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201. A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if catalyst lowers the activation energy barrier by 20 kJ mol^–1.

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Short Answer Type

202. A first order reaction has a specific reaction rate of 10^–3. How much time will it take for 10 g of the reactant to reduce to 2.5 g. Given log 2 = 0.301, log 4 = 0.6021, log 6 = 0.778.

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203. For the decomposition of azoisopropane to hexane and nitrogene at 543 K, the following data are obtained.

t(sec)	P (mm of Hg)
0	35.0
360	54.0
720	63.0

Calculate the rate constant.

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204. The following data were obtained during the first order thermal decomposition of SO₂ Cl₂ at constant volume.
SO₂ Cl₂ (g) → SO₂ (g) + Cl₂ (g)

Experiment	Time/s^-1	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of reaction when total pressure is 0.65 atm.

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Long Answer Type

205. Consider a certain reaction A → Products with K = 2.0 x 10^–2 s^–1. Calculate the concentration of A remaining after 100 S if the initial concentration of A is 1.0 mol L^–1.

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Short Answer Type

206. The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^–1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponental factor.

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Long Answer Type

207. Sucrose decomposes in acid solution into glucose and fuctose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

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Short Answer Type

208. The decomposition of hydrocarbon follows the equation
k = (4.5 x 10¹¹ s^–1) e^–28000K/TCalculate E_a.

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209. The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 x 10⁴ k/T
Calculate E_afor this reaction and at what temperature will its half period be 256 minutes.

Using arrhenius equation

\log K = \log A - \frac{E_{a}}{2.303 R} - \frac{1}{T}

Comparing this equation with

\log K = 14.34 - \frac{1.25 \times 10^{4} K}{T} gives \frac{E_{a}}{2.303 R} = 1.25 \times 10^{4} K E_{a} = 1.25 \times 10^{4} K \times 2.303 \times 8.314 J {mol}^{- 1} K^{- 1} or E_{a} = 239.339 J {mol}^{- 1} = 239.339 kJ {mol}^{- 1} .

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Long Answer Type

210. The decomposition of A into product has value of K as 4.5 x 10³ S^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would K be 1.5 x 10⁴ S^–1?

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