H_{2} + 1 / 2 I_{2}

K_{c}

K_{c}

H_{2} + I_{2} ⇌ 2 H I

H I

K_{c}

H I ⇌ H_{2} + I_{2}

Join BYJU'S Learning Program