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

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

In the reacti...

Question

In the reaction $A (g) + B (g) ⇌ 2 C (g)$ , $2 m o l e s$ of $A$ , $3.0 m o l e s$ of $B$ and $2.0 m o l e s$ of $C$ are placed in a 2 litre flask. If equilibrium concentration of $C$ is $0.5 m o l e / L$ then, the value of equilibrium constant ( $K_{c}$ ) will be:

0.073

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

0.147

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

0.05

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

0.026

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

Open in App

Solution

The correct option is D $0.05$
$A + 2 B ⟶ 2 C$

At t=0 2 3 2
At t=t 2-x 3-2x 2x+2

Now, $[C] = 0.5 m o l / L = \frac{(2 x + 2)}{2} m o l / L$
$x = - 0.5$
Put $x = - 0.5$
Thus, $A = 2.5 m o l e s$
$B = 4 m o l e s$
$C = 1 m o l e$

Now,
$K = \frac{[C]^{2}}{[A] [B]^{2}}$

$K = \frac{[1 / 2]^{2}}{[2.5 / 2] [4 / 2]^{2}}$

$K = \frac{1}{20}$

$K = 0.05$

Suggest Corrections

Similar questions

Q. In a reaction

A + 2 B ⇋ 2 C; 2 m o l e s o f A, 3

mole of

B a n d 2 m o l e o f C

are placed in a 2 L flask and the equilibrium concentration of

C i s 0.5 m o l e / L .

The equilibrium constant K for the reaction is:

Q. When

3.00 m o l e

A

and

1.00 m o l e

B

are mixed in a

1.00 l i t r e

vessel, the following reaction takes place:

A (g) + B (g) ⇋ 2 C (g)

The equilibrium mixture contains

0.5 m o l e

C

. What is the value of equilibrium constant for the reaction?

Q. For the reaction,

A (g) + 2 B (g) ⇋ 2 C (g)

, one mole of A and

1.5

moles of B are taken in a

2.0

litre vessel. At equilibrium, the concentration of C was found to be a

0.35 m o l L^{- 1}

. The equilibrium constant

K_{C}

of the reaction would be:

Q. In the reaction,

A + 2 B ⇌ 2 C

, if

2

moles of A,

3.0

moles of B and

2.0

moles of C are placed in a

2 L

flask and the equilibrium concentration of C is

0.5 m o l / L

. The equilibrium constant

(K_{c})

for the reaction is:

Q. Given the reaction:

A_{(g)} + B_{(g)} ⇌ 2 C_{(g)} + D_{(g)}

One mole of

A

and one mole of

B

are placed in a

1.0

litre vessel. After equilibrium has been established,

0.4

mole of

C

is present in the vessel. Calculate

K_{c}

for the reaction.

Join BYJU'S Learning Program

In the reaction A(g)+B(g)⇌2C(g), 2 moles of A, 3.0 moles of B and 2.0 moles of C are placed in a 2 litre flask. If equilibrium concentration of C is 0.5 mole/L then, the value of equilibrium constant (Kc) will be:

The correct option is D 0.05 A+2B⟶2CAt t=0 2 3 2At t=t 2-x 3-2x 2x+2Now, [C]=0.5mol/L=(2x+2)2mol/Lx=−0.5Put x=−0.5Thus, A=2.5 molesB=4 molesC=1 moleNow,K=[C]2[A][B]2K=[1/2]2[2.5/2][4/2]2K=120K=0.05

In the reaction $A (g) + B (g) ⇌ 2 C (g)$ , $2 m o l e s$ of $A$ , $3.0 m o l e s$ of $B$ and $2.0 m o l e s$ of $C$ are placed in a 2 litre flask. If equilibrium concentration of $C$ is $0.5 m o l e / L$ then, the value of equilibrium constant ( $K_{c}$ ) will be: