Question

Hydrogen iodide decomposes at 800 K via a second-order process to produce hydrogen and iodine according to the following chemical equation. 2 HI(g) ---> H2 (g) + I2 g) At 800 K it takes 142 seconds for the initial concentration of HI to decrease from 6.75 x 10^-2 M to 3.50 x 10^-2 M. What is the rate constant for the reaction at this temperature?

A) 5.12 x 10^-4 M^-1 s^-1
B) 9.69 x 10^-2 M^-1 s^-1
C) 10.3 M^-1 s^-1
D) 1.95 x 10^3 M^-1 s^-1

Answer 1

Answer: The rate constant for the given reaction is
`9.69* 10^(-2)m^(-1)s^(-1)`

Step-by-step explanation:

The integrated rate law equation for second order reaction follows:

`k=(1)/(t)\left ((1)/([A])-(1)/([A]_o)\right)`

where,

k = rate constant = ?

t = time taken = 142 second

[A] = concentration of substance after time 't' =
`3.50* 10^(-2)M`

`[A]_o` = Initial concentration =
`6.75* 10^(-2)M`

Putting values in above equation, we get:

`k=(1)/(142)\left ((1)/((3.50* 10^(-2)))-(1)/((6.75* 10^(-2)))\right)\\\\k=9.69* 10^(-2)m^(-1)s^(-1)`

Hence, the rate constant for the given reaction is
`9.69* 10^(-2)m^(-1)s^(-1)`