Question

Consider the general reversible reaction. Lower A upper A plus lower B upper B double-headed arrow lower C upper C plus Lower d upper D. What is the equilibrium constant expression for the given system? K e q equals StartFraction lowercase C StartBracket upper C EndBracket lowercase d StartBracket upper D EndBracket over lowercase A StartBracket upper A EndBracket lowercase B StartBracket upper B EndBracket EndFraction. K e q equals StartFraction StartBracket upper C EndBracket StartBracket upper D EndBracket over StartBracket upper A EndBracket StartBracket upper B EndBracket EndFraction. K e q equals StartFraction StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b over StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D EndFraction. K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

Answer 1

Answer:

K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

Explanation:

In order to aid understanding of the question, let me put the words into a proper chemical reaction equation.

aA + bB ⇄cC + dD

We must recall that the equilibrium constant often designated as Keq is given as;

Keq= product of the concentration of product species raised to the power of their respective molar coefficients / product of the concentration of reactant species raised to the power of their respective molar coefficients

Hence from what was written above, I can safely write;

Keq= [C]^c [D]^d / [A]^a [B]^b

This has been represented in words in the answer given above.

Answer 2

Answer:

ANSWER IS D

Explanation:

Keq= [C]^c [D]^d / [A]^a [B]^b