Question

[URGENT!] - Equilibrium Problem

Consider the following reaction:

4A + 2B ⇌ 2C + D (all gases at STP)

Initially, the reaction system is at equilibrium, with [A] = [B] = 2.60 M, and [C] = [D] = 3.10 M, making K = 0.0964. The reaction takes place in a 5.00-liter container. Moles of B are removed until the new equilibrium concentration of C is 2.70. How many moles of B were removed?

Answer 1

The equilibrium constant expression for this reaction is:

K = [C]2[D] / [A]4[B]2

We can rearrange this expression to solve for [B]:

[B] = √([A]4[C]2[D] / K)

Plugging in the given values, we get:

[B] = √((2.60 M)4(2.70 M)2(3.10 M) / 0.0964)

= √(6.76 M)

= 0.26 M

The initial concentration of B is 2.60 M, so the change in concentration of B is 2.60 M - 0.26 M = 2.34 M.

Since the reaction takes place in a 5.00-liter container, the number of moles of B that were removed is 2.34 M * 5.00 L = 11.7 moles.

Therefore, 11.7 moles of B were removed.