K to relative abundance reactants, products

Question

1. The equilibrium constant, K_c, for the following reaction is 7.00×10^-5 at 673 K.

NH₄I(s) NH₃(g) + HI(g)

This reaction is	fill in the blank 1	favored at equilibrium.
Enter PRODUCT or REACTANT.

The concentrations of NH3 and HI will be	fill in the blank 2	at equilibrium.
Enter HIGH or LOW.

2.

Write the equilibrium constant expression, K, for the following reaction:

Please enter the compounds in the order given in the reaction.

If either the numerator or denominator is 1, please enter 1

CrPO₄(s) Cr³⁺(aq) + PO₄^3-(aq)

K =

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3.

Write the equilibrium constant expression, K, for the following reaction:

If either the numerator or denominator is 1, please enter 1

NH₄I(s) NH₃(g) + HI(g)

K =

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Answer 1

Blank 1: REACTANT. Blank 2: LOW.

The size of K is the whole story. K compares products to reactants at equilibrium, so K much larger than 1 means the products dominate, and K much smaller than 1 means the reactants do. Here K = 7.00 × 10^-5, which is far below 1, so at 673 K almost all of the material is still sitting there as solid NH₄I. Very little has decomposed, so the NH₃ and HI concentrations are low. Those two blanks are not independent questions, incidentally: reactant-favoured and low product concentrations are the same fact stated twice.

You can put a number on it, which is worth doing because it turns a vocabulary answer into a real one. Starting from pure solid, the decomposition makes NH₃ and HI in a 1:1 ratio, so at equilibrium their concentrations are equal. Call each x:

K = [NH₃][HI] = x² = 7.00 × 10^-5, so x = 8.37 × 10^-3 M

About eight thousandths molar. That is what LOW means here.

2. CrPO₄(s) ↔ Cr³⁺(aq) + PO₄^3-(aq)

K = [Cr³⁺][PO₄^3-]    (denominator = 1)

3. NH₄I(s) ↔ NH₃(g) + HI(g)

K = [NH₃][HI]    (denominator = 1)

Both denominators are 1 for the same reason, and it is the thing this problem is really testing: pure solids and pure liquids never appear in an equilibrium expression. The reason is not that they are ignored arbitrarily. A pure solid has a fixed number of moles per unit volume set by its density, and that does not change as the reaction proceeds. Grinding up the NH₄I or adding a second scoop of it does not change its concentration, so it cannot shift the equilibrium. Formally its activity is 1, and multiplying by 1 changes nothing.

A practical consequence worth remembering: as long as some solid remains, adding more of it does not push the reaction forward. The only way to change where these equilibria sit is temperature, or removing one of the products.

The expression in part 2 has a name you will meet shortly, K_sp, the solubility product. It is not a special new formula, just this same rule applied to a salt dissolving.