The answer is B but do not know how they are getting to the answer. Please expla

Question

The answer is B but do not know how they are getting to the answer. Please explain process. Thank you.

Consider the equilibrium represented as A(aq)e2B(aq) which has a Ko 4.0. If 2.0 mol A and 3.0 mol B are introduced into a 1.0 L flask, what change(s) in concentrations (if any) will occur in time? CA) System is not at equilibrium so [A] increases and B increases (B) System is not at equilibrium so Al increases and B decreases (C) System is not at equilibrium so A decreases and IB] increases (D) System is not at equilibrium so A decreases and B decreases (E) System is at equilibrium so Al and B both remain unchanged.

Explanation / Answer

Initially,

[A] = number of mol of A / volume

= 2.0 mol / 1.0 L

= 2.0 M

[B] = number of mol of A / volume

= 3.0 mol / 1.0 L

= 3.0 M

Qc = [B]^2 / [A]

= (3.0)^2 / [2.0]

= 4.5

Kc = 4.0

Since Qc is greater than Kc, reaction will move in backward direction to reach equilibrium

So, [B] will decrease and [A] will increase

Answer: B

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.