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प्रश्न
Explain how and why will the rate of reaction for a given reaction be affected when the temperature at which the reaction was taking place is decreased.
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उत्तर
The rate of reaction will decrease. At lower temperatures, the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of the rate of reaction.
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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)`
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`
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(Given: R = 8.314 JK–1 mol–1)
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(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Define activation energy.
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Why in the redox titration of \[\ce{KMnO4}\] vs oxalic acid, we heat oxalic acid solution before starting the titration?
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 |
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[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
