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The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 − 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? - CBSE (Science) Class 12 - Chemistry

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Question

The rate constant for the first order decomposition of H2O2 is given by the following equation:

log = 14.34 − 1.25 × 10K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Solution

Arrhenius equation is given by,

`k = Ae^((-E_a)/"RT")`

`=>In k  = In A - E_a/"RT"`

`=>Ink = lof A - E_a/"RT"`

`=> log k = log A - E_a/(2.303 RT)`   ... i

The given equation is

log k = 14.34 - 21.25 x  104 K/T   ...  ii

From equation (i) and (ii), we obtain

`E_a/(2.303RT) = (1.25xx10^4 K)/T`

`=> E_a = 1.25 xx 10^4K xx 2.303 xxR`

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol1 (approximately)

= 239.34 kJ mol−1

Also, when t1/2 = 256 minutes,

`k = 0.693/t_(1/2)`

= 0.693/256

= 2.707 × 10−3 min−1

= 4.51 × 10−5 s−1

It is also given that, log k = 14.34 − 1.25 × 104 K/T

`=>log(4.51xx10^(-5))` = `14.34 - (1.25xx10^(4)K)/T`

`=>log(0.654 - 0.5) = 14.34 - (1.25 xx 10^4K)/T`

`=> (1.25xx10^4K)/T = 18.686`

`=> T = (1.25xx10^4K)/18.686`

= 668.95 K

= 669 K (approximately)

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Chemistry Textbook for Class 12 (2018 to Current)
Chapter 4: Chemical Kinetics
Q: 27 | Page no. 120

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Solution The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 − 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? Concept: Temperature Dependence of the Rate of a Reaction.
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