Question

Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI and the rate constant is 9.7×10−6M−1s−1. If the initial concentration of HI is 0.100 M .

What is its molarity after a reaction time of 5.00 days?

Answer 1

The reaction of decomposition of HI can be written as follows:

2HI -----> H2 + I2

The reaction is second order in HI. The Integrated second order rate law equation for this reaction can be given as follows:

1/[HI] = 1/[HI]_o + kt

Where, [HI] = molarity of HI after time t

[HI]_o = initial molarity of HI = 0.100 M

k = rate constant = 9.7×10⁻⁶ M⁻¹s⁻¹

t = time = 5.00days = 5.00 x 24 x 60 x 60 seconds = 432,000 seconds

Thus, molarity after 5.00days [HI] can be calculated as follows:

1/[HI] = 1/(0.100 M) + 9.7×10⁻⁶M⁻¹s⁻¹ x 432,000 s

1/[HI] = 10M^-1 + 4.2 M^-1

1/[HI] = 14.2M^-1

[HI] = (1/14.2)M

[HI] = 0.070 M

Thus, molarity of [HI] after five days of reaction will be 0.070 M