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Temperature dependence on chemical reaction involves studies of physical chemistry called Chemical Kinetics which entails the study of all the chemical reactions and associated processes. The food cooks slowly if the gas is kept at a low temperature while cooking. Whereas when we raise the temperature to its highest setting, the food cooks quickly. Thus, increasing temperature increases the rate of a reaction. Arrhenius equation helps explain this rate-temperature relationship.
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Key Terms: Rate of a Reaction, Temperature Dependence of the Rate of a Reaction, Arrhenius equation, Arrhenius Plot, Chemical Kinetics, Arrhenius equation
Relationship Between Rate Of A Reaction And Temperature
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With the rise in the temperature, the average kinetic energy of the particle also increases. As a result, the particles move faster and collide with each other at a faster rate per unit of time and possess greater energy. Therefore, a rise in temperature increases the rate of a reaction.
An important point to note is, for a chemical reaction with a rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation.
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Arrhenius Equation
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The Arrhenius equation was first proposed by Dutch scientist Jacobus Henricus van't Hoff, but it was physically justified and interpreted by a Swedish chemist, Arrhenius. The Arrhenius equation is based on the Collision theory.The following is the Arrhenius Equation which reflects the temperature dependence on Chemical Reaction:
k=Ae-EaRT
where,
K = The rate constant of the reaction
A = The Arrhenius Constant
Ea = Activation Energy for the reaction (in Joules mol-1)
R = Universal Gas Constant
T = Temperature in absolute scale (in kelvins)
It can be understood with the help of the following simple reaction:
H2 (g) + I2 (g) à 2HI (g)
According to Arrhenius, the above-mentioned reaction can take place only when one molecule of hydrogen and another molecule of iodine collide and form an unstable intermediate. It exists for a very short time and then breaks up further to form two molecules of hydrogen iodide. The energy required to form this intermediate is known as activation energy (Ea). And the intermediate form is called the activated complex (C).
NOTE: The minimum amount of energy required by reactant molecules to participate in a reaction is called activation energy (Ea).
Activation energy, Ea = Threshold energy – Average kinetic energy of reacting molecules.
In a graph of potential energy vs reaction coordinate, the reaction coordinate indicates the profile of energy shift as reactants transform into products in a graph of potential energy vs reaction coordinate. When the complex decomposes into products, some of the energy is released. As a result, the reaction's final enthalpy is solely determined by the nature of the reactants and products.
Temperature dependence of Rate of Reaction in Arrhenius Equation
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In Arrhenius equation, the factor e -Ea/RTcorresponds to the fraction of molecules colliding with activation energies more than Ea. Taking natural logarithms of both sides of the Arrhenius equation,
We have,
ln k = -Ea /RT + ln A ---- (1)
From the above equation, the plot of ln k vs 1/T gives a straight line.
Therefore, from the Arrhenius equation, we can conclude that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant.
In a graph of activation energy vs rate of reaction, slope = -Ea/R and intercept = ln A. So, calculating the values of Ea and A using these values,
We have,
At temperature T1, equation (1) will be
ln k1 = -Ea/RT1 + ln A ----(2)
At temperature T2, equation (2) will be,
ln k2 = -Ea/RT2 + ln A ----(3)
(At two temperatures T1 and T2, their rate constants are given by k1 and k2 respectively)
Subtracting equation (2) from equation (3), we get
ln k2 – ln k1 = Ea /RT1 – Ea /RT2
∴ ln k2 / k1 = (Ea /R) [1/T1 – 1/T2]
∴ log k2 / k1 = (Ea /2.303R) [(T2 – T1)/T1T2]
Things to Remember
- For a chemical reaction with a rise in temperature by 10°, the rate constant is nearly doubled.
- Increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant.
- Arrhenius Equation:
k=Ae-EaRT
where,
K = The rate constant of the reaction
A = The Arrhenius Constant
Ea = Activation Energy for the reaction (in Joules mol-1)
R = Universal Gas Constant
T = Temperature in absolute scale (in kelvins)
log k2 / k1 = (Ea /2.303R) [(T2 – T1)/T1T2]
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Sample Questions
Ques. Define ‘activation energy’ of a reaction. (All India 2011) (2 Marks)
Ans. The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
Ques. Define rate constant (K). (Compartment. All India 2016) (1 Mark)
Ans. The rate constant (K) is defined as the rate of reaction when the concentration of the reaction is taken as unity.
Ques. The activation energy of a chemical reaction is 100 kJ/mol and its A factor is 10 M-1s-1. Find the rate constant of this equation at a temperature of 300 K. (4 Marks)
Ans. Given,
Ea = 100 kJ.mol-1 = 100000 J.mol-1
A = 10 M-1s-1, ln(A) = 2.3 (approx.)
T = 300 K
The value of the rate constant can be obtained from the logarithmic form of the Arrhenius equation, which is
ln k = ln(A) – (Ea/RT)
ln k = 2.3 – (100000 J.mol-1)/(8.314 J.mol-1.K-1)*(300K)
ln k = 2.3 – 40.1
ln k = -37.8
k = 3.8341*10-17 M-1s-1
(from the units of the A factor, it can be understood that the reaction is a second-order reaction, for which the unit of k is M-1s-1)
Therefore, the value of the rate constant for the reaction at a temperature of 300K is approximately 3.8341*10-17 M-1s-1.
