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 Dr Roger Nix (Queen Mary, University of London) - 1.1
CHAPTER 1 : BASIC SKILLS
Mathematics with Symbols
Physical quantities are often represented by symbols - these may be the normal Latin characters ( a, b, c,
.... , A, B, C, ... etc.) or Greek characters ( α, β, γ, ... etc.). This greatly simplifies the writing out and
manipulations of equations - for example
Momentum = Mass × Velocity
Becomes
p=m×v
or more simply still
p=mv
where:
p mv -
represents momentum
represents mass
represents velocity
The above example illustrates how products of physical quantities may be written in a symbolic form in
several different ways. More generally, products may be written as
ab
or
ab
or
a⋅b
or
a×b
whilst quotients (division of quantities) may be written in any of the following ways:
a/b
or
a
b
or
a÷b
or
a b–1
1. Equations may often contain a mixture of Latin and Greek symbols.
e.g.
c = νλ
2. Symbols may be lower or upper case and may have different meanings - for example:
- p is used to represent momentum whereas upper case P is generally used for pressure.
- t is used to represent time whereas upper case T is used for temperature.
3. Symbols may be used to represent both constants (physical quantities whose values are fixed, e.g. the
speed of light, c , or the ideal gas constant, R ) and variables (physical quantities whose values may
vary, e.g. the wavelength, λ , of radiation).
4. When typed, Latin characters used to represent quantities should be italicised - however,
- the Latin v is sometimes not italicised, as it begins to look like the Greek nu , ν .
- symbols may not be italicised if the typist is careless / lazy / short of time !
 Dr Roger Nix (Queen Mary, University of London) - 1.2
Order of Evaluation of Expressions
•
In the absence of brackets, the order of evaluation is:
powers before multiplication/division before addition/subtraction.
These rules for the order of evaluating expressions also apply when using symbolic mathematics
a+b×c
e.g.
=
a + bc
the multiplication of b and c is carried out before the addition to a
To avoid any ambiguity brackets (parentheses) should be used, since the expressions inside the brackets
are always evaluated first (i.e. the brackets override the normal order of priority described above).
For example, to indicate in the above expression that a should be added to b before the multiplication by
c it should be re-written as follows:
(a + b) × c
or
(a + b)c
They may also be used, for example, to clearly differentiate
(a + b)/c
from
a + (b/c)
d [a + (b/c)]
from
d a + (b/c)
Commutative & Associative Laws
•
When several quantities are multiplied together, the multiplications may be carried out in
any order.
i.e. multiplication obeys both the commutative and associative laws:
Commutative Law :
xy = yx
Associative Law:
(x y) z = x (y z)
 Dr Roger Nix (Queen Mary, University of London) - 1.3
Brackets
Multiplication
The notation A(x + y) is shorthand for Ax + Ay , i.e.
A(x + y) = Ax + Ay = xA + yA = (x + y)A
This concept may be extended to any number of terms added or subtracted within the bracket, e.g.
A(x + y – z) = Ax + Ay – Az.
Note also the following application of this rule:
x(2y) = x(y + y) = xy + xy = 2xy
.. which confirms that multiplication is both commutative and associative, since
x(2y) =
=
=
=
x × (2 × y)
(x × 2) × y
(2 × x) × y
2xy
- assuming multiplication is associative
- assuming multiplication is commutative
... which has already been demonstrated to be true.
For two brackets multiplied together, repeated application of the rule gives:
(A + B) (x + y) = A(x + y) + B(x + y) = Ax + Ay + Bx + By
or, equally
(A + B) (x + y) = (A + B) x + (A + B) y = Ax + Bx + Ay + By
Note especially the manipulation of signs in the next two examples:
(A + B) (x – y)
=
=
A(x – y) + B(x – y)
Ax– Ay + Bx – By
(A – B) (x – y)
=
=
=
A(x – y) – B(x – y)
Ax – Ay – Bx – B(–y)
Ax – Ay – Bx + By
The complete removal of brackets from an expression is known as expansion ; the opposite process
whereby a common factor is taken out and brackets are introduced is known as factorisation .
Factorisation can always be carried out when every term has a common factor, e.g.
 Dr Roger Nix (Queen Mary, University of London) - 1.4
Ax + Ay + Az = A (x + y + z)
but can also be carried out in other cases, e.g.
(Ax + 2Ay + 3z) = A( x + 2 y + 3 z )
A
One special type of factorisation that you will come across is that associated with certain types of
quadratic function, e.g.
x2 + 2ax + a2 = (x + a)(x + a) = (x + a)2
x2 – a2 = (x + a)(x – a)
Division
Division of brackets follows the same rules because division is equivalent to multiplication by the
reciprocal of the divisor, e.g.
( x + y + z)
x y z
1
+ +
=
(x + y + z) =
5
5
5 5 5
0esting of Brackets
When expanding "brackets within brackets" (nested brackets) then the innermost brackets should be
removed by expansion first of all. When writing expressions involving nested brackets, different types
of brackets are often used to help clarify which brackets together form a pair.
For example,
x − 2[ y − 2(3 + x)]
=
x − 2[ y − 6 − 2 x)]
=
x − 2 y + 12 + 4 x
=
5 x − 2 y + 12
 Dr Roger Nix (Queen Mary, University of London) - 1.5
Fractions
•
A fraction is a ratio of two quantities - the numerator on the top line and the denominator on
the bottom line.
