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Question: Answered & Verified by Expert
What is the equation for the equilibrium constant $\left(K_c\right)$ for the following reaction?
$\frac{1}{2} A(g)+\frac{1}{3} B(g) \stackrel{T(K)}{\rightleftharpoons} \frac{2}{3} C(g)$
ChemistryChemical EquilibriumJIPMERJIPMER 2009
Options:
  • A $K_c=\frac{[A]^{1 / 2}[B]^{1 / 3}}{[C]^{3 / 2}}$
  • B $K_c=\frac{[C]^{3 / 2}}{[A]^2[B]^3}$
  • C $K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}$
  • D $K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}+[B]^{1 / 3}}$
Solution:
1905 Upvotes Verified Answer
The correct answer is: $K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}$
$\begin{aligned} & \frac{1}{2} A(g)+\frac{1}{3} B(g) \stackrel{T(K)}{\rightleftharpoons} \frac{2}{3} C(g) \\ & \text { Equilibrium constant, } K_c=\frac{\text { Rate product }}{\text { Rate of reactant }} \\ & \qquad K_c=\frac{[C]^{2 / 3}}{[A]^{1 / 2}[B]^{1 / 3}}\end{aligned}$

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