K equilibria in reversible reactions

Question

6. Consider the reversible chemical reaction A + B --> AB

You mix 1 mole of A with 1 mole of B in a 1 L container.

At equilibrium, the amount of AB will be:

(a)exactly zero moles (b)exactly one mole(c)exactly two moles(d)between 0 and 1 moles(e)between 1 and 2 moles.

If instead of being reversible, K is 5x10^32.

How much AB will form?

(a)exactly zero moles (b)exactly one mole(c)exactly two moles(d)between 0 and 1 moles(e)between 1 and 2 moles.

If instead of being reversible, K is 2x10^-19. How much AB will form? (a)exactly zero moles (b)exactly one mole(c)exactly two moles(d)between 0 and 1 moles(e)between 1 and 2 moles.

Answer 1

Short answers first: (d), then (b), then (a). Here is the reasoning, plus one honest caveat that is worth understanding.

Setup for all three: A + B in equilibrium with AB, starting at 1 M each in a 1 L container. If x moles of AB form, then at equilibrium [AB] = x and [A] = [B] = 1 - x, so

K = x / (1 - x)^2

Note the ceiling: because the stoichiometry is 1:1:1 and you started with one mole of each, x can never exceed 1. That alone eliminates choices (c) and (e) in every part of this problem - there is no way to make two moles of AB from one mole of A.

Part 1 - a reaction described only as "reversible"

Reversible means the reverse reaction is happening at a meaningful rate, so some A and B always remain. x is therefore greater than 0 but less than 1.

Answer: (d), between 0 and 1 moles.

Part 2 - K = 5 x 10^32

A huge K means the numerator dominates, so almost all of A and B is consumed. Solve for what is left over:

(1 - x)^2 = x / K, and with x very close to 1 this is (1 - x)^2 = 1 / (5 x 10^32) = 2 x 10^-33

1 - x = 4.5 x 10^-17 mol

So x = 1 - 0.000000000000000045 mol. No balance on earth could tell that apart from one mole.

Answer: (b), exactly one mole - the reaction goes to completion for all practical purposes.

Part 3 - K = 2 x 10^-19

A tiny K means the reaction barely proceeds. Here x is very small, so (1 - x) is essentially 1 and

x = K(1 - x)^2 = 2 x 10^-19 mol

Answer: (a), exactly zero moles for any practical purpose - you are left with essentially the starting materials.

The caveat worth knowing

Strictly speaking, every one of these three is mathematically (d). Equilibrium never delivers exactly zero and never delivers exactly one; those are limits that x approaches but never reaches. In part 2 there are still about 27 million molecules each of A and B floating in that liter, and in part 3 about 120,000 molecules of AB have formed. Neither number is zero.

What the question is really testing is the practical judgment chemists actually use: a K larger than about 10^3 means "goes to completion," a K smaller than about 10^-3 means "does not happen," and only in between do you need an ICE table. That is the takeaway, and it is why (b) and (a) are the intended answers even though a purist would say (d) three times.