Gibbs Free Energy And Equilibrium - Online Tutor, Practice Problems & Exam Prep

Understanding the relationship between Gibbs free energy (ΔG°) and the equilibrium constant (K) is crucial in thermodynamics. The equation:

$\mathrm{\Delta G}°=-RTln\left(K\right)$

connects these two concepts, where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. This equation applies under standard conditions, which involve standard states of reactants and products. When dealing with non-standard conditions, the Gibbs free energy change (ΔG) is calculated using the equation:

$\mathrm{\Delta G}=\mathrm{\Delta G}°+RTln\left(Q\right)$

where Q is the reaction quotient. These equations are vital for predicting the spontaneity of reactions and understanding chemical equilibria, as they link thermodynamic quantities to the position of equilibrium.

∆Gº and K

Now when we discuss Gibbs free energy and equilibrium, we're really making a comparison between Gibbs free energy and our equilibrium constant. Now here we're going to say the relationship between our Gibbs free energy which is delta G₀ and our equilibrium constant which is K and we'll say EQ. Can also be observed in the following formula.

Here we're going to say use this formula when your equilibrium constant K_AQ is given or can be calculated. Here are the formula Is that the change in our standard Gibbs free energy equals negative RT lnK. And here we're talking about K_eq. Here R would be our gas constant, which we've seen in the past. It's equal to 8.314 Joules over moles times K.

Remember, we employ this type of R constant whenever we're talking about energy, velocity, or speed. In this section, we're not really talking about velocity or speed, we're talking about energy because we're using the units of either joules or kilojoules. So just keep that in mind. When we're talking about joules or kilojoules, then we're using the R value in the form of 8.314 joules over moles times K.

ΔG = − R T ⁡ ln K

Gibbs Free Energy And Equilibrium Example

Here the example says a certain reaction takes place at 25°C and has an equilibrium constant of 2.8 * 10⁴ determinant Gibbs free energy of the reaction. All right, so we need to find Gibbs free energy, so ΔG⁰. They're giving us the equilibrium constant.

So the formula that connects them together is -RT ln K. We plug in our R value which is 8.314 Joules over moles times K. Temperature needs to be in Kelvin so add 273.15 to this degree Celsius which gives us 298.15 K and then we have ln of our equilibrium constant which is 2.8 * 10⁴.

When we multiply everything across the board, we see that Kelvin will cancel out. We'll get at the end -2.5 * 10⁴ joules per mole. So here this will be our final answer for Gibbs free energy of this reaction.

For reaction, Ag₂CO₃ (s) ⇌ Ag₂O (s) + CO₂ (g), the ∆Hº = 79.14 kJ/mol, ∆Sº = 167.2 J/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

∆G, ∆Gº and Q

Here we're going to say that the following equation relates the change in the standard Gibbs free energy which deals with standard conditions and the change with Gibbs free energy which is under non standard conditions. Here we're going to say use this formula when given Gibbs free energy under non standard conditions or dealing with the reaction quotient Q.

Here we're going to say that this version of Gibbs free energy formula is the change in the non standard Gibbs free energy equals the change in the standard Gibbs free energy plus RTLN Q. Here we're going to say R is equal to 8.314 joules over moles times K.

Remember we use this version of R anytime we're dealing with energy and we know we're dealing with energy whenever we have units of joules or kilojoules involved, right? So keep this in mind. This equation is a way of us connecting the non standard conditions with the standard conditions in relation to Gibbs free energy.

ΔG = ΔG ⁰ + R T ln Q R = 8.314 J mol ⋅ K

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

Gibbs Free Energy And Equilibrium Example

Here it says the given reaction has a change in the standard Gibbs free energy of -374 kilojoules and partial pressures of sulfur. Tetrafluoride, fluorine gas and sulfur hexafluoride are .63 atmospheres, 0.95 atmospheres and 1.7 atmospheres respectively. Here we need to calculate the change in the non standard Gibbs free energy of the reaction for this reaction.

