Chemistry, 03.08.2019 02:30 sainijasdeep27

Write the equilibrium‑constant expression for the reaction a(s)+3b(l)↽−−⇀2c(aq)+d(aq) in terms of [a], [b], [c], and [d], as needed. note that , which is sometimes symbolized as , denotes that the equilibrium constant is expressed using molar concentrations. for this question, means the same thing as .

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Answer from: mathhelpneeded66

The expression for equilibrium constant in terms of concentration is given below.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by $K_{eq}$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The $K_{eq}$ is written as:

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solid and pure liquid are taken as 1 in equilibrium expression.

For the given chemical reaction:

$A(s)+3B(l)\rightleftharpoons 2C(aq.)+D(aq.)$

The expression for $K_c\text{ or }K_{eq}$ for the given reaction follows:

$K_{c}=\frac{[C]^2[D]^1}{[A]^1[B]^3}$

Here, A and B both are pure solid and pure liquid respectively. Thus, their concentration are taken as 1.

Thus, the expression becomes:

[tex]K_{c}=[C]^2[D]^1[/tex

Hence, the expression for equilibrium constant in terms of concentration is given above.

Answer from: kiekie1986

Kc = [C]². [D]

Further explanation

For the reaction $\boxed{ \ A_{(s)} + 3B_{(l)} \rightleftharpoons 2C_{(aq)} + D_{(aq)} \ }$ , we can observe that the substances on the right-hand side have a solution phase (aq) that is allowed into the equilibrium constant K.

Remember, pure solids and liquids are disregarded and kept at 1.

Therefore, from the basic formula $\boxed{ \ K = \frac{[C]^2.[D]}{[A].[B]^3} \ }$ we get the final result $\boxed{ \ K = [C]^2.[D] \ }$

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Notes:

In 1864, chemists Cato Maximilian and Peter Waage jointly proposed and developed the law of chemical equilibrium, that at a certain temperature the chemical system would reach a state in which the ratio of the results of the reaction and reagent concentration is constant. The fixed value is formulated as the equilibrium constant (K). $\boxed{ \ K_c \ or \ K_{eq} \ }$ , denotes that the equilibrium constant is expressed using molar concentrations, i.e., $\boxed{ \ mol.dm^{-3} \ or \ mol.L^{-1} \ \ }$ . For this question, $\boxed{ \ K_c \ }$ means the same thing as $\boxed{ \ K_{eq} \ }$ .The rIght-hand side of the equation on top, left-hand side of the equation on the bottom.The square brackets show concentrations in $\boxed{ \ mol.dm^{-3} \ or \ mol.L^{-1} \ \ }$ .The indices are the numbers in front of each substance (or the coefficients) in the chemical equation.A heterogeneous balance consists of more than one phase. Typical examples include reactions involving solids and gases, or solids and liquids. Substances that can be included in formula K are substances with a gas phase (g) or aqueous phase (aq). Pure solids and liquids are unincluded in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1.Learn moreWrite the equilibrium constant for the reaction About the solution as a homogeneous mixture About electrolyte and nonelectrolyte solutions
Write the equilibrium‑constant expression for the reaction a(s)+3b(l)↽−−⇀2c(aq)+d(aq) in terms of [a

Write the equilibrium‑constant expression for the reaction a(s)+3b(l)↽−−⇀2c(aq)+d(aq) in terms of [a

Answer from: Quest

the answer is either b or c

Answer from: Quest

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Write the equilibrium‑constant expression for the reaction a(s)+3b(l)↽−−⇀2c(aq)+d(aq) in terms of [a...

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