Gibbs Free Energy And Equilibrium: Videos & Practice Problems

Gibbs free energy (ΔG) is crucial for understanding chemical equilibrium, represented by the equation ${\mathrm{\Delta G}}^{°}=-\mathrm{RT}lnKeq$. Here, R is the gas constant (8.314 J/(mol·K)). For non-standard conditions, the relationship is given by ${\mathrm{\Delta G}}^{non-standard}={\mathrm{\Delta G}}^{°}+\mathrm{RT}lnQ$, connecting standard and non-standard Gibbs free energy. Understanding these equations is essential for predicting reaction spontaneity and equilibrium behavior.

∆G° and K

Gibbs free energy is a crucial concept in thermodynamics, particularly when analyzing chemical reactions and their equilibrium states. The relationship between the standard change in Gibbs free energy (ΔG_°) and the equilibrium constant (K_eq) can be expressed through the equation:

$$\Delta G^{\circ} = -RT \ln K_{eq}$$

In this equation, R represents the gas constant, which is valued at 8.314 J/(mol·K). This constant is essential when discussing energy changes in reactions, as it connects temperature and energy units. It is important to note that when using R in this context, the units of energy are typically in joules or kilojoules.

The equation indicates that the standard Gibbs free energy change is inversely related to the natural logarithm of the equilibrium constant. A negative ΔG_° suggests that the reaction is spontaneous under standard conditions, while a positive value indicates non-spontaneity. Understanding this relationship allows for the prediction of reaction favorability and the extent to which a reaction will proceed towards equilibrium.

∆G° and K Example

To determine the Gibbs free energy change (\( \Delta G^\circ \)) for a reaction at equilibrium, we can use the relationship between Gibbs free energy and the equilibrium constant (\( K \)). The formula is given by:

\[ \Delta G^\circ = -RT \ln K \]

In this equation, \( R \) is the universal gas constant, which is approximately \( 8.314 \, \text{J/(mol·K)} \). The temperature must be in Kelvin, so to convert from degrees Celsius to Kelvin, we add \( 273.15 \) to the Celsius temperature. For a reaction occurring at \( 25 \, \text{°C} \), the temperature in Kelvin is:

\[ T = 25 + 273.15 = 298.15 \, \text{K} \]

The equilibrium constant provided is \( K = 2.8 \times 10^4 \). To find \( \Delta G^\circ \), we first calculate the natural logarithm of the equilibrium constant:

\[ \ln K = \ln(2.8 \times 10^4) \]

Now, substituting the values into the Gibbs free energy equation:

\[ \Delta G^\circ = - (8.314 \, \text{J/(mol·K)}) \times (298.15 \, \text{K}) \times \ln(2.8 \times 10^4) \]

After performing the calculations, we find that:

\[ \Delta G^\circ \approx -2.5 \times 10^4 \, \text{J/mol} \]

This negative value indicates that the reaction is spontaneous under standard conditions. Thus, the Gibbs free energy of the reactant is approximately \( -25,000 \, \text{J/mol} \).

For reaction, Ag₂CO₃ (s) ⇌ Ag₂O (s) + CO₂ (g), the ∆Hº = 79.14 kJ/mol, ∆Sº = 167.2 J/mol*K.
Determine the equilibrium constant at which the temperature is 365.1 K.

• A. 1.90 × 10⁻⁹
• B. 8.23 × 10⁻²¹
• C. 388.1
• D. 2.577 × 10⁻³

Answer: (D)

∆G, ∆G° and Q

The relationship between standard and non-standard Gibbs free energy is crucial for understanding thermodynamic processes. The equation that connects these two states is expressed as:

$\Delta G = \Delta G^\circ + RT \ln Q$

In this equation, $\Delta G$ represents the change in non-standard Gibbs free energy, while $\Delta G^\circ$ denotes the change in standard Gibbs free energy. The term $R$ is the universal gas constant, valued at 8.314 J/(mol·K), which is used when energy is measured in joules or kilojoules. The variable $T$ stands for the absolute temperature in Kelvin, and $Q$ is the reaction quotient, which reflects the ratio of the concentrations of products to reactants at any point in the reaction.

This formula is particularly useful when analyzing reactions under non-standard conditions, allowing for the calculation of Gibbs free energy changes by incorporating the effects of concentration and temperature. Understanding this relationship helps in predicting the spontaneity of reactions, as a negative $\Delta G$ indicates a spontaneous process, while a positive value suggests non-spontaneity.

∆G, ∆G°, and Q Example

In this reaction, we are tasked with calculating the change in non-standard Gibbs free energy (\( \Delta G \)) using the provided standard Gibbs free energy change (\( \Delta G^0 \)) and the partial pressures of the gases involved. The standard Gibbs free energy change is given as -374 kJ, which we convert to joules by multiplying by 1,000, resulting in \( \Delta G^0 = -374,000 \) J.

