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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Chemistry
Chapter
Chemical Equilibrium

JIPMER 2009 · Chemistry · Chemical Equilibrium

What is the equation for the equilibrium constant (K_c) for the...

What is the equation for the equilibrium constant (Kc)\left(K_c\right) for the following reaction? 12A(g)+13B(g)⇌T(K)23C(g)\frac{1}{2} A(g)+\frac{1}{3} B(g) \stackrel{T(K)}{\rightleftharpoons} \frac{2}{3} C(g)

Options

  1. AKc=[A]1/2[B]1/3[C]3/2K_c=\frac{[A]^{1 / 2}[B]^{1 / 3}}{[C]^{3 / 2}}
  2. BKc=[C]3/2[A]2[B]3K_c=\frac{[C]^{3 / 2}}{[A]^2[B]^3}
  3. CKc=[C]2/3[A]1/2[B]1/3K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}
  4. DKc=[C]2/3[A]1/2+[B]1/3K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}+[B]^{1 / 3}}

Correct answer

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

Step-by-step solution

12A(g)+13B(g)⇌T(K)23C(g) Equilibrium constant, Kc= Rate product  Rate of  reactant Kc=[C]2/3[A]1/2[B]1/3\begin{aligned} & \frac{1}{2} A(g)+\frac{1}{3} B(g) \stackrel{T(K)}{\rightleftharpoons} \frac{2}{3} C(g) \\ & \text{ Equilibrium constant, } K_c=\frac{\text{ Rate product }}{\text{ Rate\text{ of } reactant }} \\ & \qquad K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}\end{aligned}
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