Consider this reaction 290 g 2902g 02g At a certain tempera

Consider this reaction: 290 , (g) 2902(g) +02(g) At a certain temperature it obeys this rate law. (0.443 M i.s i)[SO 3]2 rate Suppose a vessel contains SO3 at a concentration of 0.800 M. Calculate how long it takes for the concentration of SO3 to decrease to 0.120 M. You may assume no other reaction is important. Round your answer to 2 significant digits

Solution

Ans. Note the unit of rate constant (M^-1 s^-1) – it is the unit of second order rate constant.

So, the reaction follows second order kinetics w.r.t. SO₃.

# Second order kinetics:      1/ [A]_t = kt + (1/ [A]₀)      - equation 1

Where, [A]₀ = Initial concentration of reactant

[A]_t = Final concentration of reactant after time t

k = rate constant

t = time of reaction

Given-

            Rate constant, k = 0.443 M^-1 s^-1

            [A]₀ = 0.800 M

            [A]_t = 0.120 M

Putting the values in equation 1-

            1/ 0.120 M = (0.443 M^-1 s^-1 x t) + (1/ 0.800 M)

            Or, 8.33 M^-1 – 1.25 M^-1 = 0.443 M^-1 s^-1 x t

            Or, t = 7.08 M^-1 / (0.443 M^-1 s^-1)

            Or, t = 15.98 s

Therefore, required time, t = 15.98 s = 16 s