Respuesta :
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
[tex]2NOBr(g)\rightarrow 2NO(g)+Br_2(g)[/tex]
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:
[tex]k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)[/tex]
where,
k = rate constant = 0.80 M⁻¹s⁻¹
t = time taken = 142 second
[A] = concentration of substance after time 't' = ?
[tex][A]_o[/tex] = Initial concentration = 0.0440 M
Putting values in above equation, we get:
[tex]0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)[/tex]
[tex][A]=0.00734M[/tex]
Hence, the concentration of after 9.0 seconds is, 0.00734 M
