Question

Hydrogen iodide decomposition via a second-order process to produce hydrogen and iodine according to the following chemical equation:
2 HI(g) -> H2 (g) + I2 (g)

If it takes 489 seconds for the initial concentration of HI to decrease from 0.0693 M to 0.0174 M, what is the rate constant for the reaction under these conditions? Units are in M-1 *s-1

Answer 1

So far i have done the following:

Ln of (concentration A divided by Concentration A at time 0) = -k * T

natural log (0.0174 M / 0.0693 M) / -489 sec
which equals to 2.826 x10^-3 s-1

Answer 2

1/A - 1/Ao = kt
A = 0.0174 M
Ao = 0.0693 M
t = 489 s
Solve for k.