Question

A certain compound, XY, undergoes a first-order reaction to form its products. At a particular temperature, the partial pressure of XY in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature?

a) 41.35 seconds
b) 15.5 seconds
c) 124 seconds
d) 62.0 seconds

Answer 1

The half-life for the first-order reaction where compound XY's partial pressure drops to one-eighth in 124 seconds is 41.35 seconds.

Step-by-step explanation:

A compound XY undergoing a first-order reaction has its partial pressure fall to one-eighth its initial value after 124 seconds. To calculate the half-life, we must understand that for a first-order reaction the time required for the concentration of a reactant to reduce to one-half (t₁/₂) is a constant and is independent of the concentration. Since the pressure falls to 1/8 (or (1/2)³), it has gone through three half-lives. Therefore, 124 seconds represent three half-lives of the reaction.

To find the half-life (t₁/₂), we divide the total time by 3:

Half-life (t₁/₂) = 124 s / 3

Half-life (t₁/₂) =41.333... seconds, which rounds to 41.35 seconds. Hence, the correct answer is (a) 41.35 seconds.

The half-life is essential in understanding how quickly the reactants become products in a reaction.