Rate of reaction is greatly affected by rise in temperature; we mark it in our daily life too. On a hotter day yoghurt gets sour; dough gets fermented and cooked food goes bad a lot faster than on a colder day. It has beenfound experimentally that rising the temperature by 10°C nearly doubles the rate constant. Before we go into complicated terms and equations, let’s try to understand what happens at the molecular level on rising the temperature? Let’s take an example of a simple reaction:
H2 + I2 ⟶2HI
H-H + I-I ⟶2(H-I)
For any reaction to happen, collision of reactant molecules is required. When reactant molecules collide, old bonds (H-H and I-I) are broken and new bonds (H-I) are formed. But the reaction doesn’t happen all of a sudden. The reaction follows a course where folloing two processes take place simultaneously.
H-H + I-I ⟶ H ⋯ I ⟶ 2 (H-I)
⋮ ⋮
H ⋯ I
An intermediate complex is formed which exists for a fraction of time. It is known as an activated complex. It creates an energy barrier that reactant molecules need to achieve to complete the reaction and form the product. The energy gap between reactant and activated complex is called as activation energy Ea.
It is not possible for every reactant molecule to jump over this barrier. Only few reactant molecules have enough kinetic energy to collide with other reactant molecules and not all collisions are effective enough to form the activated complex or intermediate complex. It means that there are limited numbers of molecules which can overcome all barriers and form the product.
Rise in temperature adds extra kinetic energy to all the molecules enabling some more molecules to collide with enough kinetic energy to form activated complex so that more product will be formed. It means that on raising the temperature rate of reaction increases.
Temperature dependence of rate of reaction has been observed by a number of scientists, such as Wilhelmy in 1850, Berthelot in 1862, J. J. Hood in 1885. They all tried to set an equation relating the rate constant to the temperature, however none of them could give a satisfactory equation.
In 1884 Van’t Hoff gave the equation for temperature dependence of equilibrium constants:
(𝜕 ln Kc)p = ΔU° .........(1)
𝜕T RT2
where kc is equilibrium constant
A + B ⇌ C + D
Rate of forward reaction = k1[A][B]
Rate of backward reaction = k-1[C][D]
So,
kc = k1/k-1
Equation 1 can be written as:
d ln k1 - d lnk-1 = ΔU° .........(2)
dT dT RT2
Van’t Hoff suggested thatthe rate constants k1 and k-1 are influenced by two different energy factors E1 and E-1, therefore he split equation 2 in two equations:
d ln k1 = E1 .........(3)
dT RT2
d ln k-1 = E-1 .........(4)
dT RT2
These two energies E1 and E-1must be such that E1 - E-1 = ΔU°.
Van’t Hoff was aware that ΔU° is not always temperature independent; therefore E1and E-1 might be temperature dependentas well. He also considered the possibility of temperature independence of E.
for an event when E1 is independent of temperature, equation 3 can be integrated
ln k = constant - E1
RT2
k = A e-E/RT .........(5)
where,
E = B + DT2
where B and D are temperature independent. And equation 4 will become:
d ln k-1 = B + DT2
dT RT2
ln k = A’ - B + DT2
RT2
k = A e-(B-DT2)/RT .........(6)
where A= eA’
This equation received much experimental support, but in most of his equations, temperature dependence was special case of the equation. In 1889 Arrhenius took Van’t Hoff’ssimplest equation 5 as starting equation and proposed a general concept of how reactions occur. That equation is now known as Arrhenius equation.
k = A e-Ea/RT
where A is known as Arrhenius factor or Frequency factor or Pre exponential factor. It gives the number of collision per second and its unit is (s-1). It is a constant, specific for a particular reaction.
Ea is activation energy, it is measured in Joules/mole and its unit is J mol-1.
R is gas constant and k is the rate constant.
It has been observed that rising the temperature causes the reaction medium to become less viscous, which gives the molecules more freedom to move and increases the chances of them to collide and react. This was a basic observation but Arrhenius gave an insight to it.
