Respuesta :
The question is incomplete, here is the complete question:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]
Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer: The concentration of [tex]SO_3[/tex] in the vessel after 0.240 seconds is 0.24 M
Explanation:
For the given chemical equation:
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]
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 = [tex]14.1M^{-1}s^{-1}[/tex]
t = time taken= 0.240 second
[A] = concentration of substance after time 't' = ?
[tex][A]_o[/tex] = Initial concentration = 1.44 M
Putting values in above equation, we get:
[tex]14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)[/tex]
[tex][A]=0.245M[/tex]
Hence, the concentration of [tex]SO_3[/tex] in the vessel after 0.240 seconds is 0.24 M
