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

Short Answer

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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

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ At \(430^{\circ} \mathrm{C}\), an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2} .\) Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

Briefly describe the importance of equilibrium in the study of chemical reactions.

Consider the following reaction at \(1600^{\circ} \mathrm{C}\) : $$ \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) $$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

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