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.