Question

What is the equilibrium constant expression for the given system?
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k subscript e q equals StartFraction StartBracket upper H subscript 2 upper O EndBracket superscript 2 over StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket EndFraction.
K subscript e q equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket EndFraction.
K subscript e q equals StartFraction StartBracket upper H subscript 2 EndBracket superscript 2 StartBracket upper O subscript 2 EndBracket over StartBracket upper H subscript 2 upper O EndBracket superscript 2 EndFraction.

Answer 1

The question is incomplete, here is the complete question:

The given chemical equation follows:

$2H_2O\rightleftharpoons 2H_2+O_2$

What is the equilibrium constant expression for the given system?

Answer: The expression of equilibrium constant for the given equation is $K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Explanation:

Equilibrium constant is the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:  

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The given chemical equation follows:

$2H_2O\rightleftharpoons 2H_2+O_2$

The expression for $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Hence, the expression of equilibrium constant for the given equation is $K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Answer 2

Answer:

Its D for edge

Explanation: