) CChieh@UWaterloo.ca

## Quiz 4 - Chemical Equilibrium

### Skills to be Tested

• Calculate equilibrium constants from a given set of conditions
• Calculate one of the unknown values in a system that is at equilibrium state
• Express the equilibrium constant for a given reaction equation
• Give a new equilibrium constant when the reaction equation is written in a different form

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.

## Chemical Equilibrium - a review or introduction

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:

a A + b B -> a C + d D
where

A, B - represent reactants
C, D - represent products
a, b, c, d - coefficients of a balanced chemical equation.
If the system is at equilibrium at a given temperature, then the following ratio is a constant.
```       [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

Mc+d-(a+b) On the other hand, the units are not emphasized for equilibrium constant, and we often do not worry about the units. The equilibrium constants are used for energy calculation. They are thermodynamic properties of chemical reactions. In those calculations, the units of K and units of R (the gas constant) must match. So, some people even considers K a unitless number.

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.

### Applications or Examples

1. You should be able to write the equilibrium expression for any reaction. As an example, the equilibrium constant for the reaction: NH3 + HOAc = NH4+ + OAc-
is
```      [NH4+] [OAc-]
---------------------  =  K
[NH3]  [HOAc]
```
2. For the ionization of an acid, H2SO4 = 2 H+ + SO42- the equilibrium constant may be written as
```       [H+]2  [SO42-]
-------------------  =  K
[H2SO4]
```
3. For the reaction: Cu2+ + 6 NH3 = Cu(NH3)62+
The equilibrium constant is expressed as
```       [Cu(NH3)62+]
-----------------  =  K
[Cu2+] [NH3]6
```
The equilibrium constant depends on the written formula.

4. The ionization of oxalic acid, H2C2O4, has two stages:

Experimentally, it has been shown that

H2C2O4 = H+ + HC2O4- - - - - - - - - - (1)
```    [HC2O4-] [H+]
-------------------  =  K1   =  0.059
[H2C2O4]
```
The second ionization constant is much smaller: HC2O4- = H+ + C2O42- - - - - - - - - - (2)
```    [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.

### Confidence Building Questions

1. If the equilibrium constant for the reaction HCOOH + CN- = HCN + HCOO- is 5e5, what is the equilibrium constant for the reaction HCN + HCOO- = HCOOH + CN-?

Hint...
Since the reaction is reversed, use the relationship K(reverse) = 1/K(forward). Try 2e-6, or 0.2e-5.

Consider...
1/(5e5) = 2e-6.

2. For reactions taking place in gas phase, the equilibrium constant is usually expressed in partial pressures of the reactants and products. If C represents the concentration, and other symbols of the ideal gas equations (P V = nR T) are used, which of the following is correct?
(a) C = RT/(PV)
(b) C = RT/P
(c) C = RT/V
(d) C = RT/(PM)
(e) C = P/(RT)
(f) C = PV/(RT)

Hint...
C = (n/V) by definition. n R T = P V

Consider...
C = (n/V) = P/(RT) P/(RT) = n/V = C which is concentration, by definition.

3. If Kc represents the equilibrium constant of concentration, and Kp represents that of partial pressure, which one of the following is correct?
(a) Kc = Kp
(b) Kc is proportional to Kp
(c) Kc = 1/Kp
(d) Kc is inversely proportional to Kp
(e) Kc * Kp = RT

Hint...
Reason with yourself before answer. Statement (c), (d), and (e) may be true for special cases, but they are not generally true.

Consider...
For a general case, Kc is proportional to Kp.

4. For which two of the following reactions are the equilibrium constants really unitless quantities?
(a) 2 H2 + O2 = 2 H2O
(b) 2 NO = N2 + O2
(c) COCl2 = CO + Cl2
(d) CO + H2O = CO2 + H2