Advertisements
Advertisements
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?
40% of a first order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction?
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Answer the following in brief.
Derive the integrated rate law for the first-order reaction.
Answer the following in brief.
Give one example and explain why it is pseudo-first-order.
Time required for 100% completion of a zero order reaction is _______.
Derive an integrated rate law expression for first order reaction: A → B + C
Derive an expression for the relation between half-life and rate constant for first-order reaction.
The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the
If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in ____________.
Give two examples for zero order reaction.
A first order reaction has a rate constant 0.00813 min-1. How long will it take for 60% completion?
The rate constant of a reaction has same units as the rate of reaction. The reaction is of ____________.
