In 1 st order reaction 60% of reactant decomposes in 4.606 min then what is t1/2
Question
In 1 st order reaction 60% of reactant decomposes in 4.606 min then what is t1/2
Solution
The half-life (t1/2) of a first-order reaction can be calculated using the formula:
t1/2 = 0.693/k
where k is the rate constant.
The rate constant can be calculated using the formula:
k = -ln(1 - x)/t
where x is the fraction of reactant that has decomposed and t is the time it took for that fraction to decompose.
In this case, 60% of the reactant has decomposed in 4.606 minutes. So, x = 0.60 and t = 4.606 minutes.
First, calculate the rate constant:
k = -ln(1 - 0.60)/4.606 = 0.150 min^-1
Then, calculate the half-life:
t1/2 = 0.693/0.150 = 4.62 minutes
So, the half-life of the reaction is approximately 4.62 minutes.
Similar Questions
A substance 'A' decomposes by a first order reaction starting initially with [A]=2.00 m and after 200 min,[ A] becomes 0.15 m. For this reaction t1/2 is53.72 min50.49 min48.45 min46.45 min
Time required for completion of 99.9% of a First order reaction is ________ times of half life (t1/2) of the reaction.
99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?
In a 1 st order reaction, reactant concentration C varies with time t as :1C increases linearly with tlogC decreases linearly with tC decreases with 1tlogC decreases with 1t
Which of the following are true for the first order reaction? t3/4 = 2t1/2 t15/16 = 4t1/2 t15/16 = 3t3/4 t7/8 = 2t3/4
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.