Final answer:
To determine how long it takes for 5 g of a reactant to reduce to 3 g in a first-order reaction with a rate constant of 1.15×10⁻³ s⁻¹, use the integrated rate law and solve for time (t), resulting in approximately 669 seconds.
Step-by-step explanation:
To calculate the time it takes for a first-order reaction with a known rate constant to reduce from 5 g to 3 g, one can use the integrated first-order rate law, which is ln([A]t/[A]0) = -kt. Here, [A]0 is the initial concentration, [A]t is the concentration at time t, and k is the rate constant. Given that the rate constant (k) is 1.15×10⁻³ s⁻¹ for the reaction, we first need to convert the masses to concentrations. Assuming the volume of the reaction does not change, we can simply use the ratios of the masses to represent the ratios of the concentrations.
To solve for t when the reaction goes from 5 g to 3 g:
- Initial concentration [A]0 = 5 g (arbitrarily assuming unit volume)
- Final concentration [A]t = 3 g
- Rate constant k = 1.15×10⁻³ s⁻¹
We plug these values into the equation:
ln(3/5) = -(1.15×10⁻³ s⁻¹)t
Solving for t gives:
t = -ln(3/5) / (1.15×10⁻³ s⁻¹) = 669 s.