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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

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0.147

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0.05

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0.026

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

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

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

$K = 0.05$

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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: