Advertisements
Advertisements
Question
Explain a graphical method to determine activation energy of a reaction.
Advertisements
Solution
The Arrhenius equation is given by
`k=Ae^((-E_a)/(RT))`
On taking logarithms of both sides, we get
`log_nk=log_nA-E_a/(RT)`
or
`log_10k=-E_a/(2.303RT)+log_10A`
`log_10k=-E_a/(2.303R) times (1/T)+log_10A`
The rate constant of a reaction is determined at various temperatures.
log10 k is plotted against the reciprocal of temperature. The graphical
representation is
The slope of the straight line graph is
`–(Ea)/(2.303R)`
from which the activation energy can be calculated.
APPEARS IN
RELATED QUESTIONS
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
The rate constant for the first-order decomposition of H2O2 is given by the following equation:
`logk=14.2-(1.0xx10^4)/TK`
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 JK–1 mol–1)
What will be the effect of temperature on rate constant?
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
The rate constant for the decomposition of N2O5 at various temperatures is given below:
| T/°C | 0 | 20 | 40 | 60 | 80 |
| 105 × k/s−1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and `1/T` and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.
Consider a certain reaction \[\ce{A -> Products}\] with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.
The decomposition of A into product has 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?
In the Arrhenius equation for a first order reaction, the values of ‘A’ of ‘Ea’ are 4 × 1013 sec−1 and 98.6 kJ mol−1 respectively. At what temperature will its half life period be 10 minutes?
[R = 8.314 J K−1 mol−1]
Calculate activation energy for a reaction of which rate constant becomes four times when temperature changes from 30 °C to 50 °C. (Given R = 8.314 JK−1 mol−1).
What is the effect of adding a catalyst on Activation energy (Ea)
Explain the following terms :
Half life period of a reaction (t1/2)
Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction. (Given : log2 = 0.3010,log 3 = 0.4771, log5 = 0.6990).
Predict the main product of the following reactions:
The chemical reaction in which reactants require high amount of activation energy are generally ____________.
Which of the following graphs represents exothermic reaction?
(a)
(b)
(c)
During decomposition of an activated complex:
(i) energy is always released
(ii) energy is always absorbed
(iii) energy does not change
(iv) reactants may be formed
The reaction between \[\ce{H2(g)}\] and \[\ce{O2(g)}\] is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.
Why does the rate of a reaction increase with rise in temperature?
Why in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?
Match the statements given in Column I and Column II
| Column I | Column I | |
| (i) | Catalyst alters the rate of reaction | (a) cannot be fraction or zero |
| (ii) | Molecularity | (b) proper orientation is not there always |
| (iii) | Second half life of first order reaction | (c) by lowering the activation energy |
| (iv) | `e^((-E_a)/(RT)` | (d) is same as the first |
| (v) | Energetically favourable reactions (e) total probability is one are sometimes slow | (e) total probability is one |
| (vi) | Area under the Maxwell Boltzman curve is constant | (f) refers to the fraction of molecules with energy equal to or greater than activation energy |
What happens to most probable kinetic energy and the energy of activation with increase in temperature?
The slope of Arrhenius Plot `("In" "k" "v"//"s" 1/"T")` of first-order reaction is −5 × 103 K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK−1mol−1]
The activation energy of one of the reactions in a biochemical process is 532611 J mol–1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x × 10–3 k310. The value of x is ______.
[Given: ln 10 = 2.3, R = 8.3 J K–1 mol–1]
An exothermic reaction X → Y has an activation energy 30 kJ mol-1. If energy change ΔE during the reaction is - 20 kJ, then the activation energy for the reverse reaction in kJ is ______.
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (Ea) for the reaction?
