The Foundations of Chemistry



0.200

x

x

7.00

x1.407.00x 8.00x1.40 x

1

8

.

.

4

0

0

0

0.175

Now we know the value of x,so the equilibrium concentrations are

[A] (0.200x) M 0.025 M;[C]x M 0.175 M

[B] (0.200x) M 0.025 M; [D] x M 0.175 M

To check our answers we use the equilibrium concentrations to calculate Qand verify that its
value is equal to Kc.

Q 49 Recall that Kc49.0

The ideas developed in Example 17-6 may be applied to cases in which the reactants
are mixed in nonstoichiometric amounts. This is shown in Example 17-7.

EXAMPLE 17-7 Finding Equilibrium Concentrations

Consider the same system as in Example 17-6 at the same temperature. If 0.600 mol of A and
0.200 mol of B are mixed in a 2.00-liter container and allowed to reach equilibrium, what are
the equilibrium concentrations of all species?

Plan

We proceed as we did in Example 17-6. The only difference is that now we have nonstoichio-
metricamounts of reactants.

Solution

As in Example 17-6, we let xmol/L of A that react; then xmol/L of B that react, and x
mol/L of C and D formed.

A  B 34 C  D

initial 0.300 M 0.100 M 0 M 0 M

change due to rxn x M x M x M x M

equilibrium (0.300x) M (0.100x) Mx Mx M

The initial concentrations are governed by the amounts of reactants mixed together. But changes
in concentrationsdue to reaction must occur in the 1
1
1
1 ratio required by the coefficients
in the balanced equation.

Kc49.0 so 49.0

We can arrange this quadratic equation into the standard form.

49.0

x^2 1.4719.6x49.0x^2

48.0x^2 19.6x1.47 0

Quadratic equations can be solved by use of the quadratic formula.

x^2



0.03000.400xx^2

(x)(x)



(0.300x)(0.100x)

[C][D]



[A][B]

(0.175)(0.175)



(0.025)(0.025)

[C][D]



[A][B]

We see that the equilibrium

concentrations of products are

much greater than those of reactants

because Kcis much greater than 1.

17-5 Uses of the Equilibrium Constant, Kc 719

The left side of this equation is nota

perfect square.