In the reaction, 3A+2B xrightarrow2C, the equilibrium constant Kc is given by

Question

In the reaction, 3A+2B xrightarrow2C, the equilibrium constant Kc is given by (A)  ([3A]× [2B]/[C])  (B)  ([A]3× [B]/[C])  (C)  ([C]2/[A]3× [B]2)  (D)  ([C]/[3A]× [2B])

Tardigrade Admin · Accepted Answer

3 A+2 B arrow 2 C KC=( text concentration of products / text conentration of reactants ) =([C]2/[A]3 ×[B]2)

Q. In the reaction, $3 A + 2 B 2 C,$ the equilibrium constant $K_{c}$ is given by

$\frac{[ 3 A ] \times [ 2 B ]}{[ C ]}$

$\frac{[ A ] ^{3} \times [ B ]}{[ C ]}$

$\frac{[ C ] ^{2}}{[ A ] ^{3} \times [ B ] ^{2}}$

$\frac{[ C ]}{[ 3 A ] \times [ 2 B ]}$

$3 A + 2 B \to 2 C$
$K_{C} = \frac{concentration of products}{conentration of reactants}$
$= \frac{[ C ] ^{2}}{[ A ] ^{3} \times [ B ] ^{2}}$

Q. In the reaction, 3A+2B​2C, the equilibrium constant Kc​ is given by

[C][3A]×[2B]​

[C][A]3×[B]​

[A]3×[B]2[C]2​

[3A]×[2B][C]​