Consider the following reaction: Ni(CO)₄(g) Ni(s)+ 4CO(g) If the initial concentration of Ni(CO)₄(g)is 1.0 M,and x is the equilibrium concentration of CO(g),what is the correct equilibrium relation?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

Kc = [Ni][CO]⁴/[Ni(CO)₄]

At equilibrium, the concentration of Ni(CO)₄ is equal to (1.0 - x) M (where x is the equilibrium concentration of CO) because one mole of Ni(CO)₄ reacts with four moles of CO. The concentration of Ni is negligible, so we can assume that its concentration at equilibrium is also equal to (1.0 - x) M.

Substituting these values in the equilibrium constant expression, we get:

Kc = x⁴/(1.0 - x)

Multiplying both sides by (1.0 - x) gives:

Kc(1.0 - x) = x⁴

Dividing both sides by 4x gives:

Kc = x/(4(1.0 - x))

This is the same as choice C.

Consider the Following Reaction: Ni(CO)4(g) Ni(s)+ 4CO(g)

Consider the Following Reaction: Ni(CO)₄(g) Ni(s)+ 4CO(g)