•
A fraction remains unaltered in value if the top and bottom lines are both multiplied by the
same quantity.
i.e.
a  a × n  an
=
=
b  b × n  bn
Note - that the entire contents of the top and bottom lines must be multiplied by the quantity
concerned.
i.e.
a
a×n
an
an
=
=
=
b + c (b + c ) × n (b + c )n bn + cn
The basic rules for combining fractions are summarized below :
Multiplication :
 a   c  ac
 ×  =
 b   d  bd
Division :
a
 
 b  =  a  ×  d  = ad
   
 c   b   c  bc
 
d 
Addition / Subtraction : for two factions to be combined by addition or subtraction they must possess
the same denominator. If so, then
 a   b  (a ± b )
 ±  =
c
c c
If the fractions do not initially have a common denominator, then they must first be manipulated (by
multiplication of top and bottom lines by the same factor) so that they do have identical bottom lines the best choice for the common denominator is usually the lowest common multiple of the denominators
of the individual fractions.
 a   c   ad   cb  (ad + bc )
 +  =  +  =
bd
 b   d   bd   db 
or
 2   1 + x   2 × x   1 + x   2 x   1 + x  (2 x + 1 + x ) 3 x + 1
=
+ 2  =  2 + 2  =
 + 2  = 
x2
x2
 x  x   x× x  x   x   x 
 Dr Roger Nix (Queen Mary, University of London) - 1.6
Manipulation of Algebraic Equations
The key things to remember about an equation are that:
•
an equation is a statement that two things are equal.
•
an equation remains correct as long as you perform the same operation to both sides.
For example, if
x = y
then
y + a
and
x – a
=
x
a
y
a
x + a
=
ax
=
ay
and
xa
=
ya
(both sides raised to the power of a )
1
x
=
1
y
(a special case of the preceding, with a = –1 )
=
y – a
Caution
As noted in the preceding section, if
x = y
then
1
x
=
1
y
but
If
1 1 1
= +
z x y
then
z ≠ x + y
(a far too common error !)
Instead, if you need to take the reciprocal, the fractions on the right-hand side must first be combined,
which (as noted just above) requires them to have a common denominator,
i.e.
1 1
y
x ( x + y)
+ =
+
=
x y xy xy
xy
hence
1 1 1
1 ( x + y)
xy
= +
⇒
=
⇒ z=
z x y
z
xy
( x + y)
 Dr Roger Nix (Queen Mary, University of London) - 1.7
Example
Consider the equation :
d
=
Mu
v
To make v the subject of the equation :
(i)
⇒
(ii)
⇒
multiply both sides by v
vd
=
Muv
v
⇒
vd
= M u
⇒
v =
Mu
d
⇒
v − u = at
⇒
a=
divide both sides by d
vd
=
d
Mu
d
Example
Consider the equation :
v = u + at
To make a the subject of the equation :
(i)
⇒
subtract u from both sides
v − u = (u + at ) − u
(ii)
divide both sides by t
⇒
v − u at
=
t
t
v−u
t
 Dr Roger Nix (Queen Mary, University of London) - 1.8
Miscellaneous Mathematical 0otation
Common Mathematical Symbols
For a table of common mathematical symbols (∞ , ⇒ , ∼ etc.) see the glossary located inside the backcover of this manual.
Subscripts
Subscripted numbers or letters have no mathematical meaning (whereas superscripts do have) ... they are
simply labels.
For example, if a molecule may occupy a series of energy levels of increasing energy then the energies of
these levels may be denoted ε1 , ε2, ε3 , etc.
In this example, εi denotes the energy of the i-th energy level.
Factorials
By definition:
n ! = n × (n – 1) × (n – 2) × ........ ×1
Example :
3! = 3×2×1 = 6
Special cases :
0! = 1
( a total of n terms)
Summations
When a quantity or function is defined by a sum of many similar terms it is often much more convenient
to write it in an abbreviated fashion: this may be done using the summation symbol, Σ , and by defining
a general term.
e.g.
Consider
F(x) = 1 + x +
x2 x3
+
+ .......
2! 3!
( = ex
- see page 1.18 )
This may be abbreviated to
F(x) =
∞
xn
∑
n = 0 n!
where the numbers above and below the summation symbol represent the limiting values of
the variable, n (i.e. n = 0, 1, 2, 3, 4, ......... - in this case the series is infinite).
xn
is the general term of the summation.
In this example,
n!
 Dr Roger Nix (Queen Mary, University of London) - 1.9
Products
When a quantity or function is defined by a product of many similar terms it is sometimes written in an
abbreviated fashion: this may be done using the product symbol, Π , and by defining a general term.
ν
e.g.
i
 Pi 
∏  P o 
where Pi is the partial pressure of component i and νi is the coefficient in the
stoichiometric chemical equation.
Equilibrium constant, K p =
Limits of Functions
A limit of a function is the value towards which it is tending, as the variable upon which the function
depends approaches some specified value.
1
as x becomes very large ?
x
.... the answer is that as x "tends to" infinity, then f(x) tends to an infinitely small value, i.e. the
"limit" (or limiting value) of the function as x → ∞ is zero.
e.g. What happens to the value of the function
f(x) =
This may be abbreviated in either of the following ways :
• as x → ∞ , f(x) → 0
• lim x →∞ (1 x) = 0
 Dr Roger Nix (Queen Mary, University of London) - 1.10
Indices (Powers)
Definition
xn = x × x × x × ......
e.g. 32 = 3 × 3 = 9;
( n terms to be multiplied out; n is called the index or exponent)
33 = 3 × 3 × 3 = 27;
34 = 3 × 3 × 3 × 3 = 81
Multiplication
In general :
x A.x B = x A+B
Proof by example :
x2 .x3 = ( x.x ) . (x.x.x) = x. x. x. x. x = x5
Numerical example :
32 × 33 = 35 = 243
Division
In general :
xA
= x A− B
xB
Proof by example :
x 3 x.x.x
=
=x
x.x
x2
Numerical example :
33 / 3 2 = 3 1 = 3
( = x1 )
Addition and Subtraction
Expressions such as x A + x B
or x A – x B
cannot in general be simplified further.
