Question

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 0.110 M. What is its molarity after a reaction time of 5.00 days? Express your answer in moles per liter to two significant figures.

Answer 1

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

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 = $9.7\times 10^{-6}M^{-1}s^{-1}$

t = time taken  = 5.00 days

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

$9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)$

$[A]=0.109M$

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M