So here is our balanced equation. It's a one to one relationship with everything. Here we're going to say that ΔG equals ΔG₀ or not plus RTlnQ. Here they're giving us these values. They're going to help us find Q. Just like an equilibrium constant is equal to products over reactants. It ignores solids and liquids, but everything here is a gas.

So it equals the pressure of sulfur hexafluoride divided by the pressure of sulfur tetrafluoride times the pressure of fluorine gas. Plug in the value, so we have 1.7 divided by 0.63 * 0.95. Q here equals 2.84, so we have what our Q is. We have what ΔG₀ is. Here R uses joules, so we're going to convert these kilojoules into joules by multiplying them by 1000.

When we do that we get -374000 Joules plus 8.314 Joules over moles times K. Here temperature were not given a temperature so we're assuming it's under standard temperature which is 298.15 Kelvin and then we have ln of Q which is 2.840. When we plug all this in, this will give us the change in the non standard Gibbs free energy which comes out to -3.71 * 10⁵ joules. OK, so this would be our final answer for this given question.

Here’s what students ask on this topic:

Which of the following correctly expresses the standard Gibbs free energy change of a reaction in terms of the reaction temperature (T) and equilibrium constant (K)?

The standard Gibbs free energy change (ΔG^°) of a reaction is related to the reaction temperature (T, in Kelvin) and the equilibrium constant (K) by the following equation:

$\mathrm{\Delta G^°}=-RTln\left(K\right)$

In this equation, R is the universal gas constant (8.314 J/(mol·K)), T is the absolute temperature in Kelvin, ln denotes the natural logarithm, and K is the equilibrium constant of the reaction at temperature T. The equilibrium constant is a dimensionless number representing the ratio of the concentration of products to the concentration of reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation.

Remember, the sign of ΔG^° tells us about the spontaneity of the reaction at standard conditions: a negative value indicates a spontaneous reaction, while a positive value indicates a non-spontaneous reaction.

What is the relationship between Gibbs free energy and equilibrium constant?

Gibbs free energy (G) is a thermodynamic quantity that can be used to predict the spontaneity of a chemical reaction. The relationship between Gibbs free energy and the equilibrium constant (K) is expressed by the equation:

$∆G_{°}=-RTlnK$

In this equation, $∆G_{°}$ is the standard change in Gibbs free energy, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant of the reaction at that temperature.

When K > 1, the reaction is product-favored at equilibrium, and $∆G_{°}$ is negative, indicating that the reaction is spontaneous in the forward direction under standard conditions. Conversely, when K < 1, the reaction is reactant-favored, and $∆G_{°}$ is positive, suggesting the reaction is non-sp

What is the relationship between Gibbs free energy and equilibrium constant?

Gibbs free energy (G) is a thermodynamic quantity that can be used to predict the spontaneity of a chemical reaction. The relationship between Gibbs free energy and the equilibrium constant (K) is expressed by the equation:

$∆G_{°}=-RTlnK$

In this equation, $∆G_{°}$ is the standard change in Gibbs free energy, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant of the reaction at that temperature.

When K > 1, the reaction is product-favored at equilibrium, and $∆G_{°}$ is negative, indicating that the reaction is spontaneous in the forward direction under standard conditions. Conversely, when K < 1, the reaction is reactant-favored, and $∆G_{°}$ is positive, suggesting the reaction is non-sp

How are Gibbs free energy and the equilibrium constant related?

Gibbs free energy (G) and the equilibrium constant (K) are fundamentally related through the equation:

$∆G°=-RTln\left(K\right)$

where ∆G° is the standard change in Gibbs free energy, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant.

At equilibrium, ∆G° is equal to zero, and thus the equation simplifies to:

$0=-RTln\left(K\right)$

Solving for K gives you:

$K=e^{-\frac{∆G°}{RT}}$