However, such expressions can often be factorized, e.g.
xA + xB
= xA ( 1 +
xB
) = x A ( 1 + x (B–A) )
A
x
Power of Zero
It is easy to prove that any quantity raised to the power zero is one:
xA/ xA = 1
(since any number divided by itself gives unity)
but
x A / x A = x(A–A) = x0
⇒
x0 = 1
 Dr Roger Nix (Queen Mary, University of London) - 1.11
0egative Indices
x −A =
In general :
–A
1
xA
( and
(0–A)
= x
1
= xA )
−A
x
x0
1
= A = A
x
x
Proof :
x
Numerical example :
3–2 = 1/32 = 1/9
1/3–2 = 1/(1/9) = 9 = 32
Powers of Powers
In general :
( x A ) B = x AB
Proof by example :
(x3)2 = x3. x3 = (x.x.x).(x.x.x) = x.x.x.x.x.x = x6
Numerical example :
(32 )3 = 36 = 729
Powers of Products
In general :
( xy ) A = x A y A
Proof by example :
(xy)3 = (xy). (xy). (xy) = x.y.x.y.x.y = (x.x.x).(y.y.y) = x 3. y 3
Numerical example :
(2 × 3)3 = 63 = 216
and
(2 × 3)3 = 23 × 33 = 8 × 27 = 216
Fractional Indices
A number raised to a fractional power is called a root. Thus:
2
x1/2 = √ x
3
x1/3 = √ x
the "square root"
the "cube root"
The operation of taking the root is the inverse of taking a number to the corresponding power, i.e. it
undoes the effect of taking the power. The opposite is also true. So, for example,
2
41/2 = √ 4 = 2
3
81/3 = √ 8 = 2
and
;
since
22 = 4
;
since
23 = 8
 Dr Roger Nix (Queen Mary, University of London) - 1.12
3
(x3) 1/3 = √ (x3 ) = x
and
A 1/A
= x
(x )
Proof :
1

 A× 
A

3
(x1/3) 3 = ( √ x ) 3 = x
= x1 = x
Other numerical examples
or
4 3/2 = (43 )1/2 = 64 1/2 = √64
4 3/2 = (41/2 )3 = (√4)3 = 2 3
= 8
= 8
It follows that the re-arrangement of equations involving powers often involves taking fractional powers
of both sides: e.g. if
a3 =
Mu
d
then to make a the subject of the equation, we raise each side to the power 1/3
3 1/3
a = (a )
M u
= 

 d 
1/ 3
Summary of Rules for Manipulating Indices
Note - even though the definition of xn given at the beginning of this section presumes
that n is an integer these rules for manipulating indices apply for all possible
values (integer and non-integer) of the indices.
x A.x B = x A+B
x −A =
1
xA
( x A ) B = x AB
xA
= x A− B
B
x
x1 A = A x
(xy) A = x A y A
 Dr Roger Nix (Queen Mary, University of London) - 1.13
Indexation of Brackets
(xy)n = xy . xy. xy. . . . n times
n. n
= (x.x.x. . . . n times) (y.y.y. . . . n times) = x y
(x + y) 2
= (x + y) (x + y) = x(x + y) + y(x + y) = xx + xy + yx + yy = x2 + 2xy + y2
(x + y) 3
= (x + y) (x + y)2 = (x + y) (x2 + 2xy + y2) = x(x2 + 2xy + y2) + y(x2 + 2xy + y2)
= x3 + 2x2y + xy2 + x2y + 2xy2 + y3
= x3 + 3x2y + 3xy2 + y3
(x + y) 4
= (x + y) (x + y)3 = (x + y) (x3 +3x2y + 3xy2 + y3)
= x(x3 +3x2y + 3xy2 + y3) + y(x3 +3x2y + 3xy2 + y3)
= x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3 + y4
= x4 + 4x3y + 6x2y2 + 4xy3+y4
Note : the pattern of the coefficients that is beginning to emerge - this enables us to predict the
coefficients for higher order expansions using Pascal's triangle.
1
(x + y)
(x + y) 2
(x + y) 3
(x + y) 4
(x + y) 5
(x + y) 6
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Each line of the triangle is obtained from the previous line by first imagining there are zeros on either
end of this previous line, and then summing up the numbers in pairs - in each case the sum is written on
the line below, half way between the two numbers of the pair from which it was derived.
 Dr Roger Nix (Queen Mary, University of London) - 1.14
Scientific 0otation -
Representation of Large and Small 0umbers in Terms of
Powers of Ten
In scientific notation a number is expressed as the product of two numbers: y × 10n. The first term, y , is
usually a number between 1 and 10 ; the second term, 10n, is some integer power of 10. Thus:
1234 = 1.234 × 10 × 10 × 10 = 1.234 × 103
Similarly, a number less than 1 can be written as a number between 1 and 10 divided by a power of 10, which
is equivalent to multiplication by a negative power of 10. Thus:
0.01234 =
1.234
1.234
=
= 1.234 × 10–2
2
10 × 10
10
Why do we do this ? - it is simply a matter of convenience when working with very large and very small
numbers. For example,
6.0 × 1023 is easier to write than 60000000000000000000000
1.2 × 10–8 is easier to write than 0.000000012
Other examples:
1000 = 1 × 103
1234 = 1.234 × 103
100 = 1 × 102
123 = 1.23 × 102
10 = 1 × 101
12 = 1.2 × 101
1 = 1 × 100
1/10 = 1 × 10–1
0.12 = 1.2 × 10–1
1/100 = 1 × 10–2
0.012 = 1.2 × 10–2
1/1000 = 1 × 10–3
0.0012 = 1.2 × 10–3
For numbers greater than 1 the value of the index n is the number of places by which the decimal point
is moved to the left to obtain the number in scientific notation:
1
2
3
4
↑ ↑ ↑
5 .
= 1.2345 × 104
↑
For numbers less than 1 the negative value of the index n is the number of places by which the decimal
point is moved to the right to obtain the number in scientific notation:
0 .0
1
↑ ↑
2
3
4 = 1.234 × 10–2
 Dr Roger Nix (Queen Mary, University of London) - 1.15
To convert a number in scientific notation to the usual form the procedures are inverted:
5 .6
7
8 × 102 = 567.8
↑ ↑
0
0
5 . 6 7 8 × 10–3 = 0.005678
↑ ↑ ↑
0ote: 1 × 10n is usually abbreviated to 10n
, e.g. 103 = 1 × 103 = 1000
Examples of Calculations using Scientific 0otation
In the following examples, the final answers are adjusted to a precision appropriate to the type of
calculation and the number of significant figures specified for the numbers involved - this topic is dealt
with in the Essential Skills for Chemists course.
