Respuesta :
The question is incomplete, the complete question is:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1[tex]M^{-1}s^{-1}[/tex] :
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]
Suppose a vessel contains [tex]SO_3[/tex] at a concentration of 1.44 M . Calculate the concentration of
Answer:
The concentration of [tex]SO_3[/tex] in the vessel 0.240 seconds later will be 0.2452 M.
Explanation:
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
k = rate constant
[tex]a_0[/tex] = initial concentration
a = Concentration after time t.
We have :
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]
Initial concentration [tex]SO_3[/tex] = [tex]a_0=1.44 M[/tex]
Final concentration [tex]SO_3[/tex] at t = a =?
t = 0.240 s
Rate constant of the reaction = k = 14.1[tex]M^{-1}s^{-1}[/tex]
[tex]\frac{1}{a}=14.1 M^{-1}s^{-1}\times 0.240 s+\frac{1}{1.44 M}[/tex]
[tex]a=0.2452 M[/tex]
The concentration of [tex]SO_3[/tex] in the vessel 0.240 seconds later will be 0.2452 M.
