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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
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
`logk=logA-E_a/2.303R(1/T)`
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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"`
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c. `-(d[I_3^-])/dt`
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`logk=14.2-(1.0xx10^4)/TK`
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(Given: R = 8.314 JK–1 mol–1)
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).
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(a)
(b)
(c)
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Activation energy of any chemical reactions can be calculated if one knows the value of:
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(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)