Addition and Subtraction
When adding or subtracting two numbers they must first be converted to the same power of ten:
[1.234 × 10–3 ] + [5.678 × 10–2 ] = [0.1234 × 10–2 ] + [5.678 × 10–2 ] = (0.1234 + 5.678) × 10–2
= 5.8014 × 10–2
= 5.801 × 10–2
[1.234 × 102 ] – [5.678 × 104 ]
(to 4 s.f)
= [0.01234 × 104 ] – [5.678 × 104 ] = (0.01234 – 5.678) × 104
= –5.66566 × 104
= –5.666 × 104
(to 4 s.f)
Multiplication
The first terms are multiplied together and the powers of ten are added together (cf. x A.x B = x A+B ):
[1.234 × 103 ] × [5.678 × 102 ]
= (1.234 × 5.678) × (103 × 102 ) = 7.006 × 105
[6.023 × 1023 ] × [2.34 × 10–2 ]
= (6.023 × 2.34) × (1023 × 10–2 )
= 14.1 × 1021 = 1.41 × 1022
 Dr Roger Nix (Queen Mary, University of London) - 1.16
Division
The first terms are divided and the powers of ten are subtracted (cf. x A / x B = x A–B ):
1.234 × 10 3 1.234 10 3
=
× 2 = 0.2173 × 101 = 2.173
2
5.678 10
5.678 × 10
6.023 × 10 23 6.023 10 23
=
×
= 4.881 × 10 25
1.234 × 10 − 2 1.234 10 − 2
Powers
The first terms are raised to the power and the powers of ten are multiplied (cf. ( x A ) B = x AB ):
[1.23 × 103 ]2
= [1.23]2 × 10(3×2) = 1.51 × 106
[5.67 × 10–4 ]3 = [5.67]3 × 10(–4×3) = 182. × 10–12 = 1.82 × 10–10
 Dr Roger Nix (Queen Mary, University of London) - 1.17
Significant Figures and Decimal Places
It is sometimes necessary (or desirable) to specify a number to only a certain precision : there are two
common methods to do this. Numbers are said to "be rounded to" or "given to" a certain number of
significant figures (s.f.) or decimal places (d.p.).
Fixed Decimal Place Format
This format may only be used when numbers are written using normal decimal notation - it cannot be
applied to numbers written in scientific notation. The "number of decimal places (d.p.)" corresponds to
the number of digits after (to the right of) the decimal point. The value of the last digit given depends
upon the value of the following digit prior to the rounding. If the following digit :
(i) falls in the range 0-4 then the last digit is left unaltered
(ii) falls in the range 5-9 then the last digit is increased by one (i.e. rounded up)
Examples:
Unrounded
123.613
3.14159
0.001356
–0.51823
To 0 d.p.
124.
3.
0.
–1.
To 2 d.p.
123.61
3.14
0.00
–0.52
To 4 d.p.
3.1416
0.0014
–0.5182
Fixed Significant Figure Format
This format may be used when numbers are written either using normal decimal notation or in scientific
notation - it is a more generally useful format. The "number of significant figures " (sig.fig, or s.f.)
corresponds to the number of significant digits given in the non-exponent part of the number (excluding
any leading zero-digits). The value of the last digit given depends upon the value of the following digit
prior to the rounding. If the following digit :
(i) falls in the range 0-4 then the last digit is left unaltered
(ii) falls in the range 5-9 then the last digit is increased by one (i.e. rounded up)
Examples:
Unrounded
123.613
3.14159
0.0013561
–0.0159352
1.38066 × 10–23
2.99792 × 108
To 2 s.f.
120
3.1
0.0014
–0.016
1.4 × 10–23
3.0 × 108
To 4 s.f.
123.6
3.142
0.001356
–0.01594
1.381 × 10–23
2.998 × 108
 Dr Roger Nix (Queen Mary, University of London) - 1.18
The Exponential Function
The exponential function is an extremely important function in science and nature; it appears in many
scientific equation including, for example, in the Boltzmann equation (which defines population
distributions of particles amongst energy levels) and in various integrated rate equations in chemical
kinetics.
Definition : the exponential function is denoted by ex
ex = 1+ x +
x2 x3
+
+ ......
2! 3!
i.e.
and is defined by the infinite series :
ex =
∞
xn
∑
n = 0 n!
Key points:
• it may also be written as exp x or exp(x)
..... but not E
• this function is also an example of a power: i.e. ex is " e taken to the power of x "
• e is a real number, approximately equal to 2.72 (see below)
The value of the real number e may be obtained by setting x = 1 in the definition.
e = e1 = 1 + 1 +
1 1
1 1
+ + ...... = 1 + 1 + + + ...... = 2.7182818 (to 8 s.f.)
2! 3!
2 6
Combinations of exponentials may therefore be manipulated in exactly the same fashion as any other
power, i.e. using the rules summarized on page 1.12.
e2x. ex = e(2x + x) = e3x
e.g.
(e2x)3 = e6x
The graphical form of the exponential function is illustrated below:
f(x) = e x = exp(x) : for x > 0
160.0
140.0
Limits :
120.0
f(0) = 1
f(x) → ∞ as x → ∞
100.0
f (x )
e0 = 1
80.0
60.0
1ote : other functions of the type
40.0
f(x) = n x (with n >1) will have a
20.0
similar form.
0.0
0.0
1.0
2.0
3.0
x
However, in chemistry it generally occurs in one of the forms illustrated overleaf.
4.0
5.0
 Dr Roger Nix (Queen Mary, University of London) - 1.19
The standard exponential decay function:
f(x) = exp(–x)
: for x > 0
1.2
1.0
Limits :
f(0) = 1
0.8
f(x) → 0 as x → ∞
f (x )
Example :
0.6
0.4
Change in reactant concentration
0.2
with time for a first order reaction.
