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CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

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Question : 13 of 26
Marks: +1, -0
The rate constant for the first order decompositon of H2O2\mathrm{H}_2\mathrm{O}_2 is given by the following equation :
log⁡k=14.2−1.0×104T K ; \log k = 14.2 - \frac{1.0 \times 10^4}{T}\,\mathrm{K} \text{ ; }
Calculate EaE_a for this reaction and rate constant kk if its half-life period be 200 minutes.
(Given : R=8.314 J K−1 mol−1R = 8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1} )
Solution:  
log⁡k=log⁡A−Ea2.303RT\log k = \frac{\log A - E_a}{2.303 RT}
Ea2.303RT=1.0×104 KT\frac{E_a}{2.303 RT} = \frac{1.0 \times 10^4\,\mathrm{K}}{T}
Ea=1.0×104×2.303×8.314E_a = 1.0 \times 10^4 \times 2.303 \times 8.314
=191471.4 J/mol= 191471.4\,\mathrm{J}/\mathrm{mol}
t12=0.693kt_{\frac{1}{2}} = \frac{0.693}{k}
k=0.693200 mink = \frac{0.693}{200}\,\mathrm{min}
=0.0034 min−1= 0.0034\,\mathrm{min}^{-1}
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