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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 is: \(HNO_2 \rightleftharpoons H^+ + NO_2^-\) \(K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}\) (b) The 螖G掳 can be calculated using the equation: \(\Delta G^慰 = -RT \ln{K_a}\) (c) At equilibrium, 螖G is 0. (d) To determine 螖G with given concentrations, use the equation: \(\Delta G = \Delta G^慰 + RT \ln{Q}\) where Q is: \(Q = \frac{(5.0 \times 10^{-2})(6.0 \times 10^{-4})}{0.20}\) Calculate Q, and then use this value, 螖G掳, R, and T to determine the 螖G value.

Step by step solution

01

Writing the Chemical Equation and Ka Expression

Write down the equation for the ionization of nitrous acid in water and the corresponding equilibrium constant (Ka): \(HNO_2 \rightleftharpoons H^+ + NO_2^-\) The equilibrium constant expression, Ka, for this reaction is: \(K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}\)
02

Calculating 螖G掳 for the Dissociation of Nitrous Acid

Use the provided \(K_a\) value and the relationship between \(K_a\) and 螖G掳, given by the equation: \(\Delta G^慰 = -RT \ln{K_a}\) where R is the gas constant (8.314 J/mol路K) and T is the temperature in Kelvin (298 K since it is 25掳C). Once we have the 螖G掳 value, it can be used to find the 螖G at the given concentrations in the later parts of the problem. Note: Be sure to convert the temperature from Celsius to Kelvin (T = 25掳C + 273.15 = 298.15 K).
03

Finding 螖G at Equilibrium

At equilibrium, the Gibbs free energy change (螖G) is zero: \(\Delta G = 0\) This is because the concentrations of products and reactants are balanced at equilibrium.
04

Determining 螖G with Given Concentrations

Now, we need to find 螖G when the solution has specific concentrations of H鈦, NO鈧傗伝, and HNO鈧. Use the provided concentrations and the reaction quotient Q, which is related to 螖G and 螖G掳 by the following equation: \(\Delta G = \Delta G^慰 + RT \ln{Q}\) The reaction quotient Q for this reaction is: \(Q = \frac{[H^+][NO_2^-]}{[HNO_2]}\) Plug in the given concentrations of the species: \([H^+] = 5.0 \times 10^{-2} M\), \([NO_2^-] = 6.0 \times 10^{-4} M\), and \([HNO_2] = 0.20 M\): \(Q = \frac{(5.0 \times 10^{-2})(6.0 \times 10^{-4})}{0.20}\) Calculate Q, and then use this value, 螖G掳 from step 2, R, and T to determine the 螖G value for these given concentrations.

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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, symbolized as \( K \), is a fundamental concept in chemistry that helps us understand the balance of a chemical reaction at a particular temperature. For an acid dissociation reaction like that of nitrous acid \( \text{HNO}_2 \), the equilibrium constant is specifically known as the acid dissociation constant \( K_a \).
This constant provides insight into the extent to which the acid dissociates into its ions in solution.Equilibrium constants are crucial because they demonstrate the ratio of product concentrations to reactant concentrations when a reaction reaches equilibrium:
  • For the dissociation of nitrous acid, we write the chemical equation as: \( \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \).
  • The expression for \( K_a \) becomes \( K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \).
Understanding \( K_a \) allows chemists to predict how a weak acid behaves in solution, revealing whether it will exist predominantly in the undissociated form or as ions.
Gibbs Free Energy
Gibbs Free Energy, denoted as \( \Delta G \), is a measure that tells us about the spontaneity of a chemical reaction at constant temperature and pressure. It combines enthalpy and entropy into one value, giving us a macroscopic understanding of the energy changes involved.
Using the relationship between the equilibrium constant \( K_a \) and Gibbs free energy change \( \Delta G^o \), we can determine whether a reaction is spontaneous.

Calculating Gibbs Free Energy

To find the standard Gibbs free energy change \( \Delta G^o \) for the dissociation of nitrous acid, we utilize the equation:
\[ \Delta G^o = -RT \ln{K_a} \]
Where \( R \) is the universal gas constant \( 8.314 \, J/mol\cdot K \) and \( T \) is the temperature in Kelvin. A negative \( \Delta G^o \) signifies that the reaction tends to proceed in the forward direction. At equilibrium, \( \Delta G \) is zero, indicating no net reaction occurring as the system is balanced.
In the context of the problem, this understanding is crucial for predicting reaction behavior under different conditions.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction's rate of forward reaction is equal to the rate of the backward reaction, resulting in no net change in the concentration of reactants and products. At this point, the reaction quotient \( Q \) equals the equilibrium constant \( K \), a state that the reaction strives towards.
Understanding equilibrium helps explain why some reactions appear static despite ongoing molecular activity.In aqueous solutions, as seen with nitrous acid, reaching equilibrium involves:
  • Balancing the formation and dissociation of molecules and ions, such as \( \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \).
  • The concentrations of each species adjust until \( K_a \) is reached.
Equilibrium is a dynamic process. It doesn't imply that reactions have stopped; rather, the rates are balanced.
This dynamic nature is essential when considering conditions like varying concentrations, temperature, and pressure, impacting the system's position at equilibrium.
Reaction Quotient
The reaction quotient, \( Q \), is a valuable tool for predicting the direction in which a chemical reaction will proceed to reach equilibrium. It is similar in form to the equilibrium constant \( K \), calculated using the initial concentrations, rather than at equilibrium.
It helps determine whether the reaction needs to move towards products or reactants to establish equilibrium.Formula for the Reaction Quotient
For the dissociation of nitrous acid, the reaction quotient is defined as:
\[ Q = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \]
Where the concentrations are the initial or existing concentrations at a given point in time rather than at equilibrium.

Interpreting \( Q \)

  • If \( Q < K \), the reaction will shift to the right, forming more products.
  • If \( Q > K \), the reaction will shift to the left, forming more reactants.
  • If \( Q = K \), the system is at equilibrium, and no shift occurs.
This concept helps chemists predict and control reactions in laboratory and industrial settings, ensuring optimal conditions for desired products.

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Most popular questions from this chapter

Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(C\) to predict how \(\Delta G\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?\) (c) Calculate \(\Delta G\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

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