Factor e-Ea/RT in his equation corresponds to the fraction of molecules that have kinetic energy greater than activation energy Ea.
taking natural log of both side of Arrhenius equation
ln k = - Ea + ln A .........(7)
RT
On plotting this equation ln k verses 1/T we get a straight line with slope -Ea/R and intercept ln A. It shows that on increasing temperature or decreasing the activation energy, rate of reaction increases.
 |
| Maxwell Boltzmann distribution curve |
Arrhenius theory gained support by James Clark Maxwell and Ludwig Boltzmann. They used statistics to predict the behaviour of large number of molecules. Usually, a number of molecules are present in a reaction mixture and they may have different levels of kinetic energy. It is difficult to measure the kinetic energy of each molecule. Boltzmann and Maxwell suggested that we can find a fraction of molecules with a given kinetic energy by plotting the fraction of molecule (NE/NT) against Kinetic energy (K.E). When they plotted the same graph at different temperatures (t) and (t+10), they found that the fraction of molecules having energy ≥ Ea was doubled at (t+10) temperature. That’s why on increasing the temperature by 10 units, rate of reaction gets doubled.
Let’s practice few problems:
Q. The rate constant of a reaction at 500K and 700K are 0.02s-1 and 0.07s-1respectively. Calculate the values of Ea and A.
At temperature T1
ln k1 = - Ea + ln A .........(8)
RT1
At temperature T2
ln k2 = - Ea + ln A .........(9)
RT2
Since Ea and A are constant for a given reaction.
Subtracting equation 8 from 9:
ln k1 – ln k2 = Ea - Ea
RT1 RT
ln k2 = Ea [1 - 1]
k1 R T1 T2
log k2 = Ea [T2 – T1] .........(10)
k1 2.303R T1 T2
on putting values in the equation 10 we can find Ea and then using Arrhenius equation we can calculate A. value of R = 8.314 JK-1mol-1.
log 0.07 = Ea [700– 500] .........(10)
0.02 2.303 × 8.314 700×500
Ea = 18230 J mol-1
k = A e-Ea/RT
0.02 = A e-18230/8.314× 500
0.02 = A 0.012
A = 1.66
Q. A first order reaction whose rate constant at 80°C was found to be 5.0 × 10-3 s-1. has an activation energy of 45 kJ mol-1. What is the value of the rate constant at 875°C?
log k2 = Ea [T2 – T1] .........(10)
k1 2.303R T1 T2
log k2 = (4.5 × 104 J mol-1) [1148K – 1073K]
k1 2.303× (8.314 J K-1 mol-1) 1148K×1073K
log k2 = 0.143
k1
k2 = 1.39
k1
value of k1 is given 5.0 × 10-3 s-1
k2= 1.39 × (5.0 × 10-3 s-1)
k2= 7.0 × 10-3 s-1
Q. the rate constant for the first order decomposition of H2O2is given by the following equation:
log k = 14.34 – 1.25×104 K/T
calculate Ea for this reaction and what temperature will its half life be 256min?
Arrhenius equation
ln k = - Ea + ln A
RT
or
log k = log A - Ea
2.303 RT
on comparing the given equation with this form of Arrhenius equation:
log A = 14.34 s-1
and
- Ea = – 1.25×104 K
2.303 RT T
Ea = 1.25×104 K × 2.303 × 8.314 J K-1 mol-1
Ea = 1.25×104 K × 2.303 × 8.314 J K-1 mol-1
Ea = 2.39×105 J K-1 mol-1
t1/2 = 0.693
k
(256× 60)s = 0.693
k
k = 4.5 × 10-5 s-1
Now calculate T by equation given in the question:
log k = 14.34 – 1.25×104 K/T
log 4.5 × 10-5 s-1 = 14.34 s-1 - 1.25×104K
T
T = 669 K
Arrhenius theory received great acceptance by scientists but as we know, there is always a possibility of improvement in science. In 1916 Max Trautz and William Lewis gave a new theory known as Collision Theory which modifies the Arrhenius equation. in the next post of Chemical Kinetics we will study Collision Theory in detail.
This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.
No comments:
Post a Comment