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Chemical Kinetics

Short Answer Type

181. State any one condition under which a bimolecular reaction may be kinetically first order.

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Long Answer Type

182. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K, if the value of A is 4 x 10¹⁰ s^–1. Calculate k at 318 k and E_a.

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183. The rate constant increases 50% when temperature is increased from 298 K to 308 K. THe value of ΔH⁰ is 15 kJ mol. Calculate the activation energies Ea^f and Ea^r.

The rate constant at 298 K = k.
When the temperature is raised from 298 k to 308 k.
Increase in rate constant

   $= k \times \frac{50}{100} = 0.5 k$
By using the equation

   $In [k_{2} / k_{1}] = [{E_{a}}^{f} / R] [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

We have,

$In (1.5 k / k) = [{E_{a}}^{f} / R] [\frac{1}{298 k} - \frac{1}{308 k}] = [{E_{a}}^{f} / R] [\frac{10}{298 \times 208}] \log (2.303 \times 1.5) = \frac{E_{a}^{f}}{8.314} \times \frac{10}{298 \times 308} \log 3.4545 = \frac{E_{a}^{f}}{8.314} \times \frac{10}{298 \times 308} 0.4460 = \frac{E_{a}^{f}}{8.314} \times \frac{10}{298 \times 308} E_{a}^{f} = \frac{0.4460 \times 8.314 \times 298 \times 308}{10} = 34.03 kJ / mol$

   $∆ rH ° = 15 kJ {mol}^{- 1}$

but $∆ rH ° = {E_{a}}^{f} - {E_{a}}^{r}$

   $∴$    ${E_{a}}^{f} = (34.03 - 15) kJ mol = 19.03 kJ {mol}^{- 1}$

100 Views

184. The energy of activation for a reaction is 100 kJ mol^–1. The presence of a catalyst lowers the enregy of activation by 75%. What will be the effect on rate of reaction at 20% C, other things being equal.

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Short Answer Type

185. The difference in energy of activation for uncatalysed reaction and catalysed reaction is 20 kJ/mol. How many times the rate constant of catalysed reaction will increase over the uncatalysed reaction?

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Long Answer Type

186.

(a) The decomposition of N₂O₅(g) is a first order reaction with a rate constant of 5 x 10^–4sec^–1 at 45°C.
i.e., 2N₂O₅(g) = 4NO₂(g) + O(g)
If initial concentration of N₂O₅ is 0.25, calculate its concentration after two minutes. Also calculate half life for the decomposition of N₂O₅(g).
(b) For an elementary reaction: 2A + B → 3C The rate of appearance of C at time ‘t’ is 1.3 x 10^–4 mol l^–1 s^–1. Calculate at this time:
(i) Rate of reaction (ii) Rate of disappearance of A.

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187. A first order reaction is 50%. Complete in 30 minutes at 27° and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in kJ/mol.

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188. In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 x 10²³ sec^–1 and 98.6 kJ mol^–1 respectively. If the reaction is of first order at what temperature will its half life period be ten times.

156 Views

Short Answer Type

189. The rate of a particular reaction triples when temperature changes from 50°C to 100°C. Calculate the activation energy of the reaction. [log 3 = 0.4771 (R = 8.314 ] K^–1mol^–1)

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190. A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is again 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. [R = 8.314 J K^–1 mol^–1, log 4 = 0.602].

2689 Views