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The standard Gibbs energy change for the reaction $$\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ is \(29.05 \mathrm{kJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{K}\). Use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of \(\mathrm{NH}_{3}(\mathrm{aq})\) \(\mathrm{NH}_{4}^{+}(\mathrm{aq}),\) and \(\mathrm{OH}^{-}(\mathrm{aq})\) are \(0.10 \mathrm{M}, 1.0 \times 10^{-3} \mathrm{M}\) and \(1.0 \times 10^{-3} \mathrm{M},\) respectively.

Short Answer

Expert verified
After comparing \( Q \) and \( K \), the direction of the spontaneous reaction is determined.

Step by step solution

01

Calculation of Equilibrium Constant from Gibbs Energy Change

Using the formula \( \Delta G = -RT \ln K \), where \(\Delta G\) is the standard Gibbs energy change (-29050 J/mol), \(R\) is the universal gas constant (8.314 J/(mol K)), and \(T\) is the temperature (298 K), one isolates \(K\) to find that \( K = \exp(-\Delta G / RT) \).
02

Calculation of Reaction Quotient

The reaction quotient is defined as \( Q = [NH_{4}^{+}][OH^{-}] / [NH_{3}][H_{2}O] \). As the concentration of water is essentially constant and large, it is usually neglected in the reaction quotient, so it simplifies to \( Q = [NH_{4}^{+}][OH^{-}] / [NH_{3}] \). The provided concentrations are then inserted into \( Q \) to calculate its value.
03

Determine Spontaneity of Reaction

The spontaneity of the reaction is determined by comparing \( Q \) and \( K \). If \( Q < K \), the reaction will proceed in the forward direction to reach equilibrium, because equilibrium lies to the right. If \( Q > K \), the reverse reaction is spontaneous, as equilibrium lies to the left. If \( Q = K \), the reaction is in equilibrium, and no net change occurs.

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Key Concepts

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

Equilibrium Constant
The Equilibrium Constant, represented as \( K \), is a vital concept when discussing chemical reactions and their dynamics. It provides a quantitative measure of the position of equilibrium for a reaction. In thermodynamic terms, the equilibrium constant relates directly to the standard Gibbs free energy change \( (\Delta G^0) \) for the reaction.

To find the equilibrium constant \( K \), we use the formula: \[ \Delta G^0 = -RT \ln K \] where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta G^0 \) is the standard Gibbs free energy change. By rearranging this formula, it allows you to calculate \( K \) as:
  • \( K = \exp(-\Delta G^0 / RT) \)
This expression shows how \( K \) is affected by the energy change of the system under standard conditions. When \( \Delta G^0 \) is negative, \( K \) will be greater than one, indicating products are favored at equilibrium. Conversely, if \( \Delta G^0 \) is positive, \( K \) will be less than one, signaling that reactants are favored.
Reaction Quotient
The Reaction Quotient, represented as \( Q \), is a snapshot measurement that tells us what the concentration of reactants and products are at any given moment before a reaction reaches equilibrium. This is useful in predicting which way a reaction will have to go to reach equilibrium.

To compute \( Q \), one must know the concentrations of the products and reactants at that specific moment in time, using the formula:
  • \( Q = \frac{[Products]}{[Reactants]} \)
Specifically, for the reaction \( \mathrm{NH}_{3} + \mathrm{H}_2\mathrm{O} \leftrightharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \),
  • \( Q = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]} \)
The value of \( Q \) is directly compared to the equilibrium constant \( K \) to determine in which direction the reaction will proceed to reach equilibrium. Because \( Q \) and \( K \) are based on the same stoichiometry, this comparison helps us to understand the immediate state of the reaction relative to equilibrium.
Spontaneity of Reactions
Spontaneity in reactions is about predicting whether a reaction is naturally favorable towards proceeding in the forward direction. The spontaneity can be deduced by comparing the values of \( Q \), the reaction quotient, and \( K \), the equilibrium constant.

The key considerations when comparing these two values are:
  • If \( Q < K \): The reaction will proceed forward to produce more products until equilibrium is achieved.
  • If \( Q > K \): The reaction will tend to reverse, forming more reactants until equilibrium is reached.
  • If \( Q = K \): The system is at equilibrium and there is no net change in the concentrations of reactants and products.
The spontaneous direction is generally towards achieving equilibrium. A reaction with \( Q < K \) suggests that it progresses spontaneously forward, ultimately favoring product formation. Meanwhile, \( Q > K \) implies that the reaction spontaneously favors the reverse direction. Understanding these conditions is essential not only in chemistry but also in industrial applications where optimizing reactions lead to desired outcomes.

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Most popular questions from this chapter

In your own words, define the following symbols: (a) \(\Delta S_{\text {univ }} ;\) (b) \(\Delta G_{f}^{0} ;\) (c) \(K\).

For the following equilibrium reactions, calculate \(\Delta G^{\circ}\) at the indicated temperature. [Hint: How is each equilibrium constant related to a thermodynamic equilibrium constant, \(K ?]\) (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\) at \(445^{\circ} \mathrm{C}\) (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(K_{c}=1.7 \times 10^{-13} \mathrm{at} 25^{\circ} \mathrm{C}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{c}=4.61 \times 10^{-3}\) at \(25^{\circ} \mathrm{C}\) (d) \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Hg}_{2}^{2+}(\mathrm{aq}) \rightleftharpoons\) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Hg}^{2+}(\mathrm{aq})\) \(K_{\mathrm{c}}=9.14 \times 10^{-6} \mathrm{at} 25^{\circ} \mathrm{C}\)

Sodium carbonate, an important chemical used in the production of glass, is made from sodium hydrogen carbonate by the reaction \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Data for the temperature variation of \(K_{\mathrm{p}}\) for this reaction are \(K_{\mathrm{p}}=1.66 \times 10^{-5}\) at \(30^{\circ} \mathrm{C} ; 3.90 \times 10^{-4} \mathrm{at}\) \(50^{\circ} \mathrm{C} ; 6.27 \times 10^{-3}\) at \(70^{\circ} \mathrm{C} ;\) and \(2.31 \times 10^{-1}\) at \(100^{\circ} \mathrm{C}\) (a) Plot a graph similar to Figure \(19-12,\) and determine \(\Delta H^{\circ}\) for the reaction. (b) Calculate the temperature at which the total gas pressure above a mixture of \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is \(2.00 \mathrm{atm}\).

At \(298 \mathrm{K},\) for the reaction \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+\) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{Br}_{2}(1)+2 \mathrm{HNO}_{2}(\mathrm{aq}), \Delta H^{\circ}=-61.6 \mathrm{kJ}\) and the standard molar entropies are \(\mathrm{H}^{+}(\mathrm{aq}), 0 \mathrm{JK}^{-1}\) \(\mathrm{Br}^{-}(\mathrm{aq}), 82.4 \mathrm{JK}^{-1} ; \mathrm{NO}_{2}(\mathrm{g}), 240.1 \mathrm{JK}^{-1} ; \mathrm{Br}_{2}(1), 152.2\) \(\mathrm{J} \mathrm{K}^{-1} ; \mathrm{HNO}_{2}(\mathrm{aq}), 135.6 \mathrm{JK}^{-1} .\) Determine (a) \(\Delta G^{\circ}\) at 298 K and (b) whether the reaction proceeds spontaneously in the forward or the reverse direction when reactants and products are in their standard states.

Explain why (a) some exothermic reactions do not occur spontaneously, and (b) some reactions in which the entropy of the system increases do not occur spontaneously.

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