Problem 72
Question
The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.
Step-by-Step Solution
Verified Answer
a. Calculate \(\Delta S^{\circ}\) at 298 K:
By using the relation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), we can calculate \(\Delta S^{\circ}\) as follows:
\(\Delta G_{rxn}^{\circ} = \Delta H_{rxn}^{\circ} - 298 \Delta S_{rxn}^{\circ}\)
Rearranging to solve for \(\Delta S_{rxn}^{\circ}\) gives:
\(\Delta S_{rxn}^{\circ} = \frac{\Delta H_{rxn}^{\circ} - \Delta G_{rxn}^{\circ}}{298}\)
Substituting the given values:
\(\Delta S_{rxn}^{\circ} = \frac{(3\times241.3 - 132.8) - (3\times191.2 - 78.2)}{298}\)
\(\Delta S_{rxn}^{\circ} = 297.6 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\)
b. Calculate \(K\) at 298 K:
Using the relation between Gibbs Free Energy and the equilibrium constant, the equation is:
\(K = e^{-(\Delta G_{rxn}^{\circ})/RT}\)
Substituting values:
\(K = e^{-(3\times191.2 - 78.2)/(8.314\times10^{-3}\times298)}\)
\(K \approx 9.54\times 10^{-7}\)
c. Estimate \(K\) at 3000 K:
Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, we can use the van't Hoff equation:
\(\ln\frac{K_2}{K_1} = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
We want to find \(K_2\), given \(K_1\) at 298 K and \(T_2 = 3000\, K\). Rearrange the equation:
\(K_2 = K_1 \times e^{\frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\)
Substituting values:
\(K_2 = 9.54\times 10^{-7} \times e^{\frac{-(3\times241.3 - 132.8)}{8.314} \left(\frac{1}{3000} - \frac{1}{298}\right)}\)
\(K_2 \approx 0.490\)
1Step 1: Calculate ∆G° for the reaction
By subtracting the standard free energies of formation of the reactants from the products, we can find the ∆G° of the reaction at 298K. Use the equation: $$\Delta G_{rxn}^{\circ} = \sum \Delta G_{f,products}^{\circ} - \sum \Delta G_{f,reactants}^{\circ}$$
Key Concepts
Free Energy of FormationStandard Enthalpy of FormationEquilibrium ConstantEntropy Change
Free Energy of Formation
Free energy of formation, denoted by \(\Delta G_f^{\circ}\), represents the change in free energy when one mole of a compound is made from its elements in their standard states. This value helps us understand how spontaneous a reaction is, under standard conditions where pressure is 1 atm and temperature is 25°C or 298 K.
When the free energy of formation is negative, the formation of the compound is spontaneous. If positive, energy is required to form the compound, making it non-spontaneous under standard conditions. In our exercise, - the \(\Delta G_f^{\circ}\) for difluoroacetylene \(\left(\mathrm{C}_2 \mathrm{F}_2\right)\) is 191.2 kJ/mol and - for hexafluorobenzene \(\left(\mathrm{C}_6 \mathrm{F}_6\right)\) is 78.2 kJ/mol.
These values assist in calculating the \(\Delta G^{\circ}_{rxn}\) (reaction free energy change) for the complete chemical reaction, which is essential for determining the equilibrium constant and spontaneity of the reaction.
When the free energy of formation is negative, the formation of the compound is spontaneous. If positive, energy is required to form the compound, making it non-spontaneous under standard conditions. In our exercise, - the \(\Delta G_f^{\circ}\) for difluoroacetylene \(\left(\mathrm{C}_2 \mathrm{F}_2\right)\) is 191.2 kJ/mol and - for hexafluorobenzene \(\left(\mathrm{C}_6 \mathrm{F}_6\right)\) is 78.2 kJ/mol.
These values assist in calculating the \(\Delta G^{\circ}_{rxn}\) (reaction free energy change) for the complete chemical reaction, which is essential for determining the equilibrium constant and spontaneity of the reaction.
Standard Enthalpy of Formation
The standard enthalpy of formation, \(\Delta H_f^{\circ}\), is defined as the heat change when one mole of a compound is formed from its elements in their standard states.
This measure allows us to understand the energy changes involved in creating a compound. A negative \(\Delta H_f^{\circ}\) indicates that the formation of the compound releases energy (exothermic), while a positive value means it absorbs energy (endothermic).
For example, in our exercise setup:
This measure allows us to understand the energy changes involved in creating a compound. A negative \(\Delta H_f^{\circ}\) indicates that the formation of the compound releases energy (exothermic), while a positive value means it absorbs energy (endothermic).
For example, in our exercise setup:
- Difluoroacetylene has a \(\Delta H_f^{\circ} = 241.3\) kJ/mol, which shows it requires energy input for formation.
- Hexafluorobenzene has \(\Delta H_f^{\circ} = 132.8\) kJ/mol, also requiring energy for formation.
Equilibrium Constant
The equilibrium constant, \(K\), is an essential concept that describes the ratio of concentrations of products to reactants at equilibrium. It tells us the extent of a reaction; whether it favors formation of products or reactants.
At standard conditions, \(\Delta G^{\circ}\) is related to \(K\) through the equation: \[\Delta G^{\circ} = -RT \ln K\]where - \(R\) is the universal gas constant (8.314 J/mol·K) and - \(T\) is the temperature in Kelvin.
In our exercise, calculating \(K\) at different temperatures allows insights into how equilibrium shifts:
At standard conditions, \(\Delta G^{\circ}\) is related to \(K\) through the equation: \[\Delta G^{\circ} = -RT \ln K\]where - \(R\) is the universal gas constant (8.314 J/mol·K) and - \(T\) is the temperature in Kelvin.
In our exercise, calculating \(K\) at different temperatures allows insights into how equilibrium shifts:
- At 298 K, this calculation uses values from \(\Delta G^{\circ}_{rxn}\).
- At 3000 K, changes in \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) can be assumed constant to estimate \(K\).
Entropy Change
Entropy change, \(\Delta S^{\circ}\), measures the disorder or randomness introduced by a reaction. It is essential because it helps determine the spontaneity of a process in conjunction with enthalpy.
In a chemical context, a positive change in entropy often corresponds to increased disorder, which on a microscopic level means more available microstates or different configurations a system can adopt.
For our exercise, calculating \(\Delta S^{\circ}_{rxn}\) involves determining the entropy contribution of both reactants and products. This value is critical as it, alongside \(\Delta H^{\circ}\), helps to assess the overall \(\Delta G^{\circ}\) of the system, ultimately influencing the equilibrium constant and the feasibility of the chemical process. Understanding \(\Delta S^{\circ}\) aids significantly in thermodynamics as it reassures which direction the reaction naturally favors.
In a chemical context, a positive change in entropy often corresponds to increased disorder, which on a microscopic level means more available microstates or different configurations a system can adopt.
For our exercise, calculating \(\Delta S^{\circ}_{rxn}\) involves determining the entropy contribution of both reactants and products. This value is critical as it, alongside \(\Delta H^{\circ}\), helps to assess the overall \(\Delta G^{\circ}\) of the system, ultimately influencing the equilibrium constant and the feasibility of the chemical process. Understanding \(\Delta S^{\circ}\) aids significantly in thermodynamics as it reassures which direction the reaction naturally favors.
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