CBSE Class 12 · 2006 · All India · Chemical KineticsShort Answer

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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)

Answer: E_a ≈ 24.2 kJ mol⁻¹

Solution

For first order,

k = 0.693/ t_{1/2}

, so

\frac{k _{2}}{k _{1}} = \frac{t _{1/2} ( T _{1} )}{t _{1/2} ( T _{2} )} = \frac{20}{5} = 4

Arrhenius equation:

lo g 4 = \frac{E _{a}}{2.303 R} (\frac{350 - 300}{300 \times 350})

0.602 = \frac{E _{a}}{19.147} \times \frac{50}{105000}

E_{a} = \frac{0.602 \times 19.147 \times 105000}{50}

E_{a} = 2.42 \times 1 0^{4} J mo l^{- 1} = 24.2 kJ mo l^{- 1}

CBSE Class 12 Board · 2006 · All India · Subjective

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