Advertisements
Advertisements
प्रश्न
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
उत्तर
According to the Arrhenius equation,
`log "k"_2/"k"_1 = ("E"_"a")/(2.303 "R") [1/"T"_1 - 1/"T"_2]`
`"K"_2/"k"_1` = 2,
T1 = 298 K,
T2 = 308 K,
R = 8.314 JK−1 mol−1
∴ `log 2 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol"^-1) [1/(298 "K") - 1/(308 "K")]`
`0.3010 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol")^-1 xx 10/(298 xx 308)`
∴ `"E"_"a" = (0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/10 "J mol"^-1`
= 52897.7 J mol−1
= 52.897 kJ mol−1
APPEARS IN
संबंधित प्रश्न
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 decomposition of hydrocarbon follows the equation k = `(4.5 xx 10^11 "s"^-1) "e"^(-28000 "K"//"T")`
Calculate Ea.
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 x 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]
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).
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
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
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 |
Total number of vibrational degrees of freedom present in CO2 molecule is
For an endothermic reaction energy of activation is Ea and enthalpy of reaction ΔH (both of there in KJ moI–1) minimum value of Ea will be
In respect of the eqn k = \[\ce{Ae^{{-E_a}/{RT}}}\] in chemical kinetics, which one of the following statement is correct?
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 10oC rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298K to 308K, what would be the value of activation energy (Ea) for the reaction?
