Question

Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?

Answer 1

A (aq) ⇌ 2B (aq) + C (aq)

Remember to use correct significant figures in your answer (round your answer to the nearest ten thousandth). Do not include units in your response.

Answer 2

Answer:

0.0119

Explanation:

There was a part missing. I think this is the whole question:

Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?

A (aq) ⇌ 2B (aq) + C(aq)

Remember to use correct significant figures in your answer. Do not include units in your response.

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

        A (aq) ⇌ 2B (aq) + C (aq)

I       0.0510       0            0

C         -x          +2x          +x

E    0.0510-x     2x            x

Since the concentration at equilibrium of A is 0.0153 M, we get

$0.0510 - x = 0.0153 \\x = 0.0357$

We can use the value of x to calculate the concentrations at equilibrium.

$[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M$

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

$Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119$

The equilibrium constant for this reaction at equilibrium is 0.0119.

You can learn more about equilibrium here: https://brainly.com/question/4289021