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 9.7×10−6M−1s−1. If the initial concentration of HI is 0.100 M .

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

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

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