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Write the equilibrium constant expression for: (a) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(g)\) \(\rightleftarrows \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\) \(\rightleftarrows \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{I}_{2}(s) \rightleftarrows \mathrm{I}_{2}(g)\) (e) \(\mathrm{TiCl}_{4}(g)+2 \mathrm{Mg}(s) \rightleftarrows \mathrm{Ti}(s)+2 \mathrm{MgCl}_{2}(s)\) (f) \(\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Ni}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\)

Short Answer

Expert verified
(a) \( K_a = \frac{[\text{HCl}]^4}{[\text{H}_2\text{O}]^2} \) (b) \( K_b = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{H}_2][\text{CO}_2]} \) (c) \( K_c = \frac{[\text{Mn}^{2+}][\text{Cl}_2]}{[\text{H}^+]^4[\text{Cl}^-]^2} \) (d) \( K_d = 1 \) (e) \( K_e = \frac{1}{[\text{TiCl}_4]} \) (f) \( K_f = \frac{[\text{Ni}^{2+}][\text{OH}^-]^2}{[\text{Ni(OH)}_2]} \)

Step by step solution

01

Write the equilibrium constant expression for reaction (a)

\( K_a = \frac{[\text{SiO}_2][\text{HCl}]^4}{[\text{SiCl}_4][\text{H}_2\text{O}]^2} \) Note that there's no need to include the solid substances (SiCl4 and SiO2) in the equilibrium constant expression, as their concentrations are considered constant due to their phase. So we can rewrite it as: \( K_a = \frac{[\text{HCl}]^4}{[\text{H}_2\text{O}]^2} \) (b) H2(g) + CO2(g) \(\rightleftarrows\) CO(g) + H2O(l)
02

Write the equilibrium constant expression for reaction (b)

\( K_b = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{H}_2][\text{CO}_2]} \) (c) MnO2(s) + 4H+(aq) + 2Cl-(aq) \(\rightleftarrows\) Mn2+(aq) + Cl2(g) + 2H2O(l)
03

Write the equilibrium constant expression for reaction (c)

\( K_c = \frac{[\text{Mn}^{2+}][\text{Cl}_2]}{[\text{H}^+]^4[\text{Cl}^-]^2} \) Ignore the solid substances (MnO2) and liquid substances (H2O) in the equilibrium constant expression. (d) I2(s) \(\rightleftarrows\) I2(g)
04

Write the equilibrium constant expression for reaction (d)

\( K_d = \frac{[\text{I}_2]}{[\text{I}_2]} \) Ignore the solid substances (I2) in the equilibrium constant expression.
05

Simplify K_d expression

\( K_d = 1 \) (e) TiCl4(g) + 2Mg(s) \(\rightleftarrows\) Ti(s) + 2MgCl2(s)
06

Write the equilibrium constant expression for reaction (e)

\( K_e = \frac{[\text{Ti}][\text{MgCl}_2]^2}{[\text{TiCl}_4][\text{Mg}]^2} \) Ignore the solid substances (Mg, Ti, and MgCl2) in the equilibrium constant expression. Therefore, the equilibrium constant is not dependent on their concentrations.
07

Simplify K_e expression

\( K_e = \frac{1}{[\text{TiCl}_4]} \) (f) Ni(OH)2(s) \(\rightleftarrows\) Ni2+(aq) + 2OH-(aq)
08

Write the equilibrium constant expression for reaction (f)

\( K_f = \frac{[\text{Ni}^{2+}][\text{OH}^-]^2}{[\text{Ni(OH)}_2]} \) Ignore the solid substances (Ni(OH)2) in the equilibrium constant expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium is a fascinating concept in chemistry that describes the state of a chemical reaction where the concentrations of reactants and products do not change over time. At this point, the forward reaction rate equals the backward reaction rate, creating a dynamic balance. This state of balance is not static, but rather, both reactions occur continuously at the same rate, maintaining a constant concentration of reactants and products.

The equilibrium state can be expressed mathematically using an equilibrium constant, denoted as \( K \). This constant indicates the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients from the balanced chemical equation. Understanding chemical equilibrium helps predict the extent of a reaction and formulate strategies to shift the equilibrium if necessary, such as by changing the temperature or concentration of reactants/products.

Consider both reversible reactions and how factors such as pressure, concentration, and temperature can affect equilibrium. By recognizing the conditions under which equilibrium occurs, chemists can control and optimize reactions for desired outcomes.
Reaction Balance
Reaction balance is an essential step in understanding equilibrium constants and chemical reactions. To write an equilibrium constant expression, one must first ensure the chemical equation is properly balanced. A balanced equation accurately reflects the conservation of atoms, where each type of atom is accounted for equally on both sides of the equation.

Balancing a reaction involves:
  • Ensuring the same number of each type of atom is present on both sides of the reaction.
  • Adjusting coefficients, which are the numbers before molecules, without changing the actual chemical formulas.
  • Considering charge balance, which is crucial for reactions involving ions in aqueous solutions.
For example, in the reaction \( \text{MnO}_2(s) + 4 \text{H}^+(aq) + 2 \text{Cl}^-(aq) \rightleftarrows \text{Mn}^{2+}(aq) + \text{Cl}_2(g) + 2 \text{H}_2\text{O}(l) \), coefficients are used to balance the oxygen, hydrogen, manganese, and chlorine atoms, ensuring both sides reflect equal numbers and maintaining charge neutrality.

This balanced equation is then used to write the equilibrium constant expression, showing the relationship between reactants and products at equilibrium.
Solid and Liquid Exclusion from Expressions
In equilibrium constants, solid and liquid substances are often excluded because their concentrations do not change during a reaction. These phases are considered pure substances with constant concentrations, so they do not appear in the equilibrium expression, which only includes substances whose concentrations can change, such as gases and aqueous ions.

This exclusion is based on the premise that:
  • Solids and liquids have fixed densities, hence their concentration is constant and does not affect the equilibrium state.
  • Only species with variable concentrations, primarily gases and solutes in solution, influence the equilibrium position.
For example, in the equilibrium expression for the reaction \( \text{Ni(OH)}_2(s) \rightleftarrows \text{Ni}^{2+}(aq) + 2 \text{OH}^-(aq) \), the solid \( \text{Ni(OH)}_2 \) is excluded. The equilibrium expression is simplified to \( K = \frac{[\text{Ni}^{2+}][\text{OH}^-]^2}{} \).

This simplification helps focus on the components that actively participate in reaching equilibrium, allowing for more straightforward calculations and predictions regarding the behavior of the reaction.

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Most popular questions from this chapter

The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)

When the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) is run at \(2000^{\circ} \mathrm{C}\), appreciable amounts of reactants and product are present at equilibrium. (a) A sealed 2.00-L container at \(2000{ }^{\circ} \mathrm{C}\) is filled with \(1.00\) mole of \(\mathrm{NO}(g)\) and nothing else. At that moment, which reaction is faster, forward or reverse? Justify your answer. (b) At equilibrium, the concentration of \(\mathrm{NO}(g)\) is \(0.0683 \mathrm{M}\) and the concentration of \(\mathrm{N}_{2}(g)\) is \(0.2159 \mathrm{M}\). What is the value of \(K_{\mathrm{eq}}\) at \(2000^{\circ} \mathrm{C} ?\)

After the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) is run, an equilibrium mixture at \(300^{\circ} \mathrm{C}\) is \(0.25 \mathrm{M}\) in \(\mathrm{N}_{2}(g), 0.15 \mathrm{M}\) in \(\mathrm{H}_{2}(g)\), and \(0.090 \mathrm{M}\) in \(\mathrm{NH}_{3}(g)\). (a) What is the value of \(K_{\text {eq }}\) ? (b) Which way does the equilibrium shift when \(\mathrm{H}_{2}(g)\) is added? (c) What happens to the value of \(K_{\text {eq }}\) when \(\mathrm{H}_{2}(g)\) is added? (d) Suppose we write this reaction as: \(2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftarrows 4 \mathrm{NH}_{3}(g)\) Now what is the value of \(K_{\text {eq }} ?\) (e) The equilibrium shifts to the right when the reaction mixture is cooled. Is this reaction exothermic or endothermic? Justify your choice.

Suppose the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB}\) proceeds via a one-step mechanism involving a collision between one \(\mathrm{A}_{2}\) molecule and one \(\mathrm{B}_{2}\) molecule. Suppose also that this reaction is reversible, and that the forward reaction is inherently much faster than the reverse reaction. (a) Does the equilibrium lie to the left or to the right? Explain your choice in terms of the reactant and product concentrations necessary to establish equal forward and reverse rates. (b) Does an analysis in terms of the relationship \(K_{\mathrm{eq}}=k_{\mathrm{f}} / k_{\mathrm{r}}\) yield the same answer as in (a)? Explain.

Sparingly soluble \(\mathrm{PbCl}_{2}\) dissolves in water to yield an equilibrium \(\mathrm{Pb}^{2+}(a q)\) concentration of \(0.039 \mathrm{M}\). (a) Write the balanced equilibrium equation for \(\mathrm{PbCl}_{2}(\) s) dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for \(\mathrm{PbCl}_{2}\). (c) What is the equilibrium concentration of chloride ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{PbCl}_{2}\) (show your calculation).

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