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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).

The gas phase decomposition of dimethyl ether follows first order kinetics:
CH₃ – O CH₃ (g) → CH₄ (g) + H₂(g) + CO(g)

we have given
Volume is constant
temperature is 500⁰ C
half life is =14.5 min

PV = nRT

\frac{n}{V} = \frac{P}{RT}

a = \frac{P}{RT} = \frac{0.4}{0.082 \times 773} = 6.31 \times 10^{- 3} mol L^{- 1} K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{14.5} = 4.78 \times 10^{- 2} \min^{- 1} K = \frac{2.303}{12} \log \frac{a}{a - x}

4.78 \times 10^{- 2} = \frac{2.303}{12} \log \frac{a}{a - x} \frac{a}{a - x} = 1.77446 a - x = \frac{6.31 \times 10^{- 3}}{1.77446} moles L^{- 1} = 3.556 \times 10^{- 3} moles L^{- 1} ∴ x = (6.310 - 3.556) \times 10^{- 3} moles / L = 2.754 \times 10^{- 3} moles / L ∴ After 12 minutes .

Total number of moles

L^{- 1}

= a + 2 x = 6.31 \times 10^{- 3} + 2 \times 2.754 \times 10^{- 3} = 11.818 \times 10^{- 3}

P = \frac{n}{V} RT = 11.818 \times 10^{- 3} \times 0.082 \times 773 = 0.749 atm .

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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.

219 Views

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.

202 Views

Short Answer Type

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

177 Views

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.

826 Views

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.

197 Views

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?

117 Views

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.

333 Views

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?

125 Views

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

695 Views