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

Law of Mass Action

Consider the ...

Question

Consider the two gaseous equilibrium involving $S O_{2}$ and the corresponding equilibrium constant at $299 K S O_{2} (g) +^{1 / 2} O_{2} (g) ⇌ S O_{3} (g); K_{1}$ $4 S O_{3} (g) ⇌ 4 S O_{2} (g) + 2 O_{2} (g); K_{2}$
The value of the equilibrium constant are related by :

K_{2} = \frac{1}{(K_{1})^{4}}

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

K_{2} = K_{1}^{4}

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

K_{2} = {(\frac{1}{K_{1}})}^{1 / 4}

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

K_{2} = \frac{1}{K_{1}}

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

Open in App

Solution

The correct option is A $K_{2} = \frac{1}{(K_{1})^{4}}$
$K_{1} = \frac{(S O_{3})}{(S O_{2}) (O_{2})^{1 / 2}} \Rightarrow K_{2} = \frac{(S O_{2})^{4} (O_{2})^{2}}{(S O_{3})^{4}} = \frac{1}{(K_{1})^{4}}$
$K_{2} = \frac{1}{(K_{1})^{4}}$

Suggest Corrections

Similar questions

{S O}_{2}

{S O}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g); K_{1}

4 {S O}_{3} (g) ⇌ 4 S O_{2} (g) + {2 O}_{2} (g); K_{2}

S O_{2}

S O_{2} (g) + 1 / 2 O_{2} (g) ⇌ S O_{3} (g); K_{1}

4 S O_{3} (g) ⇌ 4 S O_{2} (g) + 2 O_{2} (g); K_{2}

S O_{2}

S O_{2} (g) + \frac{1}{2} O_{2} (g) \Leftrightarrow S O_{3} (g); K_{1}

2 S O_{3} (g) \Leftrightarrow 2 S O_{2} (g) + O_{2} (g); K_{2}

S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

K_{1}

K_{2}

298 K

K_{1}

K_{2}

For the two gaseous reactions (i) and (ii), at $25^{0} C$

$(i) N O (g) + \frac{1}{2} O_{2} (g) ⟵ N O_{2} (g)$

$(i i) 2 N O_{2} (g) ⟵ 2 N O (g) + O_{2} (g)$ ,

The equilibrium constants are $K_{1}$ and $K_{2}$ respectively. these two constants are related as:

Join BYJU'S Learning Program