Question

For the reaction:
A ⇌ B + C
the equilibrium constant is 3.0 x 10–6. What is the concentration of B at equilibrium if A was originally 0.10 M?

Answer 1

Answer on Question #52688, Chemistry, Other

Question:

For the reaction:

$$A \rightleftharpoons B + C$$

the equilibrium constant is $3.0 \times 10^{-6}$. What is the concentration of B at equilibrium if A was originally $0.10 \, \text{M}$?

Answer:

Initial concentrations:

[A] = 0.1 M

[B] = 0 M

[C] = 0 M

When the system reaches equilibrium, we assume that X moles of A are used up but 0.1−x moles remain. This produces X moles of B and X moles of C

Equilibrium concentrations:

[A] = 0.1−X

[B] = X

[C] = X

$$K_c = [B][C]/[A] = \frac{x^2}{0.1−X} = 3 \times 10^{-6}$$

At this point I'm going to make an assumption that, since Kc is so small, then $(0.1−X) \Rightarrow 0.1$.

$$X^2 = 3 \times 10^{-6} \times 0.1 = 3 \times 10^{-7}$$ $$X = [B] = \sqrt{ (3 \times 10^{-7}) } = 5.48 \times 10^{-4} \, \text{M}$$

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