Advertisements
Advertisements
प्रश्न
Explain a graphical method to determine activation energy of a reaction.
Advertisements
उत्तर
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
संबंधित प्रश्न
Consider the reaction
`3I_((aq))^-) +S_2O_8^(2-)->I_(3(aq))^-) + 2S_2O_4^(2-)`
At particular time t, `(d[SO_4^(2-)])/dt=2.2xx10^(-2)"M/s"`
What are the values of the following at the same time?
a. `-(d[I^-])/dt`
b. `-(d[S_2O_8^(2-)])/dt`
c. `-(d[I_3^-])/dt`
The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
What is the effect of adding a catalyst on Activation energy (Ea)
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK–1 mol–1)
Explain the following terms :
Half life period of a reaction (t1/2)
The decomposition of a hydrocarbon has value of rate constant as 2.5×104s-1 At 27° what temperature would rate constant be 7.5×104 × 3 s-1if energy of activation is 19.147 × 103 J mol-1 ?
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 ____________.
Consider figure and mark the correct option.
Which of the following statements are in accordance with the Arrhenius equation?
(i) Rate of a reaction increases with increase in temperature.
(ii) Rate of a reaction increases with decrease in activation energy.
(iii) Rate constant decreases exponentially with increase in temperature.
(iv) Rate of reaction decreases with decrease in activation energy.
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?
In respect of the eqn k = \[\ce{Ae^{{-E_a}/{RT}}}\] in chemical kinetics, which one of the following statement is correct?
The activation energy in a chemical reaction is defined as ______.
The activation energy in a chemical reaction is defined as ______.
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]
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
Explain how and why will the rate of reaction for a given reaction be affected when the temperature at which the reaction was taking place is decreased.
The decomposition of N2O into N2 and O2 in the presence of gaseous argon follows second-order kinetics, with k = (5.0 × 1011 L mol−1 s−1) `"e"^(-(29000 "K")/"T")`. Arrhenius parameters are ______ kJ 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 ______.
A schematic plot of ln Keq versus inverse of temperature for a reaction is shown below
The reaction must be:
A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K−1 mol−1]
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
Activation energy of any chemical reactions can be calculated if one knows the value of:
Assertion (A): A reaction can have zero activation energy.
Reason (R): The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to the threshold value is called activation energy.
In the light of the above statements, choose the correct answer from the options given below:
