Chemistry

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Arrhenius Equation – The Effect of Temperature on Reaction Rate

Generally, the rate of a reaction increase with increasing temperature. However, there are very few exceptions. The magnitude of this increase in rate is different for different reactions. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.

A large number of reactions are known which do not take place at room temperature but occur readily at higher temperatures. Example: Reaction between H₂ and O₂ to form H₂O takes place only when an electric spark is passed.

Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to e^-(Ea/RT) and he proposed a relation between the rate constant and temperature.

k = Ae^-(Ea/RT) …………. (1)

Where A the frequency factor,
R the gas constant,
E_a the activation energy of the reaction and,

T the absolute temperature (in K)

The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant.

E_a is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react.

Taking logarithm on both side of the equation (1)

y = c + mx

The above equation is of the form of a straight line y = mx + c.

A plot of ln k Vs 1/T gives a straight line with a negative slope – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\) line with a negative slope –\(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows.

At temperature T = T₁; the rate constant k = k₁

This equation can be used to caluculate E_a from rate constants K₁ and k₂ at temperatures
T₁ and T₂.

Example 1

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s^-1 respectively. Caluculate the value of activation energy.
Solution:
According to Arrhenius equation

E_a = log(2) × 2.303 × 8.314 JK^-1mol^-1 × 400K
E_a = 2.305 J mol^-1

Example 2

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\)) Where E_a is the activation energy. When a graph is plotted for log K Vs \(\frac{1}{T}\) a straight line with a slope of – 4000k is obtained. Caluculate the activation energy.
Solution:
log k = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\))
y = c + mx
m = – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)
E_a = – 2.303 R m
E_a = – 2.303 × 8.314 JK^-1 mol^-1 × (-4000K)
E_a = 76, 589 J mol^-1
E_a = 76.589 kJ mol^-1

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Collision Theory

Collision Theory was proposed independently by Max Trautz in 1916 and William Lewis in 1918. This theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let us understand this theory by considering the following reaction.

A₂(g) + B₂(g) → 2AB(g)

Fig 7.5 Progress of the Reaction

If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.

Rate ∝ number of molecules colliding per litre per second (collision rate)

The number of collisions is directly proportional to the concentration of both A₂ and B₂.

Collision rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]

Where, Z is a constant.

The collision rate in gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 10⁹ collisions per second, i.e., 1 collision in 10^-9 second.

Thus, if every collision resulted in reaction, the reaction would be complete in 10^-9 second. In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction. In order to react, the colliding molecules must possess a minimum energy called activation energy. The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.

Fraction of effective collisions (f) is given by the following expression

f = e^-Ea/RT

To understand the magnitude of collision factor (f), Let us calculate the collision factor (f) for a reaction having activation energy of 100 kJ mol^-1 at 300K.

Thus, out of 10¹⁸ collisions only four collisions are sufficiently energetic to convert reactants to products. This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.

The figure 7.6 illustrates the importance of proper alignment of molecules which leads to reaction. The fraction of effective collisions (f) having proper orientation is given by the steric factor p.

⇒ Rate = p × f × collision rate
i.e., Rate = p × e^-Ea/RT × Z[A₂][B₂] ………….. (1)
As per the rate law,
Rate = k[A₂][B₂] ………….. (2)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
k = p Z e^-Ea/RT

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Half Life Period of a Reaction

The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.

The rate constant for a first order reaction is given by

Let us calculate the half life period for a zero order reaction.

Hence, in contrast to the half life of a first order reaction, the half life of a zero order reaction is directly proportional to the initial concentration of the reactant.

Example 1

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log 10 = 1)
Solution:
For a first order reaction
k = \(\frac{2.303}{t}\)log[latex]\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}[/latex] …………. (1)
Let [A_°] = 100M
When t = t_90%; [A]=10M (given that t_90% = 8hours)
t = t_90%; [A]=20M

Find the value of k using the given data

Substitute the value of k in equation (2)

t_80% = \(\frac{2.303}{2.303/8hours}\) log(5)
t_80% = 8hours × 0.6989
t_80% = 5.59hours

Example 2

The half life of a first order reaction x → products is 6.932 × 10⁴s at 500k. What percentage of x would be decomposed on heating at 500K for 100 min.(e^0.06=1.06).
Solution:
Given t_1/2 = 0.6932 × 10⁴s
\(\frac{\left[\mathrm{A}_{0}\right]-[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}\) × 100
We know that
For a first order reaction, t_1/2 = \(\frac{0.6932}{k}\)

Example 3

Show that case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.
Solution:
Let [A_°] = 100
When t = t_99.9%; [A] = (100 – 99.9) = 0.1

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The Integrated Rate Equation

We have just learnt that the rate of change of concentration of the reactant is directly proportional to that of concentration of the reactant. For a general reaction, A → product. The rate law is Rate = \(\frac{-d[A]}{dt}\) = k[A]^x

Where k is the rate constant, and x is the order of the reaction. The above equation is a differential equation, \(\frac{-d[A]}{dt}\), so it gives rate at any instant. However, using the above expression, we cannot answer questions such as how long will it take for a specific concentration of A to be used up in the reaction? What will be the concentration of reactant after a time ‘t’?. To answer such questions, we need the integrated form of the above rate law which contains time as a variable.

