The Foundations of Chemistry

(Marcin) #1

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.
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