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(a) How does a reaction quotient differ from an equilibrium constant? (b) If \(Q_{c}

Short Answer

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(a) The reaction quotient (Qc) represents the ratio of product concentrations to reactant concentrations at any particular moment during a reaction, while the equilibrium constant (Kc) represents this ratio when the reaction has reached equilibrium, with no further changes in concentration. (b) If \(Q_c < K_c\), the reaction will proceed in the forward direction to reach equilibrium, converting more reactants into products. (c) For \(Q_c\) to be equal to \(K_c\), the reaction must be at equilibrium, with no further changes in concentrations of reactants and products.

Step by step solution

01

(a) Differences Between Reaction Quotient and Equilibrium Constant

The reaction quotient (Qc) and equilibrium constant (Kc) both represent the ratio of product concentrations to reactant concentrations for a chemical reaction. However, Qc represents this ratio at any particular moment during the reaction's progress, while Kc represents this ratio when the reaction has reached equilibrium, and the concentrations of products and reactants no longer change with time.
02

(b) Determining Reaction Direction based on Qc and Kc values

To determine the direction of a reaction, we need to compare the values of Qc and Kc: - If \(Q_c < K_c\), it means that the concentration of the products is lower than what would be expected at equilibrium. To reach equilibrium, the reaction needs to proceed in the forward direction, converting more reactants into products. - If \(Q_c > K_c\), it means that the concentration of products is higher than what would be expected at equilibrium. To reach equilibrium, the reaction needs to proceed in the reverse direction, converting more products back into reactants. In this case, as \(Q_c < K_c\), the reaction will proceed in the forward direction to reach equilibrium.
03

(c) Condition for Qc to be equal to Kc

For the reaction quotient (Qc) to be equal to the equilibrium constant (Kc), the reaction must be at equilibrium. It means that the concentrations of the reactants and products have stopped changing with time and are in a balanced state.

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Most popular questions from this chapter

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Explain why we normally exclude pure solids and liquids from equilibrium-constant expressions.

Consider the following equilibrium, for which \(\Delta H<0\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ How will each of the following changes affect an equilibrium mixture of the three gases? (a) \(\mathrm{O}_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is. added to the mixture; \((e)\) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system.

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\)

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\) the rate constants for the forward and reverse reactionsare \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

A mixture of \(\mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a 1.0-L vessel at \(90^{\circ} \mathrm{C}\) until the following equilibrium is achieved: $$ \mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g} \mathrm{H}_{2}\). (a) Write the equilibrium- constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(S\) when doing the calculation in part \((b)\) ?

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