Question

In a second-order reaction, the rate constant is 4.00 × 10⁻⁴ M⁻¹ s⁻¹. What is the concentration of reactant after 10 min if the initial concentration is 0.800 M?

a. 0.080 M
b. 0.100 M
c. 0.160 M
d. 0.200 M

Answer 1

In a second-order reaction, the concentration of reactant after 10 min can be calculated using the rate constant and initial concentration. By substituting the given values into the equation, the final concentration is determined to be 0.080 M.

Step-by-step explanation:

In a second-order reaction, the rate constant can be calculated using the equation rate = k[A]², where [A] is the concentration of the reactant. Given a rate constant of 4.00 × 10⁻⁴ M⁻¹ s⁻¹ and an initial concentration of 0.800 M, we can calculate the final concentration using the equation:

[A]final = [A]initial / (1 + k[A]initial * t)

Substituting in the given values, we get:

[A]final = 0.800 / (1 + (4.00 × 10⁻⁴ * 0.800 * 10)) = 0.080 M

Therefore, the concentration of the reactant after 10 min would be 0.080 M.