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Chemistry Question on Equilibrium

What is the equation for the equilibrium constant (Kc) (K_{c}) for the following reaction? 12A(g)+13B(g)<=>[T(K)]23C(g)\frac{1}{2}A(g) + \frac{1}{3}B(g){ <=>[T(K)]} \frac{2}{3}C(g)

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

12A(g)+13B(g)<=>[T(K)]23C(g)\frac{1}{2}A(g) + \frac{1}{3}B(g){ <=>[T(K)]} \frac{2}{3}C(g)
Equilibrium constant, Kc= Rate product  Rate of reactant K_{c}=\frac{\text { Rate product }}{\text { Rate of reactant }}
Kc=[C]2/3[A]1/2[B]1/3K_{c}=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}