The reaction involves sulfur tetrafluoride (SF₄), fluorine gas (F₂), and sulfur hexafluoride (SF₆). The partial pressures are as follows: SF₄ = 0.63 atm, F₂ = 0.95 atm, and SF₆ = 1.7 atm. To find the reaction quotient (Q), we use the formula:

Q = \(\frac{P_{SF_6}}{P_{SF_4} \cdot P_{F_2}}\)

Substituting the values, we calculate:

Q = \(\frac{1.7}{0.63 \cdot 0.95} = 2.840\)

Next, we apply the Gibbs free energy equation:

\( \Delta G = \Delta G^0 + RT \ln Q \)

Here, R is the universal gas constant, which is 8.314 J/(mol·K), and we assume the temperature (T) is at standard conditions, 298.15 K. Plugging in the values, we have:

\( \Delta G = -374,000 \, \text{J} + (8.314 \, \text{J/(mol·K)} \cdot 298.15 \, \text{K} \cdot \ln(2.840)) \)

Calculating the natural logarithm of Q:

\( \ln(2.840) \approx 1.040 \)

Now substituting this back into the equation:

\( \Delta G = -374,000 \, \text{J} + (8.314 \cdot 298.15 \cdot 1.040) \)

Calculating the second term:

\( 8.314 \cdot 298.15 \cdot 1.040 \approx 2,570.5 \, \text{J} \)

Finally, we find:

\( \Delta G \approx -374,000 \, \text{J} + 2,570.5 \, \text{J} \approx -371,429.5 \, \text{J} \)

Thus, the change in the non-standard Gibbs free energy for the reaction is approximately \( -3.71 \times 10^5 \, \text{J} \).

Consider a hypothetical reaction at 38 ºC, X₂ (aq) + 2 Y (s) ⇌ 3 Z (aq), with a ∆G of −75.8 kJ.
Concentrations of reactants and products: [X₂] = 1.4 M, [Y] = 0.34 M, [Z] = 2.6 M. Calculate K_eq of this given reaction.

• A. 4.13 × 10¹⁵
• B. 2.67 × 10¹¹
• C. 9.60 × 10¹²
• D. 6.67 × 10¹³

Answer: (D)

Here’s what students ask on this topic:

What is the relationship between Gibbs free energy (ΔG°) and the equilibrium constant (K)?

The relationship between Gibbs free energy (ΔG°) and the equilibrium constant (K) is expressed by the equation:

$\Delta G_{°}=-RTlnK$

Here, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant. This equation shows that if K > 1 (favoring products), ΔG° is negative, indicating a spontaneous reaction under standard conditions. Conversely, if K < 1 (favoring reactants), ΔG° is positive, meaning the reaction is non-spontaneous under standard conditions.

How do you calculate Gibbs free energy (ΔG) under non-standard conditions?

To calculate Gibbs free energy (ΔG) under non-standard conditions, use the equation:

$\Delta G=\Delta G_{°}+RTlnQ$

Here, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient. Q is calculated using the concentrations or partial pressures of reactants and products at a given moment. This equation links standard conditions to real-world, non-standard conditions, helping predict whether a reaction will proceed spontaneously at a specific point in time.

What is the significance of the equation ΔG° = -RTln(K) in thermodynamics?

The equation $\Delta G_{°}=-RTlnK$ is significant in thermodynamics because it connects the standard Gibbs free energy change (ΔG°) to the equilibrium constant (K). It allows us to determine the spontaneity of a reaction under standard conditions. If ΔG° is negative, the reaction is spontaneous, and K > 1, favoring products. If ΔG° is positive, the reaction is non-spontaneous, and K < 1, favoring reactants. This equation also helps predict the position of equilibrium and the extent to which a reaction will proceed.

What is the difference between standard Gibbs free energy (ΔG°) and non-standard Gibbs free energy (ΔG)?

Standard Gibbs free energy (ΔG°) refers to the Gibbs free energy change under standard conditions, where reactants and products are in their standard states (1 atm for gases, 1 M for solutions, and pure substances for solids/liquids). Non-standard Gibbs free energy (ΔG) accounts for real-world conditions, where concentrations or pressures may differ from standard states. The relationship between them is given by:

$\Delta G=\Delta G_{°}+RTlnQ$

Here, Q is the reaction quotient, which reflects the current state of the reaction. This distinction is crucial for understanding how reactions behave under varying conditions.

How does temperature affect the relationship between ΔG° and K?

Temperature plays a critical role in the relationship between ΔG° and K, as seen in the equation:

$\Delta G_{°}=-RTlnK$

Here, T is the temperature in Kelvin. As temperature increases, the value of T directly affects the magnitude of ΔG° and K. For endothermic reactions, increasing T generally increases K, favoring products. For exothermic reactions, increasing T typically decreases K, favoring reactants. This temperature dependence is essential for predicting how equilibrium shifts with temperature changes.