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3 years ago

What is the equilibrium constant expression for the given system?

Chemistry

1 answer:

artcher [175]3 years ago

5 0

<h3>Answer:</h3>

$\displaystyle K_{eq} = \frac{[H_2]^2[O_2]}{[H_2O]^2}$

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Aqueous Solutions</u>

States of matter

<u>Kinetics</u>

Rate Laws

<u>Equilibrium</u>

Equilibrium Reactions RxN ⇄
Equilibrium Constants
Equilibrium Expression: $\displaystyle K = \frac{[Producst]^{coefficients}}{[Reactants]^{coefficients}}$

<h3>Explanation:</h3>

*Note: Only gases and liquids affect equilibrium

<u>Step 1: Define</u>

[RxN - Balanced] 2H₂O (g) ⇄ 2H₂ (g) + O₂ (g)

<u>Step 2: Identify</u>

[Reactant] H₂O

Coefficient of 2

[Product] H₂

Coefficient of 2

[Product] O₂

Coefficient of 1

<u>Step 3: Write K Expression</u>

Substitute: $\displaystyle K_{eq} = \frac{[H_2]^2[O_2]}{[H_2O]^2}$

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Maurinko [17]

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3 years ago

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Hopefully, this helps! =)

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malfutka [58]

Answer:

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

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Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

$Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)$

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