⇦ Current
⇨ Marginal Cost
Rate of Reaction
A chemical reaction results in the formation of one or more
substances (called products) from one or more starting materials (called
reactant). For instance, the “equation”
2H 2  O2  2H 2O
indicates that two molecules of hydrogen and one molecule of oxygen
form two molecules of water. Let’s consider the reaction
A B C
The concentration of a reactant A is the number of moles (6.022 x 1023
molecules) per liters and is denoted by [A]. Then concentration varies
during a reaction, so [A], [B], [C] are all functions of time t. The average
rate of reaction of the product C over a time interval from t1 to t2 is
[C ] [C ]( t 2 )  [C ]( t1 )

t
t 2  t1
The instant rate of reaction is
C d [C ]

t  0  t
dt
lim
Since the concentration of product increases as the reaction proceeds, the
d [C ]
derivative dt will be nonnegative. The concentrations of reactants,
however, decreases during reaction, so, to make the rates of reaction of A
and B positive numbers, we put minus signs in front of the derivatives
d [ A]
d [B]
dt and dt . Since each decreases at the same rate that [C] increases,
we have
d [C ]
d [ A]
d [B]


dt
dt
dt
More generally, if a reaction is given by
aA  bB  cC  dD
we have

1 d [ A]
1 d [ B ] 1 d [C ] 1 d [ D ]



a dt
b dt
c dt
d dt
The data in the following table concern the lactonization of
hydroxyvaleric acid at 25。C. They give the concentration C(t) of this acid
in moles per liter after t minutes.
T
0
2
4
6
8
C(t)
0.0800
0.0570
0.0408
0.0295
0.0210
Here is a plot of the data :
The average rate of reaction for the following time intervals:
(i) 2  t  6
C (6)  C ( 2) 0.0295  0.0570

 0.0 0 6 8 7( moles
5
/ L) m i n
62
4
(ii) 2  t  4
C ( 4)  C ( 2) 0.0408  0.0570

 0.0081 ( moles / L) min
42
4
(iii) 0  t  2
C ( 2)  C (0) 0.0570  0.0800

 0.0115 ( moles / L) min
20
2
We can find a smooth curve to fit the data by applying the least square
method. The red curve in the following graph is given by the equation
y  0.00004166666667t 3  0.001105357143t 2 0.01355119048t  0.08000285714
We can use the tangent at t = 2 to estimate the instantaneous rate of
reaction when t = 2.
So we get
dC
0.078

 0.0099  0.01 (moles/L) min
dt
7.9
which is the instantaneous rate of reaction when t = 2.
⇦ Current
⇨ Marginal Cost