Question

At a certain temperature this reaction follows second-order kinetics with a rate constant of Suppose a vessel contains at a concentration of . Calculate the concentration of in the vessel seconds later. You may assume no other reaction is important. Round your answer to significant digits.

Answer 1

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

`2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)`

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
`SO_3` in the vessel after 0.240 seconds is 0.24 M

Step-by-step explanation:

For the given chemical equation:

`2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)`

The integrated rate law equation for second order reaction follows:

`k=(1)/(t)\left ((1)/([A])-(1)/([A]_o)\right)`

where,

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

t = time taken= 0.240 second

[A] = concentration of substance after time 't' = ?

`[A]_o` = Initial concentration = 1.44 M

Putting values in above equation, we get:

`14.1=(1)/(0.240)\left ((1)/([A])-(1)/(1.44)\right)`

`[A]=0.245M`

Hence, the concentration of
`SO_3` in the vessel after 0.240 seconds is 0.24 M