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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 27 of 39
Marks: +1, -0
The rate constant for the first order decomposition of H2O2\mathrm{H}_2\mathrm{O}_2 is given by the following equation:
log⁡k=14.34−1.25×104 KT\log k = 14.34 - 1.25 \times \frac{104\ \mathrm{K}}{T}
Calculate EaE_a for this reaction and at what temperature will its half-period be 256 minutes?
Solution:  
According to Arrhenius equation, k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}
or, ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}
or, log⁡k=log⁡A−Ea2.303RT…(i)\log k = \log A - \frac{E_a}{2.303 RT} \ldots (i)
Given equation is
          log⁡k=14.34−1.25×104KT…(ii)\;\;\;\;\;\log k = 14.34 - 1.25 \times 10^{4} \frac{\mathrm{K}}{T} \ldots (ii)
Comparing (i) with (ii),
Ea2.303RT=1.25×104 KT\frac{E_a}{2.303 RT} = \frac{1.25 \times 10^{4} \ \mathrm{K}}{T}
or,     Ea=2.303R×1.25×104 K\;\; E_a = 2.303 R \times 1.25 \times 10^{4} \ \mathrm{K}
=2.303×(8.314)×1.25×104= 2.303 \times (8.314) \times 1.25 \times 10^{4}
=239.34 kJ mol−1= 239.34 \ \mathrm{kJ} \ \mathrm{mol}^{-1}
When t1/2=256 min,t_{1/2} = 256 \ \mathrm{min},
k=0.693256×60k = \frac{0.693}{256 \times 60}
=4.51×10−5 s−1= 4.51 \times 10^{-5} \ \mathrm{s}^{-1}
Substituting this value in the given equation,
log⁡(4.51×10−5)=14.34−1.25×104 KT\log (4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^{4} \ \mathrm{K}}{T}
i.e., (−5+0.6542)=14.34−1.25×104 KT(-5 + 0.6542) = 14.34 - \frac{1.25 \times 10^{4} \ \mathrm{K}}{T}
or, 1.25×104 KT=18.6858\frac{1.25 \times 10^{4} \ \mathrm{K}}{T} = 18.6858 or, T=669 KT = 669 \ \mathrm{K}
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