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 $9.69\times 10^{-2}m^{-1}s^{-1}$

Explanation:

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = ?

t = time taken  = 142 second

[A] = concentration of substance after time 't' = $3.50\times 10^{-2}M$

$[A]_o$ = Initial concentration = $6.75\times 10^{-2}M$

Putting values in above equation, we get:

$k=\frac{1}{142}\left (\frac{1}{(3.50\times 10^{-2})}-\frac{1}{(6.75\times 10^{-2})}\right)\\\\k=9.69\times 10^{-2}m^{-1}s^{-1}$

Hence, the rate constant for the given reaction is $9.69\times 10^{-2}m^{-1}s^{-1}$