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
8.0
10.0
x
The negative exponential function for the reciprocal of a variable:
f(x) = exp(–1/x)
: for x > 0
1.0
0.9
0.8
Limits :
0.7
f(x) → 0 as x → 0
f(x) → 1 as x → ∞
0.6
f (x )
0.5
0.4
Example :
Change in population of an energy
level with temperature.
0.3
0.2
0.1
0.0
0.0
2.0
4.0
x
 Dr Roger Nix (Queen Mary, University of London) - 1.20
Logarithms
Definition :
If
x = ny
then
y = log n x
and
y is said to be the logarithm of x to the base n.
(where n is some pure number)
General characteristics
The operations log n and taking " n to the power of ..." cancel each other out if applied successively the functions are therefore a pair of inverse functions.
i.e.
log n (n y ) = y
and
n log n x = x
Important : you can only take the log of a dimensionless quantity (a quantity without units) - the result is
also dimensionless.
Mathematical operations with logarithms
It is sometimes necessary to rearrange expressions involving logarithms without finding their actual values
using a calculator. The rules to use apply to all types of logarithms and they are:
log(x y)
= log(x) + log(y)
log(x / y)
= log(x) – log(y)
log(x b )
= b . log(x)
from which it also follows that ....
log(1)
= 0
log(1 / y)
= – log(y)
These rules may be used in combination to simplify expressions involving multiple log terms - for example:
C c Dd 
c log(C) + d log(D) – a log(A) – b log(B) = log  a b 
A B 
 Dr Roger Nix (Queen Mary, University of London) - 1.21
In principle n can be any number but only two types of logarithms are in common usage:
1.
Logarithms to the base 10 [written lg or log or log10 ], where n = 10.
… so from the general definition : if
log10 ( 10 y ) = y
whilst
2.
and
x = 10 y
then
y = log 10 x
10 log x = x
"Natural” logarithms [written ln], which are logarithms to the base of the number known as “e”
( e = 2.71828...., the “exponential").
… so from the general definition : if
ln ( e y ) = y
whilst
and
x = ey
then
y = ln x
e ln x = x
Given a logarithm, it is sometimes necessary to find the number which has this logarithm (sometimes
called the antilogarithm or antilog). To do this we need to apply the inverse function, i.e. we simply need
to take the power of the base number. For instance:
If
ln x = 1.234 , then what is x ?
log 10 x = 7.28 , then what is x ?
x = e ln x = e1.234 = 3.435
x = 10log x = 107.28 = 1.91 × 107
Similarly, equations involving logs can be re-arranged by application of the inverse power function
e.g.
⇒
⇒
∆G o = − RT ln K
− ∆G o
= ln K
RT
K=
− ∆G o
e RT
whilst equations involving exponents can be re-arranged by application of the inverse log function.
k = Ae
e.g.
⇒
⇒
⇒
e
− Ea
RT
− Ea
RT
=k
A
− Ea
k
= ln 
RT
 A
k
E a = − RT ln 
 A
 Dr Roger Nix (Queen Mary, University of London) - 1.22
Special characteristics of the log10 and ln functions.
As noted above if x = 10 y
then
y = log 10 x
This property is useful for relating the values of logs with the exponents of numbers written using
scientific notation. For example suppose we have a number x where 1023 < x < 1024, then
23 < log 10 x < 24.
e.g. log 10 (6.0 × 1023 ) = log 10 (6.0) + log 10 (1023 ) = 0.78 + 23 = 23.78
Natural logarithms have special properties which will emerge from calculus. The values of natural logs are
also related to the log10 values. The relationship between them is:
ln(x) = ln(10) × log10 (x) = 2.303 log10 (x)
Logarithms and hydrogen ion concentrations
As noted above, the logarithm to the base 10 is a convenient function for representing quantities which
vary over many orders of magnitude. A good chemical example is the representation of hydrogen ion
concentration by means of the pH scale. In aqueous solution the hydrogen ion is best represented as
H3O+ (as it exists almost exclusively in this form) and, making certain approximations and using the
chemists’ normal units for concentration, then the pH value is given by:
 [H 3 O + ] 

pH = − log10 
-3 
 mol dm 
Note - [H3O+ ] is a concentration and has units of mol dm–3 - the units will therefore cancel on the top
and bottom lines to give a pure number and the log of this pure number may be taken. The pH value
obtained is also dimensionless.
In aqueous solution [H3O+ ] values typically range from 1.0 mol dm–3 (in 1.0 mol dm–3 strong acid, e.g.
HCl) to 1.0 × 10–14 mol dm–3 (in 1.0 mol dm–3 strong alkali, e.g. NaOH). The corresponding pH values
range from 0 (strong acid) to 14 (strong alkali).
Other log scales (p-scales)
More generally, the "p" is used in various abbreviations to represent an operator which corresponds to
calculating the negative log of any quantity that follows the p-symbol – examples include pH, pOH, pKa,
pKb, pKw.
For example, pKa, = – log10 Ka.
 Dr Roger Nix (Queen Mary, University of London) - 1.23
Solving Equations
A single equation can generally be solved if the values of all variables except one are known - a solution
is obtained by finding the value for this last variable which satisfies the equation.
Quadratic Equations
Quadratic equations are equations of a single variable (say x ) which may be arranged into a form which
only involves positive powers of x up to order 2 , i.e. they are equations of the general form :
a x2 + b x + c = 0 , where a , b and c are constants.
There are two standard methods for solving such equations :
1.
Some (but not all) quadratic equations may be readily factorized upon inspection to give a product
of two linear expressions in x
e.g. x2 + 3 x + 2 = 0
2.
⇒
( x + 2 )( x + 1 ) = 0
⇒
(x+2) = 0
or
(x+1) = 0
⇒
x = –2
or
x = –1
All quadratics may be solved using the general solution which for a quadratic of the form :
a x2 + b x + c = 0 is given by:
x=
− b ± b 2 − 4ac
2a
Note - a polynomial equation of order n always has n solutions (e.g. a quadratic, with n = 2 , has two
solutions).
