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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)
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उत्तर
`log k=logA-E_a/2.303R(1/T)`
Ea → Activation energy
The above equation is like y = mx + c where if we plot y v/s x we get a straight line with slope ‘m’ and intercept ‘c’.
So, slope is equal to`=-E_a/2.303R`
`-E_a/2.303R=-4250k=>E_a=4250 xx 2.303 xx 8.314 =81,375.3535 j mol^(-1)`
`E_a=81.3753 KJ mol^(-1)`
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| 105 × k/s−1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
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[R = 8.314 J K−1 mol−1]
Mark the incorrect statements:
(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.
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 |
The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?
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The (i) intercept and (ii) slope of the graph are:
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[Given: ln 10 = 2.3, R = 8.3 J K–1 mol–1]
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