Chemical Equations
Some revision and some new features.
Stoichiometry
Consider a chemical reaction:
H2O2 +2HBr + 2FeBr2
→ 2H2O + 2FeBr3
The numbers (1), 2, 2 on the LHS and 2, 2 on
the RHS
of the chemical equation give the ratio of
moles
(or molecules) of the species reacting on the
LHS
and produced on the RHS. These are the
stoichiometry of the reaction.
A more ‘dramatic’ example’:
5H2O2+ 2KMnO4+6HCl → 2MnCl2+8H2O
+5O2 +2KCl
These equations are exact and can be used to
calculate the quantities reacting together.
In the above reaction, if 6.50 grams of H2O2
react, what weight of MnCl2 is formed?
Molecular Wt. H2O2 = (2x1 + 2x16) = 34 amu
Therefore 6.50 grams H2O2 contains 6.50/34
moles H2O2
=
0.191
moles H2O2
5 moles H2O2 react to produce 2 moles MnCl2,
therefore:
0.191 x (2/5) moles MnCl2 produced = 0.076
moles MnCl2
1 mole MnCl2 weighs (54.94+2x35.45) grams
= 125.84 grams, therefore
weight of MnCl2 produced = 0.076 x 125.84
grams
= 9.56 grams
Note: Be aware of limiting quantities. If 1
mole each of H2O2, KMnO4 and HCl are
reacted together then the reaction stops when
the HCl is exhausted at which point 1/3 mole
KMnO4 and 5/6 moles H2O2 have also reacted
and therefore, 2/3 mole KMnO4 and 1/6 moles
H2O2 are left un-reacted.
Measuring moles of pure compounds:
By Weight:
Number of moles of compound
= (weight in grams) / (molecular
weight)
By Volume (if density known):
Weight in grams = Volume x Density
(Units!!)
And from here to moles
In the reaction:
5H2O2+ 2KMnO4+6HCl → 2MnCl2+8H2O
+5O2 +2KCl
if 6.05 grams of KMnO4 reacts, what volume
of O2 is produced at NTP?
6.05 grams KMnO4=(6.05/
(39.10+54.94+4x16.00)) moles
= 0.0383 moles of KMnO4
2 moles KMnO4 reacts to produce 5 moles O2,
therefore
0.0383 x (5/2) moles = 0.0957 moles O2
produced
At NTP, 0.0957 moles of an ideal gas has a
volume
V = (nRT/p) = (0.0957 x 0.082 x 298)/(1) L
(Check Units!)
= 2.34 L
Try many other examples!
Solutions
Pure compounds dissolved in a solvent.
Solutions have a Concentration – number of
moles per litre
9.46 grams of NaCl dissolved in 345 ml of
H2O
9.46 grams NaCl = (9.46 / (22.99+35.45)
moles NaCl
= 0.162 moles NaCl
dissolved in
(345/1000) L = 0.345 L of H2O.
Concentration = 0.162 moles/0.345 L = 0.469
mol L-1
[NaCl(aq)] = 0.47 M
Note: [X] = conc. of X in mol L-1, M = mol L-1
The concentration is the same for any amount
of the solution.
For a solution of a given concentration, find
the number of moles in a given volume by
multiplying concentration by volume.
(Units!!!)
In the reaction:
5H2O2+ 2KMnO4+6HCl → 2MnCl2+8H2O
+5O2 +2KCl
if 250 ml of a 0.15 mol L-1 solution of HCl are
reacted, what weight of KCl is formed?
How many moles HCl? (0.15 mol L1
)x(250/1000 L) moles
= 0.0688 moles HCl
How many moles KCl formed? (0.0688 x 2/6)
moles
= 0.0229 moles KCl
What do these weigh? 0.0299 x (39.10 +
35.45) grams
= 2.23 grams of KCl
Labelling of reacting species:
O2 (g), HCl(aq), H2O(l), S(s), Na+(aq)
Reversible Reactions and Equilibrium
If there is a reaction
A+B+…  C+D+…
Clearly the reaction
C+D+…  A+B+…
also obeys the Laws of Conservation of Mass,
etc…
Some reactions can be driven in either
direction by changing the conditions.
e.g. the room temperature reaction

2NO2 (g)
N2O4 (g)
goes in the reverse direction
N2O4 (g)

2NO2 (g)
at a higher temperature. The reaction is
REVERSIBLE.
