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Derivation of Kp and Kc

Partial press...

Question

Partial pressure of $A, B, C$ and $D$ on the basis of gaseous system at equilibrium $A + 2 B ⇌ C + 3 D$ are $A = 0.20; B = 0.10; C = 0.30$ and $D = 0.50 a t m$ .
The numerical value of equilibrium constant is-

11.25

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18.75

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5

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3.75

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Solution

The correct option is A $18.75$
The equilibrium reaction is $A + 2 B ⇌ C + 3 D .$
The expression for the equilibrium constant is $K_{p} = \frac{[C] [D]^{3}}{[A] [B]^{2}} .$
Substitute the values of $A$ , $B$ , $C$ and $D$ are in the above expression.

$A = 0.20; B = 0.10; C = 0.30$ and $D = 0.50$ .

$K_{p} = \frac{(0.3) (0.5)^{3}}{(0.2) (0.1)^{2}} = 18.75.$

The correct option is $B$

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numerical value of equilibrium constant is

In the reversible gaseous reaction,
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Partial pressure of A,B,C and D on the basis of gaseous system at equilibrium A+2B⇌C+3D are A=0.20;B=0.10;C=0.30 and D=0.50atm. The numerical value of equilibrium constant is-

The correct option is A 18.75The equilibrium reaction is A+2B⇌C+3D.The expression for the equilibrium constant is Kp=[C][D]3[A][B]2.Substitute the values of A, B, C and D are in the above expression.A=0.20;B=0.10;C=0.30 and D=0.50. Kp=(0.3)(0.5)3(0.2)(0.1)2=18.75.The correct option is B

Partial pressure of $A, B, C$ and $D$ on the basis of gaseous system at equilibrium $A + 2 B ⇌ C + 3 D$ are $A = 0.20; B = 0.10; C = 0.30$ and $D = 0.50 a t m$ .
The numerical value of equilibrium constant is-