Equations may need some rearrangement before they are clearly revealed as quadratic equations which
can then be solved by one or other of the two methods described above.
e.g.
2 x = ( x − 2) 2
⇒
e.g. w4 – 3 w2 + 2 = 0
2x = x2 – 4x + 4
⇒
x2 – 3x + 2 = 0
⇒
x2 – 6x + 4 = 0
( where x = w2 )
 Dr Roger Nix (Queen Mary, University of London) - 1.24
Simple Simultaneous Equations
A set of n independent simultaneous equations may be solved to obtain the values of n variables. The
simplest case is when n = 2 i.e. two equations and two unknown variables.
2x
3x
e.g.
+
–
3y
y
=
=
7
5
To obtain the values of x and y
Step 1 : multiply the equations by constants so as to obtain the same coefficient for one of the two
variables.
2x + 3y = 7
3x – y = 5
multiply both sides by 3
multiply both sides by 2
6 x + 9 y = 21
6 x – 2 y = 10
Step 2 : subtract one equation from the other; the variable which has the same coefficient in both
equations will cancel out.
6 x + 9 y = 21
6 x – 2 y = 10
(6x – 6x) + (9y – (–2y)) = 21 – 10
⇒
11 y = 11
Step 3 : solve for the remaining variable, then substitute this value back into either of the original
equations to get the value of the other variable.
11 y = 11
⇒
⇒
⇒
y = 1
2 x + (3×1) = 7 ⇒
x = 2
i.e. the solution is x = 2 and y = 1
2x = 4
 Dr Roger Nix (Queen Mary, University of London) - 1.25
Other Equations
The best approach in general is to first rearrange the equation to make the unknown the subject of the
equation; then and only then substitute in the values for the other variables and constants to obtain the
value of the unknown (and its units !).
Example : Calculate the velocity of an electron which has a kinetic energy of 1.602 × 10–15 J.
[ me = 9.109 × 10–31 kg ]
The relevant equation is :
Ek = ½ m v2
Step 1 : make the velocity, v , the subject of the equation.
2 Ek
m
2 Ek
v=
⇒
m
Step 2 : substitute in the known values for m and Ek .
Ek = ½ m v2
v=
⇒
2 × 1.602 × 10 −15 J
=
9.109 × 10 −31 kg
v2 =
2 × 1.602 × 10 −15
J
kg m 2 s -2
15
3
.
5174
10
×
=
×
×
kg
kg
9.109 × 10 −31
⇒ v = 5.931 × 107 m s–1
Example : Calculate the hydrogen ion concentration in a solution with pH = 5.1
The relevant equation is :
 [H 3 O + ] 

pH = − log10 
-3 
 mol dm 
Step 1 : make [H3O+] the subject of the equation.
 [H 3 O + ] 
 ⇒
pH = − log10 
-3 
 mol dm 
⇒
⇒
 [H 3 O + ] 

− pH = log10 
-3 
 mol dm 
 [H 3 O + ] 
− pH

= 
10
-3 
 mol dm 
[H 3 O + ] = 10 − pH mol dm -3
Step 2 : substitute in the known value for pH .
[H 3 O + ] = 10 − pH mol dm -3 = 10 −5.1 mol dm -3
⇒ [H 3 O + ] = 10 − pH mol dm -3 = 7.9 × 10 −6 mol dm -3
 Dr Roger Nix (Queen Mary, University of London) - 1.26
Partial Fractions
If the bottom line (denominator) of a fraction is a polynomial function (e.g. it contains powers of x ) and
this polynomial may be factorized, then the fraction can be transformed into a sum of simpler fractions the so-called partial fractions.
The procedure for doing this is illustrated by the following example:
e.g.
Now
5
x + x−6
2
x 2 + x − 6 = ( x − 2)( x + 3)
so it should be possible to write the fraction as the sum of two
simpler fractions, where the denominators of these partial fractions are the factors of the polynomial.
i.e.
5
A
B
=
+
2
x + x − 6 ( x − 2) ( x + 3)
where A and B are constants whose values we need to find. We can do this, by combining the two
partial fractions using the normal procedures for addition of fractions:
A
B
A( x + 3)
B ( x − 2)
A( x + 3) + B ( x − 2) A( x + 3) + B ( x − 2)
+
=
+
=
=
( x − 2) ( x + 3) ( x − 2)( x + 3) ( x + 3)( x − 2)
( x + 3)( x − 2)
x2 + x − 6
and we know that this must be equal to
5
, whatever the value of x might be.
x + x−6
2
So,
A( x + 3) + B ( x − 2) = 5
Consider
(i) x = 2 ,
(ii) x = –3 ,
then
then
for all values of x .
⇒
⇒
5A+0B = 5
0A–5B = 5
A = 1
B = –1
i.e.
−1
5
1
1
1
=
−
=
+
x + x − 6 ( x − 2) ( x + 3) ( x − 2) ( x + 3)
2
One application of partial fractions is in simplifying the integration of functions (see section 5):
e.g.
∫x
2
5
dx =
+ x−6
1
∫ ( x − 2) dx
−
1
∫ ( x + 3)dx
 Dr Roger Nix (Queen Mary, University of London) - 1.27
Proportionality
Proportionality is concerned with expressing a simple relationship between two physical quantities
(i.e. between the two variables representing such quantities in a mathematical equation) when all other
variables which might otherwise affect their values are kept constant.
The relationship may be a linear one, i.e.
y = constant × x
y is then said to be (directly) proportional to x , a relationship which is written : y ∝ x
y = constant / x
The relationship may be a reciprocal one, i.e.
y is then said to be inversely proportional to x , a relationship which is written : y ∝ 1/x
Example 1 : The ideal gas equation ( PV = n RT ) may be re-arranged to express how the gas pressure,
P , is related to the other gas variables (volume and temperature) when the amount of gas involved is
fixed (i.e. n is constant). To do this:
•
we first make P the subject of the equation.