Integrated Rate Law for a First Order Reaction

A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following first order reaction,

A → Product

Rate law can be expressed as
Rate = k[A]¹
Where, k is the fist order rate constant.
\(\frac{-d[A]}{dt}\) = k[A]¹
⇒ \(\frac{-d[A]}{[A]}\) = k dt …………… (1)

Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A₀] to [A] at the later time.

– ln[A] – (- In[A₀]) = k (t-0)
– ln[A] + In[A₀] = kt
ln(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt ……….. (2)

This equation is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303. 2.303 log (\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) = kt
k = \(\frac{2.303}{t}\)log(\(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)) ………… (3)
Equation (2) can be written in the form y = mx + c as below
ln[A₀]-ln[A] = kt
ln[A] = ln[A₀]-kt
⇒ y = c + mx

If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated. Examples for the first order reaction

(i) Decomposition of Dinitrogen Pentoxide

N₂O₅(g) → 2NO₂(g) + \(\frac{1}{2}\)O₂(g)

(ii) Decomposition of Sulphurylchloride

SO₂Cl₂(l) → SO₂(g) + Cl₂(g)

(iii) Decomposition of the H₂O₂ in aqueous solution; H₂O₂ → H₂O(l) + \(\frac{1}{2}\)O₂(g)

(iv) Isomerisation of Cyclopropane to Propene.

Pseudo First Order Reaction:

Kinetic study of a higher order reaction is difficult to follow, for example, in a study of a second order reaction involving two different reactants; the simultaneous measurement of change in the concentration of both the reactants is very difficult.

To overcome such difficulties, A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,

Rate = k [CH₃COOCH₃] [H₂O]

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e.,concentration of water remains almost a constant.

Now, we can define k [H₂O] = k’; Therefore the above rate equation becomes
Rate = k'[CH₃COOCH₃]
Thus it follows first order kinetics.

Integrated Rate law for a Zero Order Reaction:

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.

A → product
The rate law can be written as,
Rate = k[A]°
\(\frac{-d[A]}{dt}\) = k(1) (∴[A]° = 1)
⇒ -d[A] = k dt

Integrate the above equation between the limits of [A_°] at zero time and [A] at some later time ‘t’,

Equation (2) is in the form of a straight line y = mx + c
i.e., [A] = – kt + [A_°]
⇒ y = c + mx

A plot of [A] Vs time gives a straight line with a slope of – k and y – intercept of [A_°].

Fig 7.4: A plot of [A] Vs time for a zero order reaction A → product with initial concentration of [A] = 0.5M and k = 1.5 × 10^-2mol^-1L^-1min^-1

Examples for a Zero Order Reaction:

1. Photochemical reaction between H₂ and I₂

2. Decomposition of N₂O on hot Platinum Surface

N₂O(g) ⇄ N₂(g) + \(\frac{1}{2}\)O₂(g)

3. Iodination of Acetone in Acid Medium is Zero Order With Respect to Iodine.

Rate = k [CH₃COCH₃][H⁺]

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Molecularity

Kinetic studies involve not only measurement of a rate of reaction but also proposal of a reasonable reaction mechanism. Each and every single step in a reaction mechanism is called an elementary reaction.

An elementary step is characterized by its molecularity. The total number of reactant species that are involved in an elementary step is called molecularity of that particular step. Let us recall the hydrolysis of t butyl bromide studied in XI standard. Since the rate determining elementary step involves only t-butyl bromide, the reaction is called a Unimolecular Nucleophilic substitution (S_N¹) reaction.

Let us understand the elementary reactions by considering another reaction, the decomposition of hydrogen peroxide catalysed by I^–.

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

It is experimentally found that the reaction is first order with respect to both H₂O₂ and I^–, which indicates
that I^– is also involved in the reaction. The mechanism involves the following steps.

Step: 1

H₂O₂(aq) + I^–(aq) → H₂O(l) + Ol^–(aq)

Step: 2

H₂O₂(aq) + OI^–(aq) → H₂O(l) + I^–(aq) + O₂(g)

Overall Reaction is

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

These two reactions are elementary reactions. Adding equ (1) and (2) gives the overall reaction. Step 1 is the rate determining step, since it involves both H₂O₂ and I^–, the overall reaction is bimolecular.

Differences Between Order and Molecularity: