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:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

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:

$14.1\times 0.240=\frac{1}{[A_t]}-\frac{1}{1.44}$

$[A_]t=0.24M$

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