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प्रश्न
Activation energy of a chemical reaction can be determined by ______.
विकल्प
determining the rate constant at standard temperature.
determining the rate constants at two temperatures.
determining probability of collision.
using catalyst.
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उत्तर
Activation energy of a chemical reaction can be determined by determining the rate constants at two temperatures.
Explanation:
`log k_2/k_1 = E_a/(2.303R)(1/T_1 - 1/T_2)`
Where, Ea = activation energy
T2 = higher temperature
T1 = lower temperature
k1 = rate constant at temperature T1
k2 = rate constant at temperature T2
This equation is known as Arrhenius equation.
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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)
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 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.
The decomposition of hydrocarbon follows the equation k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`
Calculate Ea.
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).
The chemical reaction in which reactants require high amount of activation energy are generally ____________.
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.
Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.
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 ______.
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]
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:
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?
Activation energy of any chemical reactions can be calculated if one knows the value of:
