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 0.140 M .
What is its molarity after a reaction time of 7.00 days?

What is the time (in days) when the HI concentration reaches a value of 8.0×10−2 M ?

Answer 1

Answer:

For 1: The concentration after the given time is 0.077 M

For 2: The time taken to reach the given value is 6.39 days

Explanation:

The integrated rate law equation for second order reaction follows:

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

where,

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

t = time taken

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

$[A]_o$ = Initial concentration = 0.140 M

• For 1:

We are given:

t = 7.00 days = (7 × 24 × 60 × 60) = 604800 seconds   (1 day = 24 hours, 1 hour = 60 mins, 1 min = 60 sec)

[A] = ?

Putting values in equation 1, we get:

$9.7\times 10^{-6}=\frac{1}{604800}\left (\frac{1}{[A]}-\frac{1}{0.140}\right)$

$[A]=0.077M$

Hence, the concentration after the given time is 0.077 M

• For 2:

We are given:

[A] = $8.0\times 10^{-2}M$

Putting values in equation 1, we get:

$9.7\times 10^{-6}=\frac{1}{t}\left (\frac{1}{(8.0\times 10^{-2})}-\frac{1}{0.140}\right)$

$t=552283s=6.39days$

Hence, the time taken to reach the given value is 6.39 days