Ques. At a temperature of 600 K, the rate constant of a chemical reaction is 2.75*10-8 M-1s-1. When the temperature is increased to 800K, the rate constant for the same reaction is 1.95*10-7M-1s-1. What is the activation energy of this reaction? (4 Marks)
Ans. Given,
T1 = 600K
k1 = 2.75*10-8 M-1s-1.
T2 = 800K
K2 = 1.95*10-7M-1s-1
When the A factor is eliminated from the Arrhenius equation, the following equation is obtained:
ln(k1/k2) = (-Ea/R) (1/T1 – 1/T2)
Substituting the given values in the equation, the value of Ea can be determined by:
Ln (2.75*10-8/1.95*10-7) = (-Ea/8.314 J.K-1.mol-1)*(0.00041K-1)
ln(0.141) = (Ea)*(-0.0000493) J-1.mol
Ea = (-1.958)/(-0.0000493)J.mol-1 = 39716 J.mol-1
The activation energy of the reaction is approximately 39716 J.mol-1.
Ques. For a decomposition reaction the values of rate constant k at two different temperatures are given below :
k1 = 2.15 × 10-8 L mol-1 s-1 at 650 K
k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K
Calculate the value of activation energy for this reaction. (R = 8.314 J K-1 mol-1) (All India 2009) (3 Marks)
Ans. Given,
k1 = 2.15 × 10-8 L mol-1 s-1, T1 = 650 K
k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K
R = 8.314 J K-1 mol -1 Ea =?
Ques. How does a change in temperature affect the rate of a reaction? How can this effect on the rate constant of a reaction be represented quantitatively? (Compartment All India 2014) (2 Marks)
Ans. The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise in temperature.
The effect can be represented quantitatively by Arrhenius equation K = Ae-Ea/RT
Where [Ea = Activation energy of the reaction and A = Frequency factor]
Ques. For a decomposition reaction, the values of k at two different temperatures are given below:
k1 = 2.15 × 10-8 L mol-1 s-1 at 650 K
k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K
Calculate the value of activation energy for this reaction (Log 11.11 = 1.046) (R = 8.314 J K-1 mol-1) (Compartment. All India 2014) (3 Marks)
Ans. Given,
k1 = 2.15 × 10-8 L mol-1 s-1, T1 = 650 K
k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K
R = 8.314 J K-1 mol-1 Ea =?
Ques. The rate constant of a reaction at 500 K and 700 K are 0.02 s-1 and 0.07 s-1 respectively. Calculate the value of activation energy, En (R = 8.314 J K-1 mol-1) (Compartment. Delhi 2015) (3 Marks)
Ans. Given, k2 = 0.07 s-1, k1, = 0.02 s-1, T1 = 500 K, T2 = 700 K, Ea = ?
Ques. The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.2 – 1.0×104TK
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K-1 mol-1) (Delhi 2016) (3 Marks)
Ans. Given,
t1/2 = 200 min Ea = ?, T = ?
Using Arrhenius equation
Ques. The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK-1 mol-1] (Compartment. All India 2016) (3 Marks)
Ans. ![The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK-1 mol-1]](https://images.collegedunia.com/public/image/3300ac310aa4d09118d5ba60f66d9ba6.png?tr=w-345,h-224,c-force?tr=w-345,h-224,c-force)
Ques. With the help of a labeled diagram explain the role of activated complex in a reaction. (Compartment. Delhi 2012) (3 Marks)
Ans. In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
This diagram is obtained by plotting a graph of potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.
Ques. (a) What is the physical significance of energy of activation ? Explain with a diagram.
(b) In general, it is observed that the rate of a chemical reaction doubles with every 10-degree rise in temperature. If the generalization holds good for the reaction in the temperature range of 295 K to 305 K, what would be the value of activation energy for this reaction?
[R = 8.314 J mol-1 K-1] (Compartment. Delhi 2012) (3 Marks)
Ans.
- The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. The less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction.
Ques. (a) The decomposition of A into products has a value of K as 4.5 × 103 s-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 104 s-1?
(b) (i) If half life period of a first order reaction is x and 3/4,th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? (Compartment. Delhi 2013) (3 Marks)
Answer:
- Given: K1 = 4.5 × 103 s-1,
T1 = 10K + 273K = 283K
K2 = 1.5 × 104 s-1, T2 = ?
Ea = 60 KJ mol-1
Using the following formula,
- (i) t1/2 = 0.693K (For first order reaction)
t3/4 = K ⇒ t3/4 = 1.3864K
According to condition
(The value 1.3864 is double of 0.693)
From the above equation it is clear that
t3/4 = 2t1/2∴ y = 2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.
Ques.(a) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(b) With the help of diagram explain the role of activated complex in a reaction. (Compartment. Delhi 2013) (3 Marks)
Ans.
- For the first order reaction,
- In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
This diagram is obtained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.
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