You visited us 1 times! Enjoying our articles? Unlock Full Access!

Characteristics of Equilibrium Constant

What is the c...

Question

What is the concentration of $C O$ in equilibrium at $25^{o} C$ in a sample of a gas originally containing $1.00 m o l$ $L^{- 1}$ of ${C O}_{2}$ ? For the dissociation of ${C O}_{2}$ at $25^{o} C$ .
$K_{c} = 2.96 \times 10^{- 92}$

Open in App

Solution

$C O_{2} (g) ⇋ 2 C O (g) + O_{2} (g)$
At equi. $(1 - 2 x)$ $(2 x)$ $(x)$
Applying law of mass action
$K_{c} = \frac{[O_{2}] {[C O]}^{2}}{{[O_{2}]}^{2}} = 2.96 \times 10^{- 92}$
It can be assumed that $1 - 2 x = 1.0$ as $K_{c}$ is very small.
So, $4 x^{3} = 2.96 \times 10^{- 92}$
or $x = 1.956 \times 10^{- 31} m o l L^{- 1}$
$[C O] = 2 x = 3.90 \times 10^{- 31} m o l L^{- 1}$

Suggest Corrections

Similar questions

C O_{2}

C O_{2}

25^{o} C

C O_{2}

25^{o} C

3.1 \times 10^{- 2}

3 O_{2} (g) ⇌ 2 O_{3} (g)

1.7 \times 10^{- 56}

25^{o} C

25^{o} C

O_{3}

O_{2}

O_{3}

O_{2}

25^{o} C

8 \times 10^{3} M

O_{3}

K_{c}

3 O_{2} (g) ⇌ 2 O_{3} (g)

2.0 \times 10^{- 50}

25^{o} C

O_{2}

25^{o} C

1.6 \times 10^{- 2}

O_{3}

{C O}_{2}

1

2.4

25^{o} C

K_{H}

{C O}_{2}

= 29.76 a t m / m o l e / l

25^{o} C

C O_{2}

1 l i t r e

25^{o} C ? (

K_{H}

C O_{2}

= 9.6 a t m / m o l / l

25^{o} C

Join BYJU'S Learning Program