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Write the \(K_{c}\) expression for the following reaction: $$ \mathrm{Ml}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons 2 \mathrm{\Lambdal}^{3+}(a q)+6 \mathrm{OH}^{-}(a q) $$

Short Answer

Expert verified
Answer: The \(K_{c}\) expression for the given reaction is \(K_{c} = [\mathrm{\Lambdal}^{3+}]^2[\mathrm{OH}^-]^6\).

Step by step solution

01

Identify the reactants and products

In the given reaction: Reactants: \(\mathrm{Ml}_{2}\mathrm{O}_{3}(s)\) and \(3\ \mathrm{H}_{2}\mathrm{O}(\ell)\) Products: \(2\ \mathrm{\Lambdal}^{3+}(aq)\) and \(6\ \mathrm{OH}^{-}(aq)\) We know the stoichiometric coefficients for each species in the reaction, so we can directly plug them into the \(K_{c}\) expression.
02

Write the \(K_{c}\) expression

Following the general form of the \(K_{c}\) expression, we will write the expression for the given reaction: $$ K_{c} = \frac{[\mathrm{\Lambdal}^{3+}]^2[\mathrm{OH}^-]^6}{[\mathrm{Ml}_2\mathrm{O}_3][\mathrm{H}_2\mathrm{O}]^3} $$ However, solids and pure liquids are not included in the equilibrium constant expression, as their concentrations do not change throughout the reaction. Since \(\mathrm{Ml}_{2}\mathrm{O}_{3}(s)\) is a solid and \(\mathrm{H}_{2}\mathrm{O}(\ell)\) is a pure liquid, we will not include these species in the \(K_{c}\) expression.
03

Simplify the \(K_{c}\) expression

Removing the solid and liquid species from the equilibrium constant expression, we have: $$ K_{c} = [\mathrm{\Lambdal}^{3+}]^2[\mathrm{OH}^-]^6 $$ This is the \(K_{c}\) expression for the given reaction.

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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 state in a reversible reaction where the rate of the forward reaction is equal to the rate of the backward reaction. At this state, the concentrations of reactants and products remain constant over time. It is important to note that this doesn't mean the concentrations are equal, just that their ratios are constant.
To identify when a system has reached equilibrium, chemical equilibrium will involve understanding both reactants and products, and often involves calculating equilibrium constants, such as the equilibrium constant ( K_{c} ), which represents the ratio of product concentrations to reactant concentrations at equilibrium. This ratio helps in understanding how far a reaction proceeds before reaching equilibrium.
In the specific exercise given, the equilibrium expression only involves species in the aqueous or gaseous state. Solids and pure liquids are excluded as their concentrations do not change. This means, in our expression for K_{c} , we did not include the solid ( ext{Ml}_{2} ext{O}_{3}( ext{s}) ) or the pure liquid ( ext{H}_{2} ext{O}( ext{l}) ), focusing only on the aqueous ions produced in the reaction.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. This involves using the stoichiometric coefficients, often from balanced chemical equations, to calculate the amount of reactants consumed and products formed.
In the provided reaction:
  • The stoichiometric coefficient for ext{H}_{2} ext{O}( ext{l}) is 3. This indicates that for every mole of ext{Ml}_{2} ext{O}_{3} consumed, 3 moles of ext{H}_{2} ext{O} are required.
  • For every mole of ext{Ml}_{2} ext{O}_{3} consumed, 2 moles of ext{ Lambdal^{3+} }(aq) and 6 moles of ext{OH}^{-}(aq) are produced.

This relationship helps to correctly write the K_{c} expression by ensuring the concentrations of the products are raised to the power of their stoichiometric coefficients. Thus, the coefficients translate into exponents in the equilibrium expression, reflecting the ratio in which the substances interact during the reaction.
Reaction Quotient
The reaction quotient, designated as Q_c , is similar to the equilibrium constant K_{c} but can be calculated at any point in time, not only at equilibrium. It is particularly useful to determine the direction in which a reaction will proceed to reach equilibrium.
To compute Q_c , you take the same form as the K_{c} expression, substituting the current concentrations of reactants and products into the expression.
  • If Q_c < K_c , the reaction will proceed in the forward direction to produce more products.
  • If Q_c > K_c , the reaction will proceed in the reverse direction, leading to the formation of more reactants.
  • If Q_c = K_c , the system is at equilibrium.

This provides a methodical way to predict and understand the shifts in the chemical reactions, clarifying whether adjustments in concentrations are necessary to achieve equilibrium. The understanding of Q_c versus K_{c} is invaluable for predicting reaction behavior in dynamic chemical systems.

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Most popular questions from this chapter

Chemical Weapon Phosgene, \(\mathrm{COCl}_{2}\), gained notoricty as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ \(K_{c}=5.0\) for this reaction at \(600 \mathrm{K} .\) What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which \(P_{\mathrm{CO}}=P_{\mathrm{Cl}_{1}}=0.265\) atm and \(P_{\mathrm{COC}_{1}}=0.000\) atm?

Making Hydrogen Gas Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ The valuc of \(K_{c}\) for the reaction at \(1000^{\circ} \mathrm{C}\) is \(3.0 \times 10^{-2}\) a. Calculate the equilibrium partial pressures of the products and reactants if \(P_{11,0}=0.442\) atm and \(P_{\mathrm{CO}}=5.0\) atm at the start of the reaction. Assume that the carbon is in excess. b. Determine the equilibrium partial pressures of the reactants and products after sufficicnt CO and \(\mathrm{H}_{2}\) are added to the cquilibrium mixture in part (a) to initially increase the partial pressures of both gascs by 0.075 atm.

How is an equilibrium constant different from a reaction quotient?

Under the appropriate conditions, NO forms \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{NO}_{2}\) : $$ 3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) $$ Use the values for \(\Delta G^{\circ}\) for the following reactions to calculate the value of \(K_{\mathrm{p}}\) for the preceding reaction at \(500^{\circ} \mathrm{C}\) $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}_{2}(g) & \Delta G^{\circ} &=-69.7 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{N}_{2}(g) & \Delta G^{\circ} &=-33.8 \mathrm{kJ} \\ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g) & \Delta G^{*} &=173.2 \mathrm{kJ} \end{aligned} $$

Enough \(\mathrm{NO}_{2}\) gas is injected into a cylindrical vessel to produce a partial pressure, \(P_{\mathrm{NO},},\) of 0.900 atm at \(298 \mathrm{K}\) Calculate the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4},\) given $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) \quad K_{\mathrm{p}}=4.0 \mathrm{at} 298 \mathrm{K} $$

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