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प्रश्न
A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
उत्तर
1. For the first order reaction k = `2.303/"t" log (["A"_0])/(["A"])`
Assume, [A0] = 100%, t = 50 minutes
Therefore, [A] = 100 – 40 = 60
k = `(2.303/50) log (100/60)`
k = 0.010216 min−1
Hence the value of the rate constant is 0.010216 min−1
2. t = ?, when the reaction is 80% completed,
[A] = 100 – 80 = 20%
From above, k = 0.010216 min−1
t = `(2.303/0.010216) log (100/20)`
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.
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