•
if the volume, V , is kept constant
i.e.
P ∝ T
⇒ P=
nRT
V
 nR 
⇒ P=
 × T = constant × T
V 
(with constant n and V )
e.g. if you double T then P will also double.
•
if the temperature, T , is kept constant
i.e.
P ∝ 1/V
⇒ P = (nRT ) ×
1
1
= constant ×
V
V
(with constant n and T )
e.g. if you double V then P will halve in value.
Example 2 : The equation for the vibrational frequency of a molecule, ν , exhibits a more complex type
of proportionality relationship.
1
2π
k
The equation for the vibrational frequency is:
ν=
It follows that if k is constant, then
ν = constant ×
or
ν ∝
i.e. ν is inversely proportional to the square root of µ .
µ
1
µ
1
µ
 Dr Roger Nix (Queen Mary, University of London) - 1.28
This second proportionality relationship may be used to calculate the effect of isotopic substitution on
vibrational frequencies. Two isotopic variants of the same molecule, with reduced masses µ1 and µ2 ,
will have the same value of k (force constant). Their vibrational frequencies (ν1 and ν2 ) are therefore
related by:
ν ∝
1
ν2
=
ν1
⇒
µ
µ1
µ2
Where does this come from?
The proportionality relationship indicates that:
ν = constant ×
ν µ = constant
i.e.
So in the comparison of two molecular states when k is constant:
ν 1 µ1 = ν 2 µ 2
and rearrangement gives:
ν2
=
ν1
µ1
µ2
So, for example, if µ2 = 2 µ1
ν2
=
ν1
⇒
ν 2 = ν1
µ1
=
2 µ1
ν
1
= 1
2
2
1
2
(≈ 0.707ν 1 )
1
µ
 Dr Roger Nix (Queen Mary, University of London) - 1.29
Equations of Classical Physics
The following is a list of some of the key equations of classical physics which are also routinely used in
chemistry.
Classical Mechanics
Velocity
=
change in distance / time
v=
dx
dt
Acceleration
=
change in velocity / time
a=
dv
dt
Momentum
=
mass × velocity
p = mv
Force
=
mass × acceleration
F = ma
Work (Energy)
=
=
force × distance moved
pressure × change in volume
W = Fx
W = P ∆V
Kinetic Energy
=
½ × mass × velocity 2
=
½ × momentum 2 / velocity
Ek = ½ m v2
p2
Ek =
2m
=
force per unit area = force / area
P=
=
wavelength × frequency
c = λν
Density
=
mass / volume
ρ=
Heat energy
=
heat capacity × change in temperature
q = C ∆T
Pressure
F
A
Electromagnetic Radiation
Velocity
Properties of Matter
m
V
Ideal Gas Equation:
Pressure × volume = no. of moles × ideal gas constant × temperature
PV = nRT
or
Pressure × volume = no. of molecules × Boltzmann constant × temperature
PV = 1kT
 Dr Roger Nix (Queen Mary, University of London) - 1.30
SI Units
(Système International D’ Unitès)
Base Units
The system depends upon certain base units. These form a set of independent units in terms of which all
physical quantities can be expressed in a consistent manner. The base units most frequently used in
chemistry are the following.
Physical Quantity
Length
Mass
Time
Electric current
Temperature
Amount of substance
0ame of SI Unit
metre
kilogram
second
ampere
kelvin
mole
Symbol for SI Unit
m
kg
s
A
K
mol
Note - by convention, the abbreviations for units named after people (e.g. Ampere, Kelvin) begin with a
capital letter.
Derived Units
Each physical quantity is denoted by only one SI unit, which is either the appropriate base unit, or the
appropriate derived unit formed by multiplication or division of two or more base units, some examples
of commonly-used derived units are given below, together with their definitions in terms of the base
units, i.e. their dimensions
Physical Quantity
Force
Energy (work)
Electric charge
Electric potential
Frequency
Pressure
Power
0ame of
SI unit
newton
joule
coulomb
volt
hertz
pascal
watt
Symbol for
SI unit
Definition in Base Units
(alternative formulation)
kg m s –2
kg m2 s–2
As
kg m2 s–3 A–1
s–1
kg m–1 s–2
( = N m –2 )
kg m2 s–3
( = J s –1)
N
J
C
V
Hz
Pa
W
Note :
(i)
different units are separated by spaces, e.g. N m –2
(ii)
we often enclose a quantity in square brackets [ ] if we wish to refer to its units, e.g.
What are the units of kT ?
⇒
[kT] = J K–1 K
not
Nm –2 ;
= J
J s –1 not Js –1
(i.e. the units of energy).
 Dr Roger Nix (Queen Mary, University of London) - 1.31
0on-SI Units
The most common non-SI units which you are likely to meet are the Ångstrom ( Å ) and the calorie
( cal ).
(i)
an Ångstrom is a unit of length , which is very convenient for bond distances.
1 Å = 10–10 m
= 100 × 10–12 m = 100 pm
= 0.1 × 10–9 m = 0.1 nm
(ii)
a calorie is a unit of energy : it is inconvenient, but it is still quite widely used in the US.
1 cal = 4.184 J
How to Work Out Dimensions
The dimensions of any quantity can be worked out from the physical law which defines it. For example,
=
[ Distance ] / [ Time ]
= m/s
= m s –1
[ Acceleration ] =
[ Velocity ] / [ Time ]
= m s –1 / s
= m s –2
[ Force ]
[ Mass ] × [ Acceleration ] = kg × m s –2
[ Velocity ]
=
= kg m s –2
[ Energy ] = [ Work ] = [ Force ] × [ Distance ] = kg m s –2 × m
[ Pressure ]
=
[ Force ] / [ Area ]
= kg m s –2 / m2
= kg m2 s –2
= kg m–1 s –2.
Prefixes
To avoid numerical values being inconveniently small or large, fractions or multiples of SI units may be
constructed using the following prefixes.
