second order rate constant from starting constant

Question

The following reaction :

2A → B + C

Is second order, following the rate law Rate = k[A]². If the starting concentration [A₀] = 0.250 M, and after 485 min, [A(t)] = 0.015 M, what is the second order rate constant k (in M⁻¹min⁻¹)?

Answer 1

second-order rate equation = k * [A]^2

reads as rate constant (k) times the concentration raises to the second power

rate = k * [A] * [A]

the time dependence of the concentration is given by

where [A] is the concentration at time t and [A]₀ is the initial concentration at zero time (the start)

the given information is rate constant ..... t = 485 min

given [A]₀ = 0.250 M, take the reciprocal using a calculator and write it down

given[A] = 0.015 M, take the reciprocal using a calculator and write it down

1 / [A]₀ = __________

1 / [A] = __________

notice that the unit M is in the denominator and you can group them on the left

as 1 / [A] minus 1 / [A]₀

divide by the time and you've solved for second-order rate constant