Problem 107
Question
Which of the following reactions has the largest cquilibrium constant at \(25^{\circ} \mathrm{C} ?\) a. \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{ClF}(g) \quad \Delta G^{*}=115.4 \mathrm{kJ}\) b. \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{ClBr}(g) \quad \Delta G^{*}=-2.0 \mathrm{kJ}\) c. \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 21 \mathrm{Cl}(g) \quad \Delta G^{*}=-27.9 \mathrm{kJ}\)
Step-by-Step Solution
Verified Answer
Answer: Reaction c has the largest equilibrium constant at \(25^{\circ}\mathrm{C}\), with a value of \(K_c = 3.073\).
1Step 1: 01. Relationship between Gibbs free energy and the equilibrium constant
The relationship between the Gibbs free energy change (\(\Delta G^{*}\)) and the equilibrium constant (K) is given by the formula:
\(\Delta G^{*} = - RT \ln K\)
Here, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.
2Step 2: 02. Convert temperature to Kelvin
To use the above formula, we need temperature in Kelvin. We can convert temperature from Celsius to Kelvin using the formula:
\(T_{K} = T_{C} + 273.15\)
So, for \(25^{\circ}\mathrm{C}\), we have:
\(T_{K}= 25 + 273.15 = 298.15\,\text{K}\)
3Step 3: 03. Calculate the equilibrium constant K for each reaction
Now, we can calculate the equilibrium constant K for each reaction using the given Gibbs free energy changes and the temperature. For each reaction, use the formula:
\(K = e^{(-\Delta G^{*}) / (RT)}\)
Reaction a:
\(K_a = e^{(-115.4 \times 10^3\,\text{J/mol}) / (8.314\,\text{J/mol·K} \times 298.15\,\text{K})} = e^{-4.61} = 0.00996\)
Reaction b:
\(K_b = e^{(2.0 \times 10^3\,\text{J/mol}) / (8.314\,\text{J/mol·K} \times 298.15\,\text{K})} = e^{0.0805} = 1.084\)
Reaction c:
\(K_c = e^{(27.9 \times 10^3\,\text{J/mol}) / (8.314\,\text{J/mol·K} \times 298.15\,\text{K})} = e^{1.122} = 3.073\)
4Step 4: 04. Compare the equilibrium constants and find the largest one
Comparing the equilibrium constants:
\(K_a = 0.00996\), \(K_b = 1.084\), \(K_c = 3.073\)
The largest equilibrium constant is \(K_c\), which belongs to reaction c. Therefore, reaction c has the largest equilibrium constant at \(25^{\circ}\mathrm{C}\).
Key Concepts
Gibbs Free Energy and EquilibriumUnderstanding Chemical EquilibriumThe Role of the Universal Gas Constant
Gibbs Free Energy and Equilibrium
The concept of Gibbs free energy, denoted \(\Delta G^{*}\), is crucial in understanding chemical reactions. It measures the spontaneity of a reaction and is connected to the equilibrium state of the reaction. When \(\Delta G^{*} < 0\), the process is spontaneous and tends toward the formation of products. Conversely, if \(\Delta G^{*} > 0\), the reverse reaction is favored. This relationship is depicted mathematically by the equation \(\Delta G^{*} = - RT \ln K\), where \(K\) is the equilibrium constant. Here, a negative \(\Delta G^{*}\) indicates a larger equilibrium constant, meaning the reaction heavily favors products.
- A negative \(\Delta G^{*}\) is linked to a large \(K\), suggesting a greater concentration of products at equilibrium.
- Conversely, a positive \(\Delta G^{*}\) implies a low \(K\), indicating less product formation.
Understanding Chemical Equilibrium
Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical system are equal. At this point, the concentrations of reactants and products remain constant over time, although not necessarily equal. The equilibrium constant \(K\) quantitatively expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
- A large \(K\) value signifies that the products are favored at equilibrium.
- A small \(K\) value indicates that reactants are favored.
The Role of the Universal Gas Constant
The universal gas constant, symbolized as \(R\), is a fundamental constant that appears in various equations in chemistry and physics, linking energy to temperature and moles. Its value is \(8.314 \, \text{J/mol·K}\). In the context of the relationship between Gibbs free energy and the equilibrium constant, \(R\) is used to incorporate temperature into the calculation. \(\Delta G^{*} = - RT \ln K\) shows how \(R\) and temperature (in Kelvin) interact to determine the sign and magnitude of \(\Delta G^{*}\);
- The constant \(R\) provides the necessary energy unit conversion, balancing the energy change with temperature.
- In equilibrium constant calculations, \(R\) ensures that the \(\Delta G^{*}\) units align with temperature, permitting accurate evaluation of \(K\).
Other exercises in this chapter
Problem 105
Starting with pure reactants, in which dircction will an equilibrium shift if \(\Delta G^{*}
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The value of \(\Delta G^{*}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ is \(68.9 \m
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At a temperature of \(1000 \mathrm{K}, \mathrm{SO}_{2}(g)\) combines with oxygen to make \(\mathrm{SO}_{3}(\mathrm{g})\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(
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