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For a reaction at \(300 \mathrm{~K}\), enthalpy and entropy changes are \(-11.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(-105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively. What is the change in Gibbs free energy? (a) \(25 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The change in Gibbs free energy is \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\), option (d).

Step by step solution

01

Write Down the Gibbs Free Energy Equation

Gibbs free energy change is given by the equation: \[ \Delta G = \Delta H - T \Delta S \]where:\( \Delta G \) is the change in Gibbs free energy,\( \Delta H \) is the change in enthalpy,\( \Delta S \) is the change in entropy, and\( T \) is the temperature in Kelvin.
02

Plug in the Given Values

The problem states the following values:\( \Delta H = -11.5 \times 10^3 \) J/mol, \( \Delta S = -105 \) J/K/mol, and \( T = 300 \) K.Substitute these into the equation:\[ \Delta G = (-11.5 \times 10^3) - 300(-105) \]
03

Calculate the Entropy Term Contribution

Calculate the term for the entropy:\[ T \Delta S = 300 \times (-105) = -31500 \] J/mol.
04

Calculate the Gibbs Free Energy

Substitute the value from Step 3 into the equation to find \( \Delta G \):\[ \Delta G = -11500 - (-31500) \]\[ \Delta G = -11500 + 31500 = 20000 \] J/mol.
05

Convert Gibbs Free Energy to kJ/mol

Since the answer choices are in kJ/mol, convert \( 20000 \) J/mol to kJ/mol by dividing by 1000:\[ \Delta G = \frac{20000}{1000} = 20 \] kJ/mol.
06

Select the Correct Answer

The calculated Gibbs free energy change is 20 kJ/mol. Cross-reference with the provided options: (d) \( 20 \mathrm{~kJ} \mathrm{~mol}^{-1} \) is the correct answer.

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented as \( \Delta H \), is a crucial concept in chemical thermodynamics. It expresses the heat absorbed or released at constant pressure during a reaction. When a reaction occurs, bonds in reactants break, and new bonds form in products. If the total energy needed to break these bonds is less than the energy released when new bonds form, the reaction is exothermic and \( \Delta H \) is negative, indicating heat is released. Conversely, if more energy is needed to break bonds than is released upon forming new ones, the reaction is endothermic, and \( \Delta H \) is positive.

For example, in our given problem, the enthalpy change is \(-11.5 \times 10^3\) J/mol. This negative value suggests an exothermic reaction, meaning the system releases heat during the reaction process.
Entropy Change
Entropy change, denoted as \( \Delta S \), measures the disorder or randomness of a system. In thermodynamics, systems tend to move towards higher entropy, or more disorder. When \( \Delta S \) is positive, it indicates an increase in disorder within the system, while a negative \( \Delta S \) suggests a decrease in disorder.

In our scenario, the entropy change of \(-105\) J/K/mol hints that the disorder of the system decreases as the reaction proceeds. Several factors like changes in physical state, molecular complexity, or the number of gas molecules can influence \( \Delta S \). An understanding of entropy is essential for predicting the spontaneity of reactions, especially when combined with enthalpy changes in Gibbs Free Energy calculations.
Thermodynamic Equations
Thermodynamic equations allow us to quantify changes in state properties like energy, enthalpy, and entropy. One of the most vital equations in this context is the Gibbs Free Energy equation:
\( \Delta G = \Delta H - T \Delta S \).

This formula connects enthalpy, entropy, and temperature to determine the spontaneity of a reaction. Spontaneity refers to whether a reaction can proceed without external influence. If \( \Delta G \) is negative, the reaction occurs spontaneously. However, a positive \( \Delta G \) means the reaction requires energy input. This equation allows scientists to predict and analyze reaction conditions and energy changes efficiently.
Temperature Dependence
Temperature plays a significant role in chemical reactions, affecting both the rate and spontaneity. In the Gibbs Free Energy equation, temperature is a multiplying factor of entropy change. This relationship implies that at different temperatures, the spontaneity of a reaction might vary even if enthalpy and entropy changes remain constant.

For instance, even if a reaction is non-spontaneous at a lower temperature, increasing the temperature could make \( T \Delta S \) sufficiently large to result in a negative \( \Delta G \). Thus, understanding how temperature impacts reactions helps in designing processes that optimize desirable outcomes.
Chemical Thermodynamics
Chemical thermodynamics involves the study of the interplay between heat, work, and chemical reactions. It encompasses principles such as enthalpy, entropy, and free energy, all essential for understanding how and why chemical reactions occur.

By using the various thermodynamic equations, scientists can predict reaction conditions and energy interactions, crafting controlled environments for industrial, laboratory, or natural processes. The principles of chemical thermodynamics guide the broader quest to harness energy efficiently and sustainably.

Overall, an appreciation of concepts like Gibbs Free Energy is crucial for mastering chemistry fundamentals and their application in solving real-world challenges.

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Most popular questions from this chapter

The standard enthalpy of formation \(\left(\Delta_{t} \mathrm{H}^{\circ}\right)\) at \(298 \mathrm{~K}\) for methane, \(\mathrm{CH}_{4}(\mathrm{~g})\) is \(-74.8 \mathrm{~kJ} \mathrm{~mol}-1\), the additional information required to determine the average energy for \(\mathrm{C}-\mathrm{H}\) bond formation would be: (a) The dissociation energy of \(\mathrm{H}_{2}\) and enthalpy of sublimation of carbon (b) Latent heat of vaporization of methane (c) The first four ionization energies of carbon and electron gain enthalpy of hydrogen (d) The dissociation energy of hydrogen molecule, \(\mathrm{H}_{2}\)

The enthalpy change of a reaction does not depend on: (a) Initial and final enthalpy change of reaction (b) State of reactants and products (c) Different intermediate reactions (d) Nature of reactants and products

The \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) are \(-393.5\), \(-110.5\) and \(-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy change (in \(\mathrm{kJ}\) ) for the reaction: \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \mathrm{is}\) (a) \(524.1\) (b) \(41.2\) (c) \(-262.5\) (d) \(-41.2\)

Two moles of an ideal gas are compressed at \(300 \mathrm{~K}\) from a pressure of 1 atm to a pressure of 2 atm. The change in free energy is: (a) \(5.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(3.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(8.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The standard enthalpy of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(58.04 \mathrm{~kJ}\) and standard entropy of this reaction is \(176.7 \mathrm{~J} \mathrm{~K}^{-1}\). The standard free energy change for this reaction at \(25^{\circ} \mathrm{C}\), is: (a) \(5.39 \mathrm{~kJ}\) (b) \(-5.39 \mathrm{~kJ}\) (c) \(539 \mathrm{~kJ}\) (d) \(53.9 \mathrm{~kJ}\)

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