Advertisements
Advertisements
प्रश्न
The rate constant of a first order reaction are 0.58 S-1 at 313 K and 0.045 S-1 at 293 K. What is the energy of activation for the reaction?
Advertisements
उत्तर
Given: Rate constant k1 = 0.58 s-1
Rate constant k2 = 0.045 s-1
T1 = 313 K
T2 = 293 K
R = 8.314 J K-1mol-1
To find: Activation energy (Ea)
Formula : `log_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R")[("T"_2 - "T"_1)/("T"_1"T"_2")]`
Calculation : From Formula,
`log10((0.045 "s"^-1)/(0.58 "s"^-1)) = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol"^-1)[(293"K" - 313 "K")/(293 "K" xx 313 "K")]`
∴ log 0.0776 = `"E"_"a"/(19.147 "J" "mol"^-1)[(-20)/(293 xx 313)]`
∴ `-1.110 = "E"_"a"/19.147[(-20)/(293 xx 313)]`
∴`"E"_"a" = (-1.110 xx 19.147 xx 293 xx 313)/-20`
= 97455.34 J `"mol"^-1`
APPEARS IN
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)
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)
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.
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?
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).
Explain the following terms :
Half life period of a reaction (t1/2)
Predict the main product of the following reactions:
The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.
Activation energy of a chemical reaction can be determined by ______.
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
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.
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
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]
The equation k = `(6.5 xx 10^12 "s"^(-1))"e"^(- 26000 " K"//"T")` is followed for the decomposition of compound A. The activation energy for the reaction is ______ kJ mol-1. (Nearest integer) (Given: R = 8.314 JK-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 ______.
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]
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
Which plot of ln k vs `1/T` is consistent with the Arrhenius equation?
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)
