Question

The decomposition of hydrogen iodide in the gas phase goes as follows at 600 K. 2 HI (g)->H2 (g) + I2 (g) Rate = k[HI]2 k = 9.7x10-6 M-1 s -1

a) Starting with an HI concentration of 0.030 M, what is the concentration of HI after 6.00 days?

b) What is the concentration of H2 in the flask after 6.00 days?

c) How much time would it take to drop the concentration from 0.030 M to 0.025 M?

d) How much time is required for a sample of HI to drop to 20.0 % of its original value

Answer 1

rate is given by

-d[HI]/dt= K[HI]² , This equation when integrated gives

1/[HI]= 1/[HI]o+ Kt

Where [HI]o= initial concentration of HI

And [HI]= concentration of HI at any time , t

K= rate constant = 9.7*10^-6 /M.Sec

When t= 6days = 6*24*60*60 seconds =129600 sec

[HI]0 =0.03M

1/[HI]= 1/[HI]o+ 129600*9.7*10^-6,

1/[HI] = 1/0.03 +1.26 =34.59

[HI] =1/34.59=0.029 M

2.For every 2 moles of HI reacted, one moles of H2 is formed. Hence concentration of H2= 0.029/2=0.0145 M

3. 1/0.025= 1/0.03 +9.7*10^-6 t

6.7= 9.7*10^-6 t

Hence t= 690721 sec

4. [HI] =0.2*[Hi]0 =0.2*0.03=0.006

1/0.006 =1/0.03 +9.7*10^-6t                    

Hence t= 1.68*10⁷ seconds