What is the equation for the equilibrium constant (Kc) for the following reaction? (1/2) A(g)+(1/3) B(g) [T(K)] (2/3) C(g)

Question

What is the equation for the equilibrium constant (Kc) for the following reaction? (1/2) A(g)+(1/3) B(g) <=>[T(K)] (2/3) C(g) (A) Kc=([A]1 / 2[B]1 / 3/[C]3 / 2) (B) Kc=([C]3 / 2/[A]2[B]3) (C) Kc=([C]2 / 3/[A]1 / 2[B]1 / 3) (D) Kc=([C]2 / 3/[A]1 / 2+[B]1 / 3)

Tardigrade Admin · Accepted Answer

(1/2) A(g)+(1/3) B(g) <=>[T(K)] (2/3) C(g) Equilibrium constant Kc=( text rate of product / text rate of reactant ) Kc=([C]2 / 3/[A]1 / 2[B]1 / 3)

Q. What is the equation for the equilibrium constant $(K_{c})$ for the following reaction?
$\frac{1}{2} A (g) + \frac{1}{3} B (g) T (K) \frac{2}{3} C (g)$

$K_{c} = \frac{[ A ] ^{1/2} [ B ] ^{1/3}}{[ C ] ^{3/2}}$

$K_{c} = \frac{[ C ] ^{3/2}}{[ A ] ^{2} [ B ] ^{3}}$

$K_{c} = \frac{[ C ] ^{2/3}}{[ A ] ^{1/2} [ B ] ^{1/3}}$

$K_{c} = \frac{[ C ] ^{2/3}}{[ A ] ^{1/2} + [ B ] ^{1/3}}$

$\frac{1}{2} A (g) + \frac{1}{3} B (g) T (K) \frac{2}{3} C (g)$
Equilibrium constant
$K_{c} = \frac{rate of product}{rate of reactant}$
$K_{c} = \frac{[ C ] ^{2/3}}{[ A ] ^{1/2} [ B ] ^{1/3}}$

Q. What is the equation for the equilibrium constant (Kc​) for the following reaction? 21​A(g)+31​B(g)T(K)​32​C(g)

Kc​=[C]3/2[A]1/2[B]1/3​

Kc​=[A]2[B]3[C]3/2​

Kc​=[A]1/2[B]1/3[C]2/3​

Kc​=[A]1/2+[B]1/3[C]2/3​