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Ronch [10]
3 years ago
7

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/mo

l·s. If the initial concentration of AB is 1.50 M, what is [AB] after 11.9 s?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
3 0

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.

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