Question #191716

Hydrogen Iodide (HI) gas decomposes into gaseous hydrogen and iodine following the second-order kinetics with a rate constant of 0.031/M.s at 400ÂºC.Â

Â

2HI(g)Â Â Â Â Â Â Â Â H2(g) + I2(g)Â

Â

a.What is the half-life of the reaction if the initial concentration of HI is 0.045 M? Will the half-life be different if the initial concentration is 0.025 M? Justify your answer.Â

b.How much HI (in M) remains after 5.0 Minutes if the reaction started with 0.024 M HI?Â

c.How many minutes will it take for the concentration of HI to decrease from 0.045 M to 0.015 M?Â

Expert's answer

**A. **Half life of second order "=\\dfrac{1}{KC_0}"

Now when "C_0=0.045" M

Half time"=" "\\dfrac{1}{0.031\\times0.045}=7.1\\times10^{2}" seconds

Now when "C_0=0.025" M

Half time"=\\dfrac{1}{0.031\\times0.025}=1.2\\times10^3" seconds

Here half time changes as the half time is inversely proportional to the initial conc.

B. Now for the second order reaction we know that

"\\dfrac{1}{C_t}-\\dfrac{1}{C_0}=Kt"** **

where t=5 minutes=300 seconds

Intial conc.=0.024 M

Rate constant=0.031

Now put in the equation

"\\dfrac{1}{C_t}=\\dfrac{1}{0.024}+0.031\\times300"

"C_t=0.02M"

C. Again putting the values in the formula

"\\dfrac{1}{0.015}-\\dfrac{1}{0.045}=0.031\\times t"

"t=143.35" seconds or "2.39" minutes approximately..

Learn more about our help with Assignments: Chemistry

## Comments

## Leave a comment