Question

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concentration of is 0.0440 , the concentration of after 9.0 seconds is ________. 0.0325 M 0.0276 M 0.0403 M 0.0334 M 0.0342 M

Answer 1

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

$2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 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 = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

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

$[A]_o$ = Initial concentration = 0.0440 M

Putting values in above equation, we get:

$0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)$

$[A]=0.00734M$

Hence, the concentration of after 9.0 seconds is, 0.00734 M