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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 \(K_{a}\). (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 \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\) \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\) and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The chemical equation for the equilibrium corresponding to K鈧 is: HNO鈧(aq) 鈬 H鈦(aq) + NO鈧傗伝(aq). (b) By using the value of K鈧, we can calculate 螖G掳 for the dissociation of nitrous acid using the formula 螖G掳 = -RT ln(K鈧), and it gives approximately 29.86 kJ/mol. (c) At equilibrium, the value of 螖G is 0. (d) With the given concentrations, we first calculate the reaction quotient Q and then find 螖G using the formula 螖G = 螖G掳 + RT ln(Q). The value of 螖G will be approximately -7.80 kJ/mol.

Step by step solution

01

a) Chemical equation for equilibrium corresponding to K鈧

Firstly, we should understand what K鈧 is. K鈧 is the acid dissociation constant for an acid, which shows the strength of the acid. We're dealing with nitrous acid (HNO鈧) as our given acid in this exercise. The chemical equation for the dissociation of HNO鈧 in water can be written as: HNO鈧(aq) 鈬 H鈦(aq) + NO鈧傗伝(aq) From the equation, we can express K鈧: K鈧 = \(\frac{[H^+][NO_2^-]}{[HNO_2]}\)
02

b) Calculate the standard Gibbs free energy change (螖G掳)

Next, we will relate K鈧 with the standard Gibbs free energy change (螖G掳) via the following equation: 螖G掳 = -RT ln(K鈧) where R is the gas constant (8.314 J/mol路K), T is the temperature in Kelvin, and K鈧 is the acid dissociation constant for nitrous acid. First, we must convert the temperature given (25掳C) to Kelvin: T = 25 + 273.15 = 298.15 K Now, use the given K鈧 value (4.50 x 10鈦烩伌) to calculate 螖G掳: 螖G掳 = - (8.314 J/mol路K) (298.15 K) ln(4.50 x 10鈦烩伌) This will give you the value for the standard Gibbs free energy change.
03

c) Value of 螖G at equilibrium

At equilibrium, the reaction quotient Q is equal to the equilibrium constant K鈧. Therefore, the value of 螖G at equilibrium is 0, as it indicates that the reaction is in equilibrium and there is no net change in the system.
04

d) Calculate the value of 螖G with given concentrations

Now, we are given concentrations for H鈦, NO鈧傗伝, and HNO鈧 in the solution. We first need to calculate the reaction quotient Q: Q = \(\frac{[H^+][NO_2^-]}{[HNO_2]}\) Q = \(\frac{(5.0 \times 10^{-2}M)(6.0 \times 10^{-4}M)}{0.20 M}\) Next, we calculate 螖G based on the reaction quotient Q using the following formula: 螖G = 螖G掳 + RT ln(Q) By substituting the values of 螖G掳, R, and T calculated earlier, and the newly calculated value of Q, we can now find the value of 螖G.

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Key Concepts

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

Acid Dissociation Constant
The acid dissociation constant, commonly represented as \(K_{a}\), is a measure of the strength of an acid in a solution. It specifically pertains to the equilibrium process where an acid donates a proton (\(H^+\)) to water. In the context of nitrous acid (HNO鈧), the dissociation into hydrogen ions and nitrite ions (NO鈧傗伝) is key. When HNO鈧 dissociates in water, the chemical equation is:
  • HNO鈧(aq) 鈬 H鈦(aq) + NO鈧傗伝(aq)
This equilibrium expression is quantified by \(K_{a}\), which is calculated using the concentrations of these ions at equilibrium:\[K_{a} = \frac{[H^+][NO_2^-]}{[HNO_2]} \]A larger \(K_{a}\) value indicates a stronger acid as it more readily donates protons to the solution. In practical terms, knowing the \(K_{a}\) helps predict how an acid will behave in various chemical environments.
Gibbs Free Energy
Gibbs Free Energy, denoted as \(\Delta G\), is a fundamental concept that describes the amount of energy available to do work during a chemical reaction. It incorporates both enthalpy and entropy to indicate the spontaneity of a process. In equilibrium processes, particularly those involving acid dissociation, it's related to \(K_{a}\) through the equation:\[\Delta G^\circ = -RT \ln(K_{a})\]where:
  • \(R\) is the gas constant (8.314 J/mol路K)
  • \(T\) is the temperature in Kelvin
  • \(K_{a}\) is the acid dissociation constant
The standard Gibbs Free Energy change (\(\Delta G^\circ\)) reflects how spontaneous the reaction is under standard conditions (1 atm pressure, 298 K). When \(\Delta G^\circ\) is negative, the reaction proceeds spontaneously. At equilibrium, \(\Delta G\) is zero since no net change occurs. By knowing \(K_{a}\), you can directly calculate \(\Delta G^\circ\), providing insight into the thermodynamics of the dissociation reaction.
Reaction Quotient
The reaction quotient, \(Q\), provides a snapshot of the current state of a reaction, much like \(K\) (equilibrium constant) but at any given point in the reaction's progress. For the dissociation of nitrous acid, \(Q\) is expressed similarly to \(K_{a}\):\[Q = \frac{[H^+][NO_2^-]}{[HNO_2]} \]Whereas \(K_{a}\) describes the state at equilibrium, \(Q\) can be calculated at any point to determine if the reaction is at, before, or beyond equilibrium. By comparing \(Q\) to \(K_{a}\):
  • If \(Q < K_{a}\), the reaction will shift right, indicating more products will form.
  • If \(Q > K_{a}\), it will shift left, showing excess products reverting to reactants.
  • If \(Q = K_{a}\), the system is at equilibrium.
Understanding \(Q\) helps predict directionality and assess what needs to change to reach equilibrium.
Nitrous Acid Dissociation
Nitrous acid (HNO鈧) dissociates in water to form hydrogen ions (H鈦) and nitrite ions (NO鈧傗伝). This dissociation process can be depicted by the equation:
  • HNO鈧(aq) 鈬 H鈦(aq) + NO鈧傗伝(aq)
This equilibrium reaction is essential in understanding the behavior of weak acids, as nitrous acid is not completely dissociated in solution. Its \(K_{a}\) value is a testament to its partial ionization, reflecting both its acid strength and its equilibrium in aqueous solution.
To calculate how spontaneity changes in response to specific concentrations at a given time, you employ the Gibbs Free Energy equation:\[\Delta G = \Delta G^\circ + RT \ln(Q)\]This equation bridges the gap between standard conditions and the actual reaction conditions described by \(Q\). By considering both \(\Delta G^\circ\) and \(Q\), you gain a complete picture of the thermodynamic viability and progress of nitrous acid dissociation in real-time conditions.

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Most popular questions from this chapter

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous or nonspontaneous?

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K,\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{Hl}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

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