Predicting the Direction of Reaction -
the Reaction Quotient, Q
We
have already been introduced to the equilibrium constant, Kc,
which indicates the ratio of products to reactants for a reaction at equilibrium.
However, sometimes we need the ratio of
products to reactants at some point other than equilibrium (usually initially).
That is where the reaction quotient, Q,
comes in. Like Kc, Q is
reported as a unitless value.
ex. If 2.00 mol H2, 1.00 mol N2
and 2.00 mol of NH3 are placed in a 1.00 L flask at 472°C,
will N2 and H2 react to form more NH3? (Kc = 0.105)
Answer
Since
the data given is all initial information, we can calculate Q, using the Haber process reaction equation, N2(g) + 3H2(g) ↔
2NH3(g):
[NH3]2
Q
= --------------
[N2][H2]3
2.002
=
---------------
1.00(2.00)3
=
0.500
Remembering
that both Q and Kc are simply ratios of products to reactants:
[P]
Q
or Kc = ------
[R]
Since
Kc = 0.105, the [P]/[R] must change from 0.500 to 0.105 (get smaller), for the
reaction to proceed from its initial state to the one required at equilibrium:
Initial
[P]/[R] = 0.500/1 ↠↠↠↠ Equilibrium [P]/[R] = 0.105/1
Thus,
within this ratio, [P] must decrease while [R] increases.
Therefore,
the reaction will proceed to the left (←) to produce more reactants to reach
equilibrium.
- if Q > K – reaction proceeds left (←) toward reactants
- if Q < K – reaction proceeds right (→) toward products
- if Q = K – reaction is at equilibrium and has no need to shift (−)
Calculating Equilibrium Constants
ex. A mixture of 5.00 x
10-3 mol H2 and 1.00 x
10-2 mol I2 is placed in a 5.00 L flask at 448°C
and allowed to equilibrate. At
equilibrium, [HI] = 1.87 x 10-3 M. Calculate Kc at 448°C.
Answer
This question is just like the one we saw in a prior lesson. So, let's remind ourselves how this works:
Since
the ICE table requires concentration values:
[H2] = 5.00 x 10-3 mol/5.00 L = 1.00 x
10-3 M & [I2] = 1.00 x
10-2 mol/5.00 L = 2.00 x 10-3 M
H2(g) + I2(g) ↔ 2HI(g)
initial [ ] 1.00 x 10-3 2.00 x 10-3 0
change in [
] -0.935 x
10-3 -0.935 x
10-3 +1.87 x
10-3
equilibrium [ ]
0.065 x 10-3 1.065 x
10-3 1.87 x 10-3
[HI]2
Kc = ----------
[H2][I2]
Kc = ----------
[H2][I2]
(1.87 x
10-3)2
= -------------------------------------
(0.065 x 10-3)(1.065 x 10-3)
= -------------------------------------
(0.065 x 10-3)(1.065 x 10-3)
=
50.5
Calculating Equilibrium Concentrations
ex. A 2.000 L flask is filled with 2.000
mol H2 and 4.000 mol I2 at 448°C. At this temperature, Kc =
50.5. What are all the equilibrium
concentrations?
Answer
In
this question, unlike the one above, we aren’t given enough information to fill
in the ICE table completely with numerical data. So, we do what any good mathematician would
do and fill in the unknowns with the “x” variable. We must still bear in mind the stoichiometric
relationships between substances when using “x” variable.
Also, I will not show the calculations for the concentrations we will place in the ICE table from here on out.
H2(g) + I2(g) ↔ 2HI(g)
initial [
] 1.000 2.000 0
change in [ ] -x -x +2x
equilibrium [ ]
1.000 - x 2.000 - x 2x
[HI]2
Kc = ----------
[H2][I2]
Kc = ----------
[H2][I2]
(2x)2
50.5 = --------------------------
(1.000 –x)(2.000-x)
50.5 = --------------------------
(1.000 –x)(2.000-x)
Simplifying
the above equation (FOILing, collecting like terms), gives a quadratic
equation:
50.5
(2 – 3.000x + x2) = 4x2
101.0
– 151.5 x + 50.5x2 = 4x2
46.5x2
– 151.5x + 101.0 = 0
Now,
we need the roots of the quadratic equation.
Let’s hear it for the quadratic formula:
-b ±
(b2 – 4ac)½
x = ----------------------
2a
x = ----------------------
2a
=
2.323 or 0.935
So,
which root do we use?
If we sub 2.323
into the ICE table at the appropriate spots (for instance, [H2]eq
= 1.000 – 2.323 = -1.323 M), we get some weird answers.
So, 2.323 must be the extraneous (bad) root
and 0.935 M is the one we want.
Therefore,
at equilibrium,
[H2]
= 1.000 – 0.935 = 0.065 M
[I2]
= 2.000 – 0.935 = 1.065 M
[HI]
= 2(0.935) = 1.87 M
A Couple of Mathematical Tips
Tip
#1
If
you arrive at a point in your calculation that looks like this:
x2
1.60
= ----------------
(0.100 – x)2
(from the answer to HW # 27, which I take you through completely in this video)
Notice
that you can take the square root of both sides of the equation, which
eliminates the need to use the quadratic formula:
x
1.27
= ----------------
0.100 – x
Much
easier to solve for “x” now.
Tip
#2
If
you arrive at a point in your calculation that looks like this:
x2
3.2
x 10-34 = ----------------
(2.00 – 2x)2
(from the answer to HW
#28, which I take you through completely in this video)
Since,
3.2 x 10-34 (which is the Kc value) is a very small number;
we know that the ratio of [P]/[R] is a very small number. This allows us to assume that x, which
represents the concentration of the products at equilibrium ([H2]
and [I2]), is negligibly small.
Even when multiplied by two (2x), the value is still extremely small. Since this negligible 2x value is being subtracted
from a non-negligible 2.00 in the above equation, we can just remove the 2x
term and the equation becomes easier to solve.
x2
3.2
x 10-34 = ----------
(2.00)2
For
this particular example, we could have used Tip#1 instead. Either way works, just don’t try to use Tip
#1 AND Tip#2 on the same question. One or
the other, please.
I also take you through the answer to HW #30 in this video.
*****Need more help? Check out this video. And this one. This one, too.*****
