The concentration of SO3 in the vessel 0.240 seconds later is 0.25 M.
Second-order kinetics is a type of chemical reaction in which the rate of the reaction is proportional to the square of the concentration of a single reactant. The rate law for a second-order reaction is:
rate = k[A]^2
where:
rate is the rate of the reaction
k is the rate constant
[A] is the concentration of the reactant
The rate constant for this reaction is 14.1·M−1s−1. The initial concentration of SO3 is 1.44M. We can use the following equation to calculate the concentration of SO3 after 0.240 seconds:
[A] = [A]0 / (1 + kt)
where:
[A] is the concentration of SO3 after 0.240 seconds
[A]0 is the initial concentration of SO3
k is the rate constant
t is the time
Plugging in the values, we get:
[A] = 1.44M / (1 + (14.1·M−1s−1)(0.240s)) = 0.25M
Therefore, the concentration of SO3 in the vessel 0.240 seconds later is 0.25 M.
Question
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g 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 .