Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

Question

3 years ago

15

Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

per O subscript 2 (g). What is the equilibrium constant expression for the given reaction? K subscript e q equals StartFraction bracket upper C a upper O EndBracket over upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartBracket upper C upper O subscript 2 EndBracket. K subscript e q equals StartFraction 1 over StartBracket upper C upper O subscript 2 EndBracket EndFraction.

Chemistry

2 answers:

stiks02 [169] · Answer 1 · 2022-03-29T09:08:42+03:00

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

Len [333] · Answer 2 · 2022-03-29T11:37:26+03:00

Len [333]3 years ago

8 0

Answer: is A

Explanation:pij