Question

For the simple decomposition reaction: AB(g) LaTeX: \longrightarrow⟶ A(g) + B(g), the rate = k{AB}2 ({ = [) and k = 0.85 1/MLaTeX: \cdot⋅s. How long (in seconds) will it take for the concentration of AB to reach one-third of its initial concentration of 2.01?

Answer 1

Answer:

1.169s

Explanation:

k = 0.851 M-1s-1

The unit of the rate constant, k tells us this is a second order reaction.

From the question;

Initial Concentration  [A]o = 2.01M

Final Concentration  [A] = One third of 2.10 = (1/3) * 2.10 = 0.67M

Time = ?

The integrated rate law for second order reactions is given as;

1 / [A] = (1 / [A]o) + kt

Making t subject of interest, we have;

kt = (1 / [A] ) - (1 / [A]o )

t = (1 / [A] ) - (1 / [A]o ) / k

Inserting the values;

t = [ (1 / 0.67 ) - (1 /  2.10) ] / 0.851

t = ( 1.4925 - 0.4975 ) / 0.851

t = 0.995 / 0.851

t = 1.169s

[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)