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18 Dec 2019

The decomposition of hydrogen iodide:

2HI(g) ® H2(g) + I2(g)

Is second order and has a rate constant of The rate constant is 0.838 M-1 day-1 at 600 K. Assuming an initial concentration of 0.100 M, determine at this temperature

The half-life (in days)

The number of days needed for the HI concentration to reach a value of 0.088 M

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Related questions

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


Any help would be greatly appreciated:

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 0.120 M:

(A) what is the molarity after 6 days (_______M)

(B) how many days will it take for the reaction to reach a value of 8.5 x 10^-2M (_________days)

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