Question

The reaction 2NOBr (g) → 2 NO (g) + Br2 (g) is a second-order reaction with a rate constant of 0.80 M-1s-1 at 11 °C. If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 6.0 seconds is ________.

Answer 1

Answer : The concentration of NOBr after 6.0 seconds is, 0.036 M

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time = 6.0 s

$[A_t]$ = final concentration = ?

$[A_o]$ = initial concentration = 0.0440 M

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

$0.80\times 6.0=\frac{1}{[A_t]}-\frac{1}{0.0440}$

$[A_t]=0.036M$

Therefore, the concentration of NOBr after 6.0 seconds is, 0.036 M