Consider the equilibria 1 and 2 with equilibrium constants K 1 and K 2 respectively- S O 2 g -12 O 2 g - S O 3 g - 1 2S O 3 g - 2S O 2 g - O 2 g - 2 K 1 and K 2 are related as-

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

Consider the ...

Question

Consider the equilibria (1) and (2) with equilibrium constants $K_{1}$ and $K_{2}$ respectively.
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$ ...... (1)
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ ...... (2)
$K_{1}$ and $K_{2}$ are related as:

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is B $K_{1}^{2} = \frac{1}{K_{2}}$
Reaction (2) is twice the reverse reaction of reaction (1)

Therefore equilibrium constants are related as:

$K_{2} = \frac{1}{K_{1}^{2}}$ or $K_{1}^{2} = \frac{1}{K_{2}}$

Because on doubling by changing the coefficient the equilibrium constant term gets squared and on reversing the reaction, it becomes reciprocal.

Suggest Corrections

Similar questions

Q. Two gaseous equilibria

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

and

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

have equilibrium constants

K_{1}

and

K_{2}

respectively at

298 K

. Which of the following relationships between

K_{1}

and

K_{2}

is correct:

Two gaseous equilibria $S O_{2} (g) + 0.5 O_{2} (g) ⇌ S O_{3} (g)$ and

$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ have equilibrium constants $K_{1}$ and $K_{2}$ respectively

at 298K. Which of the following relationships between $K_{1}$ and $K_{2}$ is correct

Q. At

298 K

equilibrium constant

K_{1}

and

K_{2}

of following reaction

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

----- (1)

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

-------- (2)
The relation between

K_{1}

and

K_{2}

is :

Q. The equilibrium constants of the reactions,

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

and

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

are

K_{1}

and

K_{2}

respectively. The relationship between

K_{1}

and

K_{2}

is:

Q. If

K_{1}

and

K_{2}

are the equilibrium constants of the equilibriums

(A)

and

(B)

respectively, then what is the relationship between the two constants?

(A)

S O_{2} (g) + 1 / 2 O_{2} (g) ⇋ S O_{3} (g), K_{1}

(B)

2 S O_{3} (g) ⇋ 2 S O_{2} (g) + O_{2} (g), K_{2}

Join BYJU'S Learning Program

Consider the equilibria (1) and (2) with equilibrium constants K1 and K2 respectively.SO2(g)+12O2(g)⇌SO3(g) ...... (1)2SO3(g)⇌2SO2(g)+O2(g) ...... (2)K1 and K2 are related as:

The correct option is B K21=1K2Reaction (2) is twice the reverse reaction of reaction (1)Therefore equilibrium constants are related as:K2=1K21 or K21=1K2Because on doubling by changing the coefficient the equilibrium constant term gets squared and on reversing the reaction, it becomes reciprocal.

Consider the equilibria (1) and (2) with equilibrium constants $K_{1}$ and $K_{2}$ respectively.
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$ ...... (1)
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ ...... (2)
$K_{1}$ and $K_{2}$ are related as: