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Chapter 15: Problem 27

Define reaction quotient. How does it differ from equilibrium constant?

Short Answer

Expert verified
The reaction quotient (\(Q_c\)) is the ratio of the concentrations of the products of a reaction to the concentrations of the reactants at any given moments during a reaction, each raised to their stoichiometrically appropriate powers. The equilibrium constant (\(K_c\)) is the value of \(Q_c\) when the reaction has reached equilibrium. The key difference between them is that \(Q_c\) can be calculated at any point during a reaction, while \(K_c\) is only calculated when the reaction has reached equilibrium.

Step by step solution

01

Definition of Reaction Quotient

The Reaction Quotient, \(Q_c\), for a reaction is a ratio that is equal to the concentrations of the products of a reaction, raised to the respective stoichiometric coefficients, divided by the concentrations of the reactants, also raised to their stoichiometric coefficients, at any given moments during a reaction. The subscript \(c\) indicates that the concentrations are used to define \(Q\). For a general reaction, \(aA + bB \rightarrow cC + dD\), the reaction quotient is given by: \(Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)
02

Definition of Equilibrium Constant

The equilibrium constant is a special case of the reaction quotient. It is defined as the value of the reaction quotient when the reaction has reached equilibrium (when the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products). The equilibrium constant, \(K_c\), is given by: \(K_c = \frac{[C]^c_{eq} [D]^d_{eq}}{[A]^a_{eq} [B]^b_{eq}}\)
03

Distinguishing between Reaction Quotient and Equilibrium Constant

While the reaction quotient and the equilibrium constant may seem similar in form, the key difference between them lies in when they are used. The reaction quotient can be calculated at any point during a reaction, while the equilibrium constant is calculated only when the reaction has reached equilibrium. Also, while the equilibrium constant \(K_c\) remains constant for a given reaction at a specific temperature, regardless of the initial concentrations of reactants and products, \(Q_c\) varies with the concentrations of reactants and products at any given point in time during the reaction.

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Most popular questions from this chapter

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$ \begin{aligned} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) &+3 \mathrm{H}_{2}(g) \\ \Delta H^{\circ} &=206 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol} $$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

In the uncatalyzed reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ at \(100^{\circ} \mathrm{C}\) the pressures of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377 \mathrm{~atm}\) and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm} .\) What would happen to these pressures if a catalyst were present?

A mixture of 0.47 mole of \(H_{2}\) and 3.59 moles of \(H C l\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. The \(K_{P}\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) is 193 at \(2800^{\circ} \mathrm{C}.\)

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)
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