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प्रश्न
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?
उत्तर
(i) log k = log A − `"E"_"a"/(2.303 "RT")`
The given equation is as follows:
log k = `14.34 - (1.25 xx 10^4)/"T"`
On comparing both the equations,
`"E"_"a"/(2.303 "R")` = 1.25 × 104
or, Ea = 1.25 × 104 × 2.303 × 8.314 J mol−1
= 239.34 kJ mol−1
(ii) `t_(1/2)` = 256 min = 256 × 60 s
k = `0.693/(256 xx 60)`
= 4.51 × 10−5 s−1
∴ log 4.51 × 10−5 = `14.34 - (1.25 xx 10^4)/"T"`
−4.35 = `14.34 - (1.25 xx 10^4)/"T"`
∴ T = 669 K
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