chemistry homework question

Question

The rate constant for this second‑order reaction is  0.820 M−1⋅s−1 at 300 ∘C

A⟶products

How long, in seconds, would it take for the concentration of A to decrease from 0.990 M to  0.400 M?

Answer 1

Second order reactions

1/(CONC) _t = 1/(conc) _initial + kt

t stands for the time and k is the rate constant

1/0.4 = 1/0.990 + 0.820 X time

2.5 = 1.0111 + 0.820 X time

2.5 - 1.0111= 0.820 X time

Time = 1.82 seconds