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प्रश्न
Solve
A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life.
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उत्तर
Given:
[A]0 = 100%, [A]t = 100 - 30 = 70%, t = 40 min
To find:
Half life of reaction (t1/2)
Formula:
`"k" = 2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`
Calculation:
Substitution of these in above
`"k" = 2.303/"t" "log"_10 100/70`
= `2.303/"40 min" "log"_10 (1.429)`
= `2.303/"40 min" xx 0.155`
= 0.008924 min-1
t1/2 = `0.693/"k" = 0.693/(0.008924 "min"^-1) = 77.66 "min"`
The half life of reaction is 77.66 min.
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