75% of a first-order reaction is completed in 30 minutes. The time required for a 93.75% completion of the same reaction (in hours) is: 

CONCEPT:

Half-Life of a First-Order Reaction

• The half-life (t_1/2) of a first-order reaction is independent of the initial concentration of the reactant and is given by:

  t_1/2 = 0.693 / k

• The time required for a certain percentage of the reaction to be completed can be calculated using the first-order rate law:

  t = (2.303 / k) log(1 / (1 - fraction completed))

EXPLANATION:

• From the problem, 75% of the reaction is completed in 30 minutes. For a first-order reaction, 75% completion means 3/4 of the reactant is consumed, and 1/4 remains.
• Using the first-order rate law:

  t = (2.303 / k) log(1 / (1 - fraction completed))

  Substituting the given values:

  30 = (2.303 / k) log(1 / (1 - 0.75))

  30 = (2.303 / k) log(1 / 0.25)

  30 = (2.303 / k) log(4)

  30 = (2.303 / k) × 0.6021

  k = (2.303 × 0.6021) / 30

  k ≈ 0.0463 min^-1

• Next, calculate the time required for 93.75% completion (fraction completed = 0.9375):

  t = (2.303 / k) log(1 / (1 - 0.9375))

  t = (2.303 / 0.0463) log(1 / 0.0625)

  t = (2.303 / 0.0463) log(16)

  t = (2.303 / 0.0463) × 1.2041

  t ≈ 60 minutes

• Since 60 minutes equals 1 hour, the time required for 93.75% completion is 1 hour.

Therefore, the correct answer is 1 hour.