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Dmitrij [34]
2 years ago
6

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

ncentration of A at equilibrium is 0.0153 M?
Chemistry
2 answers:
Lilit [14]2 years ago
7 0

Answer:

0.0119

Explanation:

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

<em>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?</em>

A (aq) ⇌ 2B (aq) + C(aq)

<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

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: brainly.com/question/4289021

djyliett [7]2 years ago
5 0
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.
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