There is obviously are temperatures at which
the reaction is going in both directions:
2NO2 (g) 
N2O4 (g) (This is the wrong
symbol but it’s the only one available in
standard WORD)
and there is a balance between reactants and
products so that both are present and their
‘amounts’ (in fact concentrations ) are not
changing. The reaction is in EQUILIBRIUM.
For a general reaction:
mA + nB+ …→ xC + yD + …
define a quantity known as the Reaction
Quotient:
[C ]x [ D] y ...
Q
[ A]m [ B]n ...
concentrations to the power of the
stoichiometry coefficients.
Consider the reaction: 2SO2 + O2  2SO3
If initially [SO2] = 2.0 mol L-1 and [O2] = 1.0
mol L-1
the concentrations will vary with time:
The concentrations and the Reaction Quotient:
[ SO3 ]2
Q
[ SO2 ]2 [O2 ]
becomes a constant when equilibrium is
reached.
The value of Q when equilibrium is reached is
called the EQUILIBRIUM CONSTANT.
For a reaction mixture, if Q < K, the reaction
is going from LEFT  RIGHT, if Q > K the
reaction is going from
RIGHT  LEFT. IF Q = K the reaction is at
equilibrium.
Examples:
For the following reactions with the initial
concentrations shown, and the value of the
equilibrium constant, in what direction is the
reaction going?
2NH3 (g)
 N2 (g) + 3 H2 (g)
[NH3] = 0.200 M, [N2] = 0.100 M, [H2] =
0.100 M, K=17
Q = [N2] [H2]3 / [NH3]2 = 0.1 x 0.13/0.22 =
2.5x10-3
Q < K, reaction going L  R
2NO(g) + Cl2 (g)  2 NOCl (g)
in a 2 liter flask, there are 0.1 mol of NO(g),
2.20 g of Cl2(g) and 4.5 x 1023 molecules of
NOCl(g), K = 4.6 x 104
Find concentrations:
[NO(g)] = 0.1 mol/ 2L = 0.05 mol L-1
2.20 g Cl2(g) = (2.20/(2x35.45)) mol Cl2(g)
=0.031 mol Cl2(g)
[Cl2(g)] = 0.031mol / 2L = 0.016 mol L-1 Cl2(g)
4.5 x 1023 molecules of NOCl(g) =
(4.5 x 1023 / 6.022 x 1023) mol NOCl(g)
= 0.747 mol NOCl(g)
[NOCl(g)] = 0.747 mol / 2L = 0.373 mol L-1
NOCl(g)
Q = [NOCl(g)]2 / [NO(g)]2 [Cl2(g)] = 0.3732/
(0.052x0.016)
= 3478.5 < K , therefore the reaction is
going L  R.
Equilibrium is reached only if the reactants
and products do not ‘escape’.
For example, at 900oC
CaCO3 (s)  CaO (s) + CO2 (g) (the
manufacture of lime) is industrially achieved
by drawing off the CO2 (g).
Calculating Equilibrium Constants
Given the concentrations of reactants and
products at equilibrium – substitute in
Reaction Quotient expression.
e.g.
At a certain temperature the reaction:
2SO2(g) + O2(g)  2SO2(g)
has equilibrium concentrations [SO2(g)] =
0.590 M,
[O2(g)] = 0.045 M, [SO3(g)] = 0.260 M. The
value of the equilibrium constant
K = [SO3(g)]2 / ([SO2(g)]2 [O2(g)]) = 0.2602 /
(0.5902 x 0.045)
= 4.315
Units – apparently for the above case, M-1 =
mol-1 L. The apparent units of Equilibrium
Constant depend on the stoichiometry of the
reaction.
H2 + I2  2HI , K – unitless
N2 + 3H2  2NH3 , K – M-2
2Cl2 + 2H2O  4HCl + O2 , K - M
Generally no Units are quoted for K.