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प्रश्न
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
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उत्तर
(i) energy is always released
(iv) reactants may be formed
Explanation:
Activation Energy is the amount of energy released when reactant molecules collide and form an activated complex. When energy is released, the complex decomposes into a product.
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संबंधित प्रश्न
Explain a graphical method to determine activation energy of a reaction.
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
`logk=logA-E_a/2.303R(1/T)`
Where Ea is the activation energy. When a graph is plotted for `logk Vs. 1/T` a straight line with a slope of −4250 K is obtained. Calculate ‘Ea’ for the reaction.(R = 8.314 JK−1 mol−1)
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 of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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.
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?
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.
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]
The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.
Which of the following graphs represents exothermic reaction?
(a)
(b)
(c)
Mark the incorrect statements:
(i) Catalyst provides an alternative pathway to reaction mechanism.
(ii) Catalyst raises the activation energy.
(iii) Catalyst lowers the activation energy.
(iv) Catalyst alters enthalpy change of the reaction.
Why does the rate of a reaction increase with rise in temperature?
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 |
The activation energy in a chemical reaction is defined as ______.
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.
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]
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
Activation energy of any chemical reactions can be calculated if one knows the value of:
The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation.
(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)
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:
