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=(1)/(t)\left ((1)/([A])-(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)=(1)/(142s)\left ((1)/([A])-(1)/((0.0440M))\right)`

`[A]=0.00734M`

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