If 50% if reactant is converted to product in first order in 25 minutes.how much of it would react in 100 minutes?
it is reduced by 1/2 each 25 min.
100 min = 4 * 25
so, (1/2)^4 remains
To find out how much of the reactant would be converted in 100 minutes, we need to use the concept of first-order kinetics.
First-order kinetics follows an exponential decay model, which means that the rate of reaction is directly proportional to the concentration of the reactant. The rate constant for a first-order reaction, denoted as k, remains constant throughout the reaction.
The general equation for a first-order reaction is given by:
ln([A]t/[A]0) = -kt
Where:
[A]t = concentration of reactant at time t
[A]0 = initial concentration of reactant
k = rate constant for the reaction
t = time
Given that 50% of the reactant is converted to product in 25 minutes, we can calculate the rate constant (k):
ln([A]t/[A]0) = -kt
For 50% conversion, [A]t is equal to 0.50 and [A]0 is equal to 1.00 (100%):
ln(0.50/1.00) = -k * 25 minutes
Simplifying this equation, we find:
ln(0.50) = -k * 25 minutes
Calculating the natural logarithm of 0.50 gives:
-0.6931 = -k * 25 minutes
Dividing both sides of the equation by -25 minutes:
k ≈ 0.0277 min⁻¹
Now that we have the rate constant (k), we can use it to calculate the amount of reactant remaining after 100 minutes.
Using the same equation as before:
ln([A]t/[A]0) = -kt
Let's denote [A]t as the concentration of reactant remaining after 100 minutes. We can solve for [A]t:
ln([A]t/1.00) = -0.0277 min⁻¹ * 100 minutes
Simplifying this equation, we find:
ln([A]t/1.00) = -2.77
Calculating the exponent of both sides, using the natural logarithm as the inverse of the exponential function:
[A]t/1.00 = e^(-2.77)
Solving for [A]t, we find:
[A]t ≈ 0.0623
Therefore, approximately 6.23% of the reactant would react in 100 minutes.