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प्रश्न
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
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उत्तर
Thermodynamics feasibility of a reaction depends on Gibbs free energy i.e., AG must be negative for spontaneous process. Kinetic feasibility depends on the activation energy of reaction, the lesser is the activation energy, the greater is the feasibility of reaction, i.e.,
i.e., Diamond `->` Graphic ΔG = – ve
This process is thermodynamically feasible but it is very slow due to its high activation energy.
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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 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)
What will be the effect of temperature on rate constant?
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 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?
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?
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)
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(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.
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Why in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?
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The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?
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)
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 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?
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)
