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Law of Mass Action

The equilibri...

Question

The equilibrium concentrations of x, y, and z are 4, 2, and $2 m o l L^{- 1}$ , respectively, at equilibrium of the reaction $2 x + y ⇌ z$ . The value of K is:

\frac{1}{16} m o l^{- 2} L^{- 2}

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16 m o l^{- 2} L^{- 2}

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32 m o l^{- 2} L^{- 2}

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None of these

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Solution

The correct option is D $\frac{1}{16} m o l^{- 2} L^{- 2}$
The equilibrium concentrations of x, y, and z are 4, 2, and $2 m o l L^{- 1}$ , respectively, at equilibrium of the reaction $2 x + y ⇌ z$ . The value of K is $\frac{1}{16} m o l^{- 2} L^{- 2}$
The expression for the equilibrium constant is $K = \frac{[z]}{[x]^{2} [y]}$
Substitute values in the above expression.
$K = \frac{2 m o l / L}{(4 m o l / L)^{2} \times 2 m o l / L} = \frac{1}{16} m o l^{- 2} L^{- 2}$

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The equilibrium concentrations of x, y, and z are 4, 2, and 2molL−1, respectively, at equilibrium of the reaction 2x+y⇌z. The value of K is:

The equilibrium concentrations of x, y, and z are 4, 2, and $2 m o l L^{- 1}$ , respectively, at equilibrium of the reaction $2 x + y ⇌ z$ . The value of K is: