Question

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹ : ---> 2SO₃g + 2SO₂g O₂g Suppose a vessel contains SO₃ at a concentration of 1.44M . 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 1

Answer : The concentration of SO₃ in the vessel 0.240 seconds later is, 0.24 M

Explanation :

The expression used for second order kinetics is:

`kt=(1)/([A_t])-(1)/([A_o])`

where,

k = rate constant =
`14.1M^(-1)s^(-1)`

t = time = 0.240 s

`[A_t]` = final concentration = ?

`[A_o]` = initial concentration = 1.44 M

Now put all the given values in the above expression, we get:

`14.1* 0.240=(1)/([A_t])-(1)/(1.44)`

`[A_]t=0.24M`

Therefore, the concentration of SO₃ in the vessel 0.240 seconds later is, 0.24 M