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151. The gas phase decomposition of dimethyl ether follows first order kinetics:
CH₃ – O CH₃ (g) → CH₄ (g) + H₂(g) + CO(g)
The reaction is carried out in a constant volume container at 500°C and has a half life of 14.5 minutes. Initially only dimethyl ether is present at a present of 0.40 atmosphere. What is the total pressure of the system after 12 minutes?
(Assume ideal gas behaviour).

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152. The half-life for radioactive decay of

14_{C}

is 5730 years. An archaeological artifact containing wood had only 80% of the

{}^{14}C

activity found in living tree. Estimate the age of the sample or calculate the age of the artefact.

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153. During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 Years. If 1 μg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

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Short Answer Type

154. A first order decomposition reaction takes 40 minutes for 30% of decomposition. Calculate t_1/2 value for it.

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Long Answer Type

155. For first order reaction show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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Short Answer Type

156. A first order reaction has a specific reaction rate of 2.31 x 10^–3 sec^–1. Calculate the half-life period of the reaction.

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157. 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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158. Show that the time required for the completion of 3/4th reaction of first order is twice the time required for completion of 1/2 of the reaction.

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159. If half-life of a first order reaction involving reactant A is 5 min. How long will it take [A] to react 25% of its initial concentration?

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160. A first order reaction is 15% complete in 20 minutes. How long will it take to be 60% completed ?

$For the first order reaction t = \frac{2.303}{k} \log \frac{a}{a - x} let, a = 100 initally reaction is 15 % completed, so x = 15, t = 20 \min so, thus applying in the above equation 20 = \frac{2.303}{k} \log \frac{100}{100 - 15} 20 = \frac{2.303}{k} \log \frac{100}{85} 20 = \frac{2.303}{k} \log 1.1764 20 = \frac{2.303}{k} 0.0706 k = 0.00813$

to completed 60% reaction

$For the first order reaction t = \frac{2.303}{k} \log \frac{a}{a - x} let, a = 100 initally reaction is 15 % completed, so x = 60, k = 0.00813 so, thus applying in the above equation t = \frac{2.303}{0.00813} \log \frac{100}{100 - 60} t = \frac{2.303}{0.00813} \log \frac{100}{40} t = \frac{2.303}{0.00813} \log 2.5 t = \frac{2.303}{0.00813} \times 0.3979 k = 283.27 x 0.3979 t = 112.71 m i n$

Ans. 112.7 min

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