Question

# Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order...

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 ?

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

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

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