Skills to be Tested
When enter numerical values, use a form accepted by computers. For example, 0.000123 and 1.23e-4 are both fine. For further information, read the module CACT conventions.
Energy drives the chemical reactions. At some point, a reaction can go in either direction. A system at this condition is said to be at equilibrium.
For chemical equilibrium, let us consider a general chemical reaction of the type:
[C]c [D]d
---------------- = Keq
[A]a [B]b
The square brackets "[ ]" around the chemical species represent their
concentrations. This is the ideal law of chemical equilibrium or law
of mass action.
The unit for K depends on the units used for concentration. If M is used for all concentrations, K has a unit of
If the system is not at equilibrium, the ratio may be different from the equilibrium constant. In such cases, the quotient is designated as Q.
[C]c [D]d
---------------- = Q
[A]a [B]b
A system will change in such a way that Q approaches the
equilibrium constant Keq.
[NH4+] [OAc-]
--------------------- = K
[NH3] [HOAc]
[H+]2 [SO42-]
------------------- = K
[H2SO4]
[Cu(NH3)62+]
----------------- = K
[Cu2+] [NH3]6
The equilibrium constant depends on the written formula.
Experimentally, it has been shown that
[HC2O4-] [H+]
------------------- = K1 = 0.059
[H2C2O4]
The second ionization constant is much smaller:
[H+] [C2O42-]
-------------------- = K2 = 0.000064
[HC2O4-]
The overall ionization can be obtained by adding (1) and (2) to give (3).
H2C2O4 = 2 H+ + C2O42- - - - - - - - - - (3)
and the equilibrium constant is
[H+]2 [C2O42-]
--------------------- = K3
[H2C2O4]
It is obvious that
K3 = K1 * K2
= 3.8E(-6)
You should confirm the above obvious relationship
to satisfy yourself.
There are questions in the dialogue for practice.
In the DOS version, you press the (Q) key to enter the quiz mode.
Hint...
Since the reaction is reversed, use the relationship
K(reverse) = 1/K(forward).
Try 2e-6, or 0.2e-5.
Answer 2e-6
Consider...
1/(5e5) = 2e-6.
Hint...
C = (n/V) by definition. n R T = P V
Answer e
Consider...
C = (n/V) = P/(RT)
P/(RT) = n/V = C which is concentration, by definition.
Hint...
Reason with yourself before answer. Statement (c), (d), and (e)
may be true for special cases, but they are not generally true.
Answer b
Consider...
For a general case, Kc is proportional to Kp.
Hint...
Two of the statements are correct.
There is another reaction for the said condition.
Read the question carefully. Give all reactions for the said condition.
Answer bd
Consider...
You are right. Both reactions have equal numbers of reactants
and products.