Advertisements
Advertisements
Question
The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Advertisements
Solution
1. Order of a reaction = 1
`"t"_(1/2)` = 60 seconds,
k = ?
We know that, k = `0.6932/"t"_(1/2)`
k = `2.303/60` = 0.01155 s−1
2. [A0] = 100%
t = 180 s
k = 0.01155 seconds−1
[A] = ?
For the first order reaction k = `2.303/"t" log (["A"_0])/(["A"])`
0.01155 = `2.303/180 log (100/(["A"]))`
`(0.01155 xx 180)/2.303 = log (100/(["A"]))`
0.9207 = log 100 – log [A]
log [A] = log 100 – 0.9207
log [A] = 2 – 0.9207
log[A] = 1.0973
[A] = antilog of (1.0973)
[A] = 12.5%
APPEARS IN
RELATED QUESTIONS
Answer the following in brief.
Obtain the relationship between the rate constant and half-life of a first-order reaction.
For a first-order reaction, the rate constant is 6.909 min−1 the time taken for 75% conversion in minutes is
The rate constant of a reaction is 5.8 × 10−2 s−1. The order of the reaction is ____________.
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?
A reaction that is of the first order with respect to reactant A has a rate constant 6 min−1. If we start with [A]0 = 0.5 mol dm−3, when would [A] reach the value 0.05 mol dm−3?
If time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 hours, the half life of this first order reaction is ____________.
The expression to calculate time required for completion of zero order reaction is ______.
Which of the following is correct for a first-order reaction?
Which is the relation between half life and rate constant for a zero order?
