Question

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentration of 0.86 M, calculate the concentration of NOBr after 95 s.

Answer 1

Answer : The concentration of NOBr after 95 s is, 0.013 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 = $0.80M^{-1}s^{-1}$

t = time taken  = 95 s

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

$[A]_o$ = Initial concentration = 0.86 M

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

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M