Partial pressures of $A, B, C$ and $D$ on the basis of gaseous system

$A + 2 B$ $⇌$ $C + 3 D$ 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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3.75

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Solution

The correct option is B
18.75

$A$ + $2 B$ $⇌$ $C$ + $3 D$

$K$ = $\frac{[p C] [p D]^{3}}{[p A] [p B]^{2}} = \frac{0.3 \times 0.5 \times 0.5 \times 0.5}{0.2 \times 0.1 \times 0.1}$ = $18.75$

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Similar questions

Q. 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-

In the reversible gaseous reaction,
A + 2B ⇌ C + 3D
The partial pressure of A, B, C and D are 0.20, 0.10, 0.30 and 0.50 atm. respectively at equilibrium. The numerical value for K_p is:

Q. For a system of gases A, B, C, and D at equilibrium

A + 2 B ⇌ C + 3 D

, the partial pressures are found to be

A = 2.0, B = 2.0, C = 3.

0, and

D = 5.0 a t m

. The value of equilibrium constant is :

For the reaction A+2B $⇌$ C, the expression for equilibrium constant is

For the system 3A + 2B $⇌$ C, the expression for equilibrium constant is

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

The correct option is B 18.75 A + 2B ⇌ C + 3D K = [pC][pD]3[pA][pB]2 = 0.3×0.5×0.5×0.50.2×0.1×0.1 = 18.75

Partial pressures of $A, B, C$ and $D$ on the basis of gaseous system

$A + 2 B$ $⇌$ $C + 3 D$ are $A = 0.20; B = 0.10; C = 0.30$ and $D = 0.50 a t m$ . The

numerical value of equilibrium constant is

The correct option is B
18.75

$A$ + $2 B$ $⇌$ $C$ + $3 D$

$K$ = $\frac{[p C] [p D]^{3}}{[p A] [p B]^{2}} = \frac{0.3 \times 0.5 \times 0.5 \times 0.5}{0.2 \times 0.1 \times 0.1}$ = $18.75$