Question

Enter your answer in the provided box. For the simple decomposition reaction AB(g) → A(g) + B(g) rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 11.9 s?

Answer 1

Answer:

The answer to the question is;

The concentration of AB, which is written as [AB] after 11.9 s is 0.328 M.

Explanation:

The form of the equation for the second order integrated rate law is

y = mx +b

Where:

y = 1/A

m = k

x = t and

b = 1/A₀

Therefore for the reaction

AB(g) → A(g) + B(g)

rate = k[AB]² and k = 0.20 L/mol·s we have

$\frac{1}{[AB]} = \frac{1}{[AB]_0} + kt$

Since the initial concentration of AB is [AB]₀ = 1.50 M,

t = 11.9 s

k = 0.20 L/mol·s

we have

$\frac{1}{[AB]} = \frac{1}{1.50 M/L} + 0.20 L/(mol)(s)*11.9 s$

$\frac{1}{[AB]}$ = 3.047 L/mol

[AB] = 0.328 M

The concentration of AB after 11.9 s is 0.328 M.