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Calculate \(K_{\mathrm{eq}}\) for a reaction with \(\Delta G^{\circ}=-17.1 \mathrm{~kJ} / \mathrm{mol}(-4.09 \mathrm{kcal} / \mathrm{mol})\) at \(328 \mathrm{~K}\). Compare this value to the \(1 \times 10^{3}\) seen at \(298 \mathrm{~K}\).

Short Answer

Expert verified
In this exercise, we calculated the equilibrium constant (K_eq) for a reaction at 328 K with a given standard Gibbs free energy change (∆G°) of -17.1 kJ/mol, using the relationship ∆G° = -RT ln(K_eq). We found K_eq at 328 K to be approximately 6.51 x 10^3. Comparing this value to the given K_eq at 298 K (1 x 10^3), we observed that the equilibrium constant increased approximately 6.5 fold as the temperature increased from 298 K to 328 K.

Step by step solution

01

Convert the given values to appropriate units

The given standard Gibbs free energy change is -17.1 kJ/mol. For calculation purposes, we will need to convert this value to the same energy unit as the gas constant R, which is in J/mol. To convert from kJ/mol to J/mol, simply multiply by 1,000: $$-17.1 \ (kJ/mol) = -17,100 \ (J/mol)$$
02

Rearrange the equation to solve for \(K_{eq}\)

Now we need to rearrange the equation to solve for the equilibrium constant: $$K_{eq} = \mathrm{e}^{\frac{-\Delta G^{\circ}}{RT}}$$
03

Calculate the equilibrium constant at 328 K

Now that we have the equation and the units are consistent, we can plug in the given values for \(\Delta G^{\circ}\), \(R\), and \(T\) to calculate \(K_{eq}\): $$K_{eq} = \mathrm{e}^{\frac{-(-17,100 \ J/mol)}{(8.31 \ J/(mol \cdot K)) \cdot (328 \ K)}}$$ $$K_{eq} \approx 6.51 \times 10^{3}$$
04

Compare the equilibrium constant at 328 K to the given value at 298 K

The given equilibrium constant at 298 K is \(1 \times 10^3\). To compare this value to the calculated equilibrium constant at 328 K, we can observe that the equilibrium constant increases as the temperature increases: $$K_{eq}(298 \ K) = 1 \times 10^3$$ $$K_{eq}(328 \ K) \approx 6.51 \times 10^3$$ The equilibrium constant has increased approximately 6.5 fold as the temperature increased from 298 K to 328 K. This comparison helps us in understanding how the equilibrium constant varies with temperature for this particular reaction.

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy, represented by \( \Delta G^{\circ} \), is a crucial concept in chemistry, particularly when discussing equilibrium and spontaneity of reactions. It is a measure of the maximum reversible work that a system can perform. In simple terms, it helps us understand whether a chemical reaction will occur spontaneously without external energy input.
A negative value of Gibbs Free Energy change, like \( \Delta G^{\circ} = -17.1 \text{ kJ/mol} \) in our example, indicates that the reaction tends to proceed in the forward direction spontaneously under standard conditions. This is because a negative \( \Delta G^{\circ} \) implies that the system is releasing energy, making it more stable.
Ultimately, Gibbs Free Energy links energy with equilibrium through its relation to the equilibrium constant \( K_{eq} \). The relationship is given by the equation:
\[ K_{eq} = \mathrm{e}^{\frac{-\Delta G^{\circ}}{RT}} \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.
Equilibrium Reactions
Equilibrium reactions are reactions that reach a state where the forward and reverse reaction rates are equal, resulting in constant concentrations of reactants and products.
This state is characterized by the equilibrium constant, \( K_{eq} \), which is a measure of the relative concentrations of products and reactants at equilibrium. It provides insight into the position of equilibrium – whether it favors products or reactants.
The equilibrium constant is affected by Gibbs Free Energy through the relationship \( K_{eq} = \mathrm{e}^{\frac{-\Delta G^{\circ}}{RT}} \). This relationship indicates that as \( \Delta G^{\circ} \) becomes more negative, \( K_{eq} \) becomes larger, signifying a greater tendency for the reaction to favor product formation at equilibrium.
Therefore, when analyzing or predicting which direction an equilibrium reaction will shift, consider both the sign and magnitude of Gibbs Free Energy.
Temperature Dependence
Temperature plays a significant role in influencing the equilibrium position of reactions. It can alter the value of the equilibrium constant, \( K_{eq} \).
According to the equation \( K_{eq} = \mathrm{e}^{\frac{-\Delta G^{\circ}}{RT}} \), temperature inversely affects the exponential term.
For example, in our exercise, as the temperature increases from 298 K to 328 K, \( K_{eq} \) increases approximately sixfold from \( 1 \times 10^3 \) to \( 6.51 \times 10^3 \).
This increase in \( K_{eq} \) signifies that higher temperatures favor the forward direction of an endothermic reaction, or equivalently, a reaction where more energy is absorbed than released. On the other hand, for exothermic reactions, increasing temperature may decrease \( K_{eq} \).
Thus, understanding temperature dependence is crucial for controlling reactions in industrial and laboratory settings.
Thermodynamics
Thermodynamics is the study of energy changes and transfers that accompany chemical reactions. Three important laws guide this field and help us understand concepts like Gibbs Free Energy, equilibrium, and reaction spontaneity.
  • The First Law of Thermodynamics, the law of energy conservation, states that energy cannot be created or destroyed. Instead, it is transformed from one form to another.
  • The Second Law of Thermodynamics introduces the concept of entropy, a measure of disorder or randomness of a system. It indicates that for spontaneous processes, the total entropy of the universe increases.
  • The Third Law states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.

These laws help us understand the balance between enthalpy (heat content), entropy, and free energy in chemical reactions. Gibbs Free Energy, specifically, is a direct application of these principles, combining enthalpy and entropy to predict reaction spontaneity and equilibrium.
Therefore, thermodynamics is indispensable in explaining how energy transformations and constraints influence chemical reactions in both physical and applied chemistry.

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Most popular questions from this chapter

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