Question

A drug with concentration 0.83 M decomposes via acid-catalysed hydrolysis with a rate constant of 0.026 M−1 h−1. What is the drug concentration after 55 hours? Report your answer to 2 decimal places. How to solve this?

Answer 1

The rate law for the acid-catalyzed hydrolysis of the drug is given by:

rate = k * [drug] * [H+]

where k is the rate constant, [drug] is the concentration of the drug, and [H+] is the concentration of the acid.

Assuming that the concentration of the acid is constant and that the initial concentration of the drug is [drug]0, the concentration of the drug after time t can be calculated using the integrated rate law for a first-order reaction:

[drug] = [drug]0 * e^(-kt)

where e is the mathematical constant approximately equal to 2.71828, and t is the time in hours.

Substituting the given values into the equation, we have:

[drug]0 = 0.83 M (given)

k = 0.026 M^-1 h^-1 (given)

t = 55 h (given)

Substituting these values into the integrated rate law, we get:

[drug] = 0.83 M * e^(-0.026 M^-1 h^-1 * 55 h)

Simplifying the expression inside the exponential, we get:

[drug] = 0.83 M * e^(-1.43)

Calculating the exponential using a calculator or spreadsheet software, we get:

[drug] ≈ 0.34 M

Therefore, the concentration of the drug after 55 hours of acid-catalyzed hydrolysis is approximately 0.34 M, rounded to 2 decimal places.