If it takes 84.2 min for the concentration of a reactant to drop to 20.0% of its initial value in a first-order reaction, what is the rate constant for the reaction in the units min-1?
ln(No/N) = kt
No = 100 arbitrary number
N = 20
k = unknown
t = 84.2
Solve for k.
ln[80/100]=-k[84.2]
-2.2=-k[84.2]
-2.2/-84.2=k
k=0.0261min^-1
we know that
k=2.303/t log[a/a-x]
here a-x=20
so k=2.303/84.2 log[100/20]
To find the rate constant for a first-order reaction, we can use the formula:
k = ln(2) / t1/2
Where:
k - rate constant
ln - natural logarithm
2 - value representing the change in concentration (in this case, the concentration dropping to 20% is equivalent to a change of 80%, which is approximately 2 in terms of the natural logarithm)
t1/2 - half-life of the reaction (the time in which the concentration drops to half of its initial value)
In this case, the given time is 84.2 minutes for the concentration to drop to 20% of its initial value. To find the half-life, we can use:
t1/2 = ln(2) / k
We need to rearrange this equation to find the rate constant:
k = ln(2) / t1/2
Now, substituting the given values:
k = ln(2) / 84.2 min
Using a calculator or a scientific calculator, calculate the natural logarithm of 2 and divide it by 84.2 minutes to find the rate constant in units min-1.