IIT JEE past question Kinetics 6

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The rate constant for the first order decomposition of a dcertain reaction is described by the equation

log_{10} k(s^{-1})=14.3-\frac{1.25\times10^4K}{T}

A) What is the energy of activation for this reaction?
B) At what temperature will its half life period be 256 minutes?

#Chemistry #Physical Chemistry

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6 Answers

skygirl ·2009-03-27 23:02:31

rate = Ae^-Ea/RT

lnk = lnA - Ea/RT

=> logk = logA - Ea/2.303RT

comparing the given eqn with the above one...

Ea/2.303R = 1.25E4

=> Ea= 2.303X8.314X1.25E4 J

half life = 256 mins = 256X60 s.

k = 0.693/[256X60]

wen we put this in the above eqn, we get T.

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tapanmast Vora ·2009-03-27 23:03:20

Are v expecting the same level of tuffness for the questions ???

Sir, pl. reply!!!!!!!

A) 1.25*10⁴ = E_a/2.303R

B) K = ln2 / 256*60

now sub. in given eqtn to get T

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eureka123 ·2009-03-27 23:03:53

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eureka123 ·2009-03-27 23:04:18

fir se late......

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tapanmast Vora ·2009-03-27 23:05:52

hum dono late [2]

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Lokesh Verma ·2009-03-27 23:07:05

dont worry all pinks :)

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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IIT JEE past question Kinetics 6

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