Factor
Prefix
Symbol
10–15
femto
f
10–12
pico
p
10–9
nano
n
10–6
micro
µ
10–3
milli
m
10–2
centi
c
10–1
deci
d
103
kilo
k
106
mega
M
Note :
1)
Most of the prefix symbols are lower case (e.g. m ) but a few are upper case (e.g. M )
109
giga
G
 Dr Roger Nix (Queen Mary, University of London) - 1.32
2)
Prefixes are written directly attached to the unit they modify,
mg = milligram = 10–3 g
whereas products of separate units are written with a space between them. Thus
m g = metre × gram
3)
The base unit for mass (kg) is unusual in that it contains a prefix. Fractions and multiples of it are
formed by adding the appropriate prefix to the word gram and the symbol g .
Thus
µkg and
mg not
Mg not
kkg
Representation of Physical Quantities
In general,
Physical Quantity = 1umerical Value × Unit
e.g. Mass = 760 × kg
which is normally just written as
Mass = 760 kg
Without its units, a physical quantity is meaningless (unless it happens to be dimensionless). It is
conventional to leave a small space between the numerical value and the unit.
e.g. Length = 12 m
not
12m
In calculations, you should normally convert all quantities into their standard SI units (without prefixes)
before entering them into the equation. For a “well-behaved” equation the answer will then also be in
the appropriate SI units. However, if you are uncertain, you can also derive (check) the units for the
quantity to be calculated using the relevant equation at the same time as you perform the numerical
calculation, and always give the units as well as a numerical value for your answer.
e.g.1
If :
Density = mass / volume
m = 3261 kg , V = 0.431 m3 - what is the density ?
ρ=
m 3261 kg
3261 kg
=
=
×
= 7566 kg m -3
3
V 0.431 m
0.431 m 3
Kinetic Energy = ½ × mass × velocity 2
e.g.2
If :
( Ek = ½ m v2 )
m = 9.109 × 10–31 kg , v = 2.04 × 107 m s–1 - what is the kinetic energy ?
Ek =
⇒
( ρ =m/V)
1
9.109 × 10 -31 × (2.04 × 10 7 ) 2
× (9.109 × 10 -31 kg) × (2.04 × 10 7 m s -1 ) 2 =
× kg × (m s -1 ) 2
2
2
E k = 3.79 × 10 −16 kg m 2 s -2 = 3.79 × 10 −16 J
 Dr Roger Nix (Queen Mary, University of London) - 1.33
Amount of Substance and Molecular Masses
The Mole
1 mole (SI symbol mol ) of a substance contains as many specified elementary entities (atoms,
molecules, etc ) as there are atoms in exactly 12 g of the pure 12C isotope of carbon.
Avogadro Constant
The number of elementary entities in a mole is the Avogadro constant, 1A (or L ) = 6.022 × 1023 mol–1.
Example
3 moles (3 mol) of hydrogen ( H2 molecules) contain 3 × 6.022 × 1023 H2 molecules.
3 moles (3 mol) of hydrogen atoms contain 3 × 6.022 × 1023 H atoms.
Masses from Moles and Moles from Masses
The molar mass of substance X is conventionally denoted M and given in the non-SI units of g mol–1 ;
it is then numerically equal to the relative molecular mass (RMM, Mr ) of X.
The RMM is the sum of the relative atomic masses (RAM) of the atoms in the molecule – where the
values of the RMM and RAM are based on the assumption that all isotopes of elements are present in
their natural abundances.
An amount, n moles , of substance X will have a mass m (in g ) where
m = n×M
[ Check: units of m = mol × g mol-1 = g ]
Example
2.0 mol of sodium has a mass of 2.0 mol × 23.0 g mol-1 = 46 g
Conversely, a mass m (in g ) of substance X contains n moles, where
n =
m
M
[ Check: units of n =
g
1
=
= mol ]
-1
g mol
mol -1
Example
1.6 g of oxygen ( O2 molecules) contain
1.6 g
= 0.050 mol of dioxygen molecules.
32 g mol -1
Masses of Individual Molecules
A molecular mass (m) is specific to one molecule and dependent upon the actual isotopes of the elements
present in the molecule – thus the molecular mass of 12C16O is different from 13C16O.
 Dr Roger Nix (Queen Mary, University of London) - 1.34
Calculation of molecular masses from molar masses
If, and only if, a substance is “isotopically pure” (e.g. F2, which is 100% 19F2 ) then the mass of one
molecule of X (the molecular mass mx ) is given by
mX =
molar mass of X per mole
number of molecules of X per mole
mX =
M
1A
=
M
6. 022 × 10 23 mol −1
where mx will have units of g if Mr has its conventional units of g mol–1. To get the mass m in kg
it will be necessary to further divide the value obtained by 1000 in order to convert g to kg .
Calculation of molecular masses using the atomic mass unit
A more reliable method of calculating an accurate value of molecular mass uses the values of the
isotopic masses expressed in atomic mass units.
The atomic mass unit (u) is defined to be 1/12th of the mass of one atom of 12C , i.e.
m(12 C)
1u =
12
and has a value of 1 u = 1.66054 × 10–27 kg (to 6 s.f.)
The molecular mass is then simply given by:
m=
∑ (isotopic masses of atoms)
atoms
where it is necessary to use accurate, tabulated values for the isotopic masses.
Example
The molecular mass of a 1H35Cl molecule is:
m(1H35Cl) = m(1H) + m(35Cl) = 1.0078 u + 34.9688 u = 35.9766 u
= 35.9766 × ( 1.66054 × 10–27 kg ) = 5.97405 × 10–26 kg
If the molar mass of pure 1H35Cl is required then this can be obtained by multiplying by the Avogadro
constant.
Mass of one mole of 1H35Cl = 5.97405 × 10–26 kg × 6.02214 × 1023 mol–1 = 0.035977 kg mol–1
= 35.977 g mol-1
(This is about 0.5 g lower than the molar mass value for “normal” HCl, which contains around 25% of
1 37
H Cl )