Question

Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI and the rate constant is 9.7×10^−6 M^−1s^−1. If the initial concentration of HI is 0.130 M...

What is its molarity after a reaction time of 7.00 days?

What is the time (in days) when the HI concentration reaches a value of 7.5×10^−2 M ?

Answer 1

1)

we have:

[HI]o = 0.13 M

t = 7 days = 7*24*3600 s = 604800 s

k = 9.7*10^-6 M-1.s-1

use integrated rate law for 2nd order reaction

1/[HI] = 1/[HI]o + k*t

1/[HI] = 1/(0.13) + 9.7*10^-6*6.048*10^5

1/[HI] = 7.692 + 9.7*10^-6*6.048*10^5

1/[HI] = 13.56

[HI] = 7.375*10^-2 M

Answer: 7.38*10^-2 M

2)

we have:

[HI]o = 0.13 M

[HI] = 7.5*10^-2 M

k = 9.7*10^-6 M-1.s-1

use integrated rate law for 2nd order reaction

1/[HI] = 1/[HI]o + k*t

1/(7.5*10^-2) = 1/(0.13) + 9.7*10^-6*t

13.33 = 7.692 +9.7*10^-6*t

9.7*10^-6*t = 5.641

t = 5.815*10^5 s

= (5.815*10^5)/(24*3600) days

= 6.73 days

Answer: 6.73 days