Calculating Gibbs Free Energy Change ΔG For Reaction AB(s) ⇌ A⁺(aq) + B⁻(aq)
#Introduction
Hey guys! Today, we're diving deep into a fascinating chemistry problem that involves calculating the Gibbs free energy change (**ΔG**) for a reversible reaction. Specifically, we're looking at the reaction where a solid compound AB dissociates into its ions A⁺ and B⁻ in an aqueous solution. This is a classic example of a chemical equilibrium, where the forward and reverse reactions occur at the same rate, and we can use some cool thermodynamic principles to understand what's going on. So, buckle up, and let's get started!
