Problem 99
Question
The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C} .\) Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\)
Step-by-Step Solution
Verified Answer
The value of ΔG for this reaction at 25°C with the given concentrations of OH⁻ and H⁺ ions is 47.7 kJ/mol.
1Step 1: Convert temperature to Kelvin
To convert the temperature from Celsius to Kelvin, we will use the formula:
T(K) = T(°C) + 273.15
For T = 25°C,
T(K) = 25 + 273.15 = 298.15 K
2Step 2: Calculate the Reaction Quotient, Q
The Reaction Quotient for this reaction is given by:
Q = \(\frac{[H^{+}][OH^{-}]}{[H_{2}O]}\)
Since we are given the concentrations of H⁺ and OH⁻ ions, and given that the concentration of liquid water does not change, we can simply use the concentrations of H⁺ and OH⁻ ions in the expression for Q:
Q = \(\frac{(0.71)(0.15)}{1}\) = 0.1065
3Step 3: Calculate ΔG
Now we have all the values needed to calculate ΔG using the equation:
ΔG = ΔG° + RT ln Q
Where ΔG° = 79.9 kJ/mol, R = 8.314 J/(mol K), T = 298.15 K, and Q = 0.1065.
ΔG = \(79.9 \times 10^3 + (8.314)(298.15) \ln (0.1065)\)
ΔG ≈ 79.9 kJ/mol - 32.2 kJ/mol = 47.7 kJ/mol
The value of ΔG for this reaction at 25°C with the given concentrations of OH⁻ and H⁺ ions is 47.7 kJ/mol.
Key Concepts
ThermodynamicsReaction QuotientChemical EquilibriumEnthalpy
Thermodynamics
Thermodynamics is a field of physics that explains how energy moves and changes within a system. It's important to understand this topic because it helps us determine how energy is transformed and transferred during chemical reactions.
In the context of chemical reactions, such as the one provided in the exercise, thermodynamics allows us to investigate the energetic feasibility of a reaction, primarily through concepts like Gibbs Free Energy (\(\Delta G\).).
In the context of chemical reactions, such as the one provided in the exercise, thermodynamics allows us to investigate the energetic feasibility of a reaction, primarily through concepts like Gibbs Free Energy (\(\Delta G\).).
- If \(\Delta G\) is negative, the reaction is spontaneous, meaning it can occur without any additional input of energy.
- Conversely, if \(\Delta G\) is positive, the reaction is non-spontaneous under the given conditions.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is a measure of the concentrations of reactants and products in a reaction at any point in time. It's similar to the equilibrium constant \(K\), but \(Q\) is used when the system is not at equilibrium.
In the original exercise, \(Q\) is calculated using the initial concentrations of \([H^+]\) and \([OH^-]\). The formula is:\[Q = \frac{[H^+][OH^-]}{[H_2O]}\]Here, water is deemed constant due to its liquid state, simplifying the calculation.
In the original exercise, \(Q\) is calculated using the initial concentrations of \([H^+]\) and \([OH^-]\). The formula is:\[Q = \frac{[H^+][OH^-]}{[H_2O]}\]Here, water is deemed constant due to its liquid state, simplifying the calculation.
- If \(Q < K\), the reaction will move forward to reach equilibrium.
- If \(Q > K\), the reaction shifts in reverse to balance again.
- When \(Q = K\), the system is at equilibrium.
Chemical Equilibrium
Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time. In this state, the forward and reverse reactions occur at the same rate.
At equilibrium, the reaction quotient \(Q\) equals the equilibrium constant \(K\). This means there's a state of balance where no net change is observed in the concentrations of reactants and products. Even though these reactions continue at the molecular level, their rates are equal.
At equilibrium, the reaction quotient \(Q\) equals the equilibrium constant \(K\). This means there's a state of balance where no net change is observed in the concentrations of reactants and products. Even though these reactions continue at the molecular level, their rates are equal.
- For the reaction given in the exercise, reaching equilibrium involves a balance between the formation of \(H^+\) and \(OH^-\) ions from \(H_2O\).
- Understanding how equilibrium works helps in determining the possible concentration of each component in the reaction mixture once equilibrium is achieved.
Enthalpy
Enthalpy, represented by \(H\), is a measure of the total energy of a thermodynamic system. It includes internal energy as well as energy stored as pressure and volume.
Although enthalpy was not a focal point in the given exercise, it heavily influences reactions along with entropy. Both affect Gibbs Free Energy since \(\Delta G = \Delta H - T\Delta S\), where \(\Delta S\) is entropy change.
Although enthalpy was not a focal point in the given exercise, it heavily influences reactions along with entropy. Both affect Gibbs Free Energy since \(\Delta G = \Delta H - T\Delta S\), where \(\Delta S\) is entropy change.
- When \(\Delta H\) is negative, a reaction is exothermic, releasing heat to its surroundings.
- If \(\Delta H\) is positive, it's endothermic, meaning it absorbs heat.
Other exercises in this chapter
Problem 96
For rubidium \(\Delta H_{\text {vap }}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C},\) its boiling point. Calculate \(\Delta S^{\circ},
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Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\)
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Consider the dissociation of a weak acid HA \(\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a
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Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=-92.50 \ma
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