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Chapter 4: Problem 8

Calculate \(\Delta G^{\circ}\left(25{ }^{\circ} \mathrm{C}\right)\) for \(K_{\mathrm{eq}}=0.001,0.01,0.1,1,10,100\), and 1000 .

Short Answer

Expert verified
Calculate \( \Delta G^{\circ} \) using \( \Delta G^{\circ} = -RT \ln K_{eq} \) for each value of \( K_{eq} \). The results are: 17.1, 11.4, 5.7, 0, -5.7, -11.4, -17.1 kJ/mol respectively.

Step by step solution

01

Understand the formula

To find the standard Gibbs free energy change (\( \Delta G^{\circ} \)) at 25掳 C, we use the relation \( \Delta G^{\circ} = -RT \ln K_{\mathrm{eq}} \), where \( R \) is the ideal gas constant \( 8.314 \text{ J/mol路K} \) and \( T \) is the temperature in Kelvin.
02

Convert temperature to Kelvin

The temperature is given as 25掳C. Convert this to Kelvin using the formula: \( T = 25 + 273.15 = 298.15 \text{ K} \).
03

Calculate \( \Delta G^{\circ} \) for each \( K_{\mathrm{eq}} \) value

Calculate \( \Delta G^{\circ} \) using \( \Delta G^{\circ} = -RT \ln K_{\mathrm{eq}} \) for each given \( K_{\mathrm{eq}} \):- For \( K_{\mathrm{eq}} = 0.001 \): \[ \Delta G^{\circ} = -(8.314 \text{ J/mol路K})(298.15 \text{ K}) \ln(0.001) = 17.1 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 0.01 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(0.01) = 11.4 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 0.1 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(0.1) = 5.7 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 1 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(1) = 0 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 10 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(10) = -5.7 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 100 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(100) = -11.4 \text{ kJ/mol} \]- For \( K_{\mathrm{eq}} = 1000 \): \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(1000) = -17.1 \text{ kJ/mol} \]

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. It's crucial to understanding how energy is transferred in chemical and physical processes. One key concept in thermodynamics is the idea of Gibbs Free Energy, denoted as \( \Delta G \). It helps predict whether a chemical reaction will occur spontaneously. Spontaneity in thermodynamics means that a process can proceed without requiring energy input from the surroundings.
  • If \( \Delta G < 0 \), the process is spontaneous.
  • If \( \Delta G > 0 \), the process is non-spontaneous.
  • If \( \Delta G = 0 \), the system is in equilibrium, meaning the reaction is perfectly balanced, and no net change occurs over time.
Thermodynamics gives insight into the energy changes in reactions, and understanding these changes can be crucial for fields like chemistry and engineering. It not only predicts the direction of a reaction but also its extent, which helps in various applications such as predicting the efficiency of engines or the yield of chemical reactions.
Equilibrium Constant
The equilibrium constant, \( K_{\mathrm{eq}} \), is a key concept in chemical equilibrium. It quantitatively expresses the ratio of the concentrations of products to reactants for a reaction at equilibrium. This state is reached when the reaction proceeds without any further net change in reactant and product concentrations.

The value of \( K_{\mathrm{eq}} \) gives vital information on the position of equilibrium:
  • A \( K_{\mathrm{eq}} \) greater than one indicates that products are favored at equilibrium.
  • A \( K_{\mathrm{eq}} \) less than one means reactants are favored.
  • A \( K_{\mathrm{eq}} \) equal to one signals that neither reactants nor products are favored, and the equilibrium lies in the middle.
In Gibbs free energy calculations, \( K_{\mathrm{eq}} \) is used in the equation \( \Delta G^{\circ} = -RT \ln K_{\mathrm{eq}} \). This formula links the concept of equilibrium with the free energy change, bridging thermodynamics and chemical equilibria. Understanding \( K_{\mathrm{eq}} \) is vital for predicting how changes in conditions like temperature or pressure can shift the position of equilibrium, thus altering the balance and rates of reactions.
Ideal Gas Constant
The ideal gas constant, \( R \), is a fundamental parameter in the ideal gas law, which is commonly used in chemistry and physics to describe the behavior of gases. In the context of Gibbs free energy, it's crucial as it provides the link between energy, temperature, and equilibrium.

The value of \( R \) is 8.314 J/mol路K, which is utilized in various equations involving thermodynamic calculations. For instance, in the equation \( \Delta G^{\circ} = -RT \ln K_{\mathrm{eq}} \), \( R \) serves as a constant to relate temperature (in Kelvin) and the natural logarithm of the equilibrium constant to calculate the standard Gibbs free energy change.
  • It helps in achieving a quantitative understanding of how changes in temperature and other conditions affect reactions.
  • Despite being a constant, \( R \) enables the effective calculation of energy changes in reactions across different scenarios.
The ideal gas constant is essential in connecting the macro world of observable phenomena to the micro world of molecular interactions by consistently relating pressure, volume, and temperature in gas reactions.

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