Factor Affecting Reaction Rate
Objectives
1. Explain factors affecting reaction rate;
-Concentration / Pressure -Temperature
-Catalyst -Particle size
2. Explain the effect of temperature on reaction
rate using Maxwell-Boltzmann distribution
curve.
3. Explain the effect of catalyst on activation
energy based on energy profile diagram for
exothermic and endothermic reactions.
4. Relate rate constant to temperature and activation energy using the Arrhenius
equation
k = Ae-Ea/RT or ln k = ln A –Ea/RT
5. Determine k, Ea, T and A using Arrhenius
equation by calculation and graphical method.
Ln (k1 / k2) = Ea/R (1/T2 – 1/T1) |
Factors Affecting Reaction rate
a) Effect of Concentration
o When concentration of reactant increases, frequency
of collision also increases.
o More particles present in the same volume, they are more likely to collide
o The probability of effective collisions increase
frequency of effective collision = collision frequency x fraction of molecules with sufficient energy
o Rate of reaction increases.
[reactants] increases,the frequency of collision increases
o This observation correlates with the RATE LAW
Reaction rate = k [ A ]x [ B ]y …
(A & B = reactants)
(x & y = rate order)
o Based upon this equation,
Reaction rate directly proportional to the concentration of reactants
o REMINDER!
Only in zero order reactions, the rate of reaction is not
dependant upon the concentration of the reactants.
(depending on its rate order)
Reaction rate = k [ A ]0 = k (constant)
b) Effect of Temperature
o At a higher temperature, molecules have higher kinetic energy and move at higher speed
o more collisions will occur in a given time
o Furthermore, the higher the KE, the higher the energy of the collisions
o So more molecules will have energy greater than Ea
o effective collision also will be increased
o thus the reaction rate increases.
Distribution of Kinetic of Molecules(Maxwell-Boltzmann Distributions)
The collision frequency is higher temperature because the fraction of reactant molecules with activation energy is higher.
Maxwell-Boltzmann Distributions
• The figure shows the distribution KE of gaseous molecule at temperature T1 and T2.
• At higher temperature, the fraction of molecules with energy greater than Ea increases.axwel-Boltzmann Distributions
• Thus, rate of reaction increase with increase of
temperature.
Take note:
• Area under the curve ∝ to the total number of molecules
area under the curve is the same, bcoz no. of molecules is the same.
Maxwel-Boltzmann Distributions
• There are a wide range of molecular energy
A few molecules have low and high KE, most have value in the middle.
• The shape of the graph changes with temperature
At higher temperature the peak of the graph moves towards the right to a higher KE value as the curve broadens out.
• At higher temperatures, the fraction of molecules with higher energy increase
Average energy increase at higher temperature, but fraction of molecules with low energy decreases. At higher temperature, fraction of molecules with energy
greater than Ea increase.
c). Effect Of Temperature
ARRHENIUS EQUATION
• the effect of temperature on the rate constant, k:
k = A e -Ea⁄RT
Where…
k = rate constant
A = frequency factor
(is a measure of the probability of a favorable collision)
e = natural log exponent
Ea = activation energy for the reaction (kJ/mol)
R = universal gas constant (8.314 J mol-1 K-1)
T = absolute temperature (T in Kelvin)
Arrhenius Equation - Derivation
k = A e^ -Ea⁄RT
ln k = ln (A.e^ –Ea/RT)
ln k = ln A – Ea/RT |
ln k = ln A - Ea / R (1/T)
y = C + mx
Example
The table below gives the rate constants, k for the reaction between potassium hydroxide and bromoethane at different temperatures.
K(M^-1s^-1) | T(k) |
0.63 | 322 |
2.50 | 331 |
10.0 | 347 |
22.6 | 353 |
a) Using a graphical method, calculate the activation energy (kJmol-1) for this reaction.
b) What is the overall order of reaction? Explain
c) Calculate the initial rate of reaction at 330 K when the concentrations for both KOH and CH3CH2Br are 0.1M.
Solution……
1/T | 0.0031 | 0.0030 | 0.0029 | 0.0028 |
ln k | -0.46 | 0.92 | 2.30 | 3.12 |
a) Slope = Ea/R
= 12492
Ea = 12492 × 8.314
= 1.04 ×105 Jmol-1
= 104 kJmol-1
b) Second order, unit of k = M-1s-1
c) ln k = -Ea/R(1/T) + ln A
ln k = -12492 (1/330) + 38.45
k = 1.81 M^-1s^-1
Rate = k [KOH][CH3CH2Br]
= 1.81 x 0.1 x 0.1
= 1.81 x 10-2 Ms-1
Arrhenius Equation – Further Derivation
Rate constant, k varies with T.
The ratio of rate constant at two different T can be calculated.
ln k1 = ln A – Ea / RT1……………………….(1)
ln k2 = ln A – Ea / RT2……………………….(2)
Equation (1) minus (2) gives :
ln ( k1/k2) = Ea / R (1/T2-1/T1)
d). Effect Of Catalyst
• A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
• It increases the reaction rate by providing an alternative reaction pathway of which having lower activation energy.
• A catalyst provides a different reaction mechanism.
A catalyst provides an alternative pathway for the reaction to occur (----curve) which has a lower activation energy.
When Ea decreases, k increases, reaction rate increases
e). Effect Of Surface Area
• For reactions that occur on a surface that is for solid, the rate increases as the surface
area is increased.
• A larger surface area increases the contact area between the reactants thus increases
the frequency of collision and the probability of effective collision between the molecules
of the reactants.
