At 25 ° C, hydrogen iodide decomposes very slowly to hydrogen and iodine in a second order reaction.
2 HI (g) →H₂ (g) + I₂ (g)
If the initial concentration of HI is 0.0100 M, how long will it take for the concentration of HI to reduce to 0.00500 M?
K = 2.40×10^{- 21} L/mol × s

Question

Quizplus · Accepted Answer

We can use the integrated rate law for a second-order reaction to solve for time: $$\frac{1}{[HI]}-\frac{1}{[HI]_0}=kt$$ where [HI] is the concentration at time t, [HI]0 is the initial concentration, k is the rate constant, and t is time. Plugging in the given values: $$\frac{1}{0.00500}-\frac{1}{0.0100}=2.40 imes10^{-21}~\mathrm{L/mol\cdot s}~t$$ Solving for t: $$t=\frac{\frac{1}{0.00500}-\frac{1}{0.0100}}{2.40 imes10^{-21}~\mathrm{L/mol\cdot s}}=4.17 imes10^{22}~\mathrm{s}$$ Since the answer choices are in seconds, we need to convert this to scientific notation with 2 decimal places: $$t=4.17 imes10^{22}~\mathrm{s}=4.17 imes10^{22-21}~\mathrm{s}=4.17 imes10^1~\mathrm{s}=41.70~\mathrm{s}=4.17 imes10^{22\pm2}~\mathrm{s}$$ The closest answer choice to this is E: 4.17×10²² seconds.

At 25 ° C, Hydrogen Iodide Decomposes Very Slowly to Hydrogen