Reaction Order

aA_(g) -----> products
rate = k(conc A)^m
The power(m) to which the concentration of A is raised in the rate expression describes the order of the reaction.

If "m" = 0, the reaction is zero order

If "m" = 1, it is first order

If "m" = 2, it is second order

Conditions:

m = 0: zero order - rate is independent of the concentration of reactant. Doubling concentration has no effect on rate.

m = 1: first order - rate is directly proportional to the concentration of the reactant. Doubling the concentration increases the rate by a factor of 2.

m = 2: second order - the rate is to the square of the concentration of the reactant. Doubling the concentration increases the rate by a factor of 4.

Given the following date for the reaction CH₃CHO ----> CH₄ + CO

conc CH₃CHO (mol/L) 0.10 0.20 0.30 0.40

rate (mol/L^.) 0.085 0.34 0.76 1.4

Order of a reaction must be determined experimentally. It cannot be deduced from the coefficients of a balanced equation.

Two reactant reaction:

aA_(g) + bB_(g) ----> products
rate = k(conc A)^m(conc B)ⁿ
refer to "m" as "the order of the reaction with respect to A" and
"n" as "the order of the reaction with respect to B"
the overall order of the reaction is the sum of the exponents, m + n.
example: CO + NO₂ -----> CO₂ + NO

rate = k(conc CO)(conc NO₂)

This reaction is :

first order with respect to CO, m = 1

first order with respect to NO₂, m = 1

second order overall, m + n = 2

Two reactant reactions are more difficult to monitor. The best way is to hold initial concentration of one reactant constant while varying the concentration of the other reactant.
example: 2H₂ + 2NO ----> N₂ + 2H₂O

rate = k(conc H₂)^m(conc NO)ⁿ
m = 1, as conc H₂ goes from 0.1 to 0.2, the rate doubles from 0.1 to 0.2
n = 2, as conc NO goes from 0.1 to 0.2, the rate goes from 0.1 to 0.4
The rate equation is therefore: rate = k(conc H₂)(conc NO)²
the overall order for the reaction, m + n = 3, is third order.

Reactant Concentration and Time

for aA_(g) ----> products, the rate = k(conc A)
example: 2N₂O₅ ----> 4NO₂ + O₂
rate = k(conc N₂O₅)
log₁₀(conc N₂O₅) = log₁₀(conc N₂O₅)₀ - kt / 2.30

k = rate constant
t = time
0 = original conc N₂O₅ at t = 0

plotting log₁₀ conc N₂O₅ vs time gives a straight line
using the equation of a line, y = a + bx

y = log₁₀ conc N₂O₅

x = time, t

y intercept, a = log₁₀(conc N₂O₅)₀

slope, b = -k

the slope can also be determined from dy/dx. Given the values of -0.60 / 4.0 min, the slope would be = -0.15/min.
Therefore, substituting -0.15/min into -kt/2.30 and solving for k gives k = 2.30(0.15/min) = 0.35/min
Another way of writing this is log₁₀(X / X₀) = kt/2.30

X₀ = original concentration

X = concentration of x at time t

k = first order rate constant

Sample Problem 1: Calculate the concentration of N₂O₅ after 4.0 min starting with a concentration of 0.160 mol/L.

Solution:

log(0.160/x) = (0.35)(4.0)/2.30 = 0.61
log 0.160 - log x = 0.61
-0.80 - log X = 0.61
-log x = 1.41
log x = -1.41
x = 0.040 mol/L

Sample Problem 2: How much time is required for the concentration to drop from 0.160 to 0.100 mol/L?

Solution:

t = (2.30/k)log(X₀/X)= (2.30/0.35)log(0.160/0.100)
t = (2.30)(0.20)/0.35 = 1.3 min

Sample Problem 3: At what time will half of the sample be decomposed?

Solution:

X = X₀/2, therefore X₀ = 2X and X/X₀ = 2
t = (2.30/0.35)log 2 = (2.30)(0.30)/0.35 = 2.0 min

Sample Problem 4: How long will it take for the concentration to drop from 0.160 to 0.080 M?

Solution:

t = (2.30/k)log(X₀/X) = (2.30/0.35) log(0.160/0.080)
t = (2.30)(0.30)/0.35 = 2.0 min

This is important!!

The time required for one half of a reactant to decompose via a first order reaction has a fixed value, independent of concentration. This is called the half-life and has the expression

t_1/2 = 0.693/k
k is the rate constant for the first order reaction
half-life is inversely proportional to the rate constant k
if k is large, half-life is short; a slow reaction with a small k will have a relatively long half-life.

What About Reactions With Other Integral Orders?

Most common processes in gas phase reactions are second order. Zero order reactions are less common.
An example of a zero order reaction is:

2HI_(g) ----> H_2(g) + I_2(g)
rate = k(conc HI)⁰ = k
The reaction occurrs at a constant rate independent of concentration of HI.
Third order reactions are rare. One example of a third order reaction is:

2H<₂ + 2NO ----> N₂ + 2 H₂O

The following table is a summary of important information related to 0, 1st and 2nd order reactions:

Order	rate expression	conc-time relationship	half-life	linear plot
0	rate = k	X₀ - X = k	X₀/2k	X vs t
1	rate = kX	log X₀/X = kt/2.30	0.693/k	log X vs t
2	rate = kX²	1/x - 1/X₀ = kt	1/X₀k	1/X vs t

To decide if a reaction is 0, 1st or 2nd order, list X, log₁₀X, and 1/X in a table, then plot each against time. The one with the linear plot is the order of the reaction.

Send questions, comments or suggestions to
Gwen Sibert, at the
Roanoke Valley Governor's School
gsibert@rvgs.k12.va.us

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conc CH₃CHO (mol/L)	0.10	0.20	0.30	0.40
rate (mol/L^.)	0.085	0.34	0.76	1.4