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The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a \cdot}\) (b) By using the value of \(K_{a},\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} M\) \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M,\) and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The chemical equation for the acid dissociation equilibrium of nitrous acid (HNO2) is: \(HNO_{2} (aq) \rightleftharpoons H^{+} (aq) + NO_{2}^{-} (aq)\) (b) Calculate ΔG° using the equation: \(\Delta G^{\circ} = -RT \ln K_{a}\) and the given Ka value. (c) At equilibrium, ΔG = 0. (d) To find ΔG under the provided concentrations, first calculate Q, then use \(\Delta G = \Delta G^{\circ} + RT \ln Q\) to find ΔG.

Step by step solution

01

(a) Write the chemical equation for the equilibrium associated with Ka

The chemical equation for the acid dissociation equilibrium of nitrous acid (HNO2) in an aqueous solution is written as: \(HNO_{2} (aq) \rightleftharpoons H^{+} (aq) + NO_{2}^{-} (aq)\)
02

(b) Calculate ΔG° for the dissociation of nitrous acid using Ka

To find the standard Gibbs free energy change (ΔG°) for the dissociation of HNO2, we can use the following equation: \(\Delta G^{\circ} = -RT \ln K_{a}\) Where R is the gas constant (8.314 J/mol.K), T is the temperature in Kelvin (298K, since we have 25°C), and Ka is the given acid dissociation constant for HNO2. Using the provided Ka value from Appendix D, we can calculate ΔG°.
03

(c) Find the value of ΔG at equilibrium

At equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (Ka). Since ΔG is related to Q through the following equation: \(\Delta G = \Delta G^{\circ} + RT \ln Q\) When Q = Ka, ΔG is equal to zero. Therefore, at equilibrium, ΔG = 0.
04

(d) Calculate the value of ΔG when given concentrations of H+, NO2-, and HNO2

To find the value of ΔG under given conditions, we must first find the reaction quotient (Q) by substituting the provided concentrations: \(Q = \frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]}\) After calculating Q, we can use the same equation mentioned earlier (\(\Delta G = \Delta G^{\circ} + RT \ln Q\)) to determine the value of ΔG for the non-equilibrium situation. Simply input the values of ΔG°, R, T, and Q into the equation, and calculate ΔG.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial when studying the behavior of chemical reactions, especially reversible ones. Chemical equilibrium is the state in a reversible reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. It is important to note that this does not mean the reactants and products are equal in concentration, but rather that their concentrations have stabilized at a particular ratio.

For the dissociation of nitrous acid (HNO_2), the equilibrium is represented by the equation: \( HNO_2 (aq) \rightleftharpoons H^+ (aq) + NO_2^{-} (aq) \). At equilibrium, the concentrations of these species do not change, despite the continuous exchange of ions between the reactant and the products. The point at which this equilibrium is achieved can be quantified by the Ka value, known as the acid dissociation constant, which is specific to each acid in solution.

In this context, knowing the Ka allows us to understand the extent of the acid's dissociation under equilibrium conditions and to predict the concentrations of the ions produced.
Gibbs Free Energy Change
The Gibbs free energy change (ΔG) of a reaction determines the spontaneity of a process at constant temperature and pressure. A negative ΔG signifies a spontaneous process, while a positive ΔG means the reaction is nonspontaneous. At equilibrium, ΔG is equal to zero, indicating the system is in its most stable state and no net reaction occurs.

For the dissociation of nitrous acid, we calculate the standard Gibbs free energy change (ΔG°) using the equation \(ΔG^{ocircle} = -RT ln Ka\), where R is the gas constant and T is the temperature in Kelvin. This value provides insight into the energetics of the acid dissociation at standard conditions (1 molar concentration and 1 atmosphere pressure). By understanding the link between ΔG° and Ka, we gain an understanding of how the equilibrium position (extent of dissociation) affects the energy change of the reaction.
Reaction Quotient
The reaction quotient (Q) is a measure that tells us how far a reaction has progressed towards equilibrium at any given point in time and under any conditions. It has the same form as the equilibrium constant (K), but unlike K, it is not restricted to equilibrium conditions and can utilize any concentrations or partial pressures of reactants and products.

For the nitrous acid dissociation, Q is calculated by \(Q = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}\) using the concentrations at a given moment. If Q is equal to Ka, the system is at equilibrium, and ΔG will be zero as mentioned before. If Q is greater or lesser than Ka, it tells us in which direction the reaction will proceed to reach equilibrium, and this influences the calculation of ΔG for the reaction.

In practice, calculating Q and comparing it to Ka can predict whether the reaction will proceed forward or reverse to achieve equilibrium, offering a dynamic view of chemical processes.
Equilibrium Constant
The equilibrium constant (K), such as the acid dissociation constant (Ka) for weak acids, is a dimensionless value representing the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficient in the balanced equation.

For nitrous acid (HNO_2), the Ka value is obtained through experimental measurements, and it reflects the strength of the acid in terms of its tendency to donate protons to water. Small Ka values indicate weaker acids, which dissociate less in solution, while larger Ka values correspond to stronger acids. This constant is an essential part of understanding the chemical behavior of acids and bases in solution and plays a key role in calculating the ΔG° of acid dissociation reactions.

In summary, Ka provides a quantitative measure of the position of equilibrium, aiding in the prediction of the concentrations of ions at equilibrium state for a particular acid in solution. It is crucial for chemists to make accurate predictions about reactions involving weak acids, such as nitrous acid in aqueous solutions.

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Most popular questions from this chapter

Would each of the following changes increase, decrease, or have no effect on the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{J} / \mathrm{K}\) . (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?(c) Calculate \(\Delta G^{\circ}\) for the reaction at 298 \(\mathrm{K}\) . (d) Is the reaction spontaneous at 298 \(\mathrm{K}\) under standard conditions?

(a) Using data in Appendix \(C,\) estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. (b) Use a reference source, such as Web Elements (www. webelements.com), to find the experimental melting and boiling points of \(I_{2}\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

Predict the sign of the entropy change of the system for each of the following reactions: $$\begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (b) } \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)} \\ {\text { (c) } 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)} \\ {\text { (d) } \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)}\end{array}$$

A certain constant-pressure reaction is barely nonspontaneous at \(45^{\circ} \mathrm{C}\) . The entropy change for the reaction is 72 \(\mathrm{J} / \mathrm{K} .\) Estimate \(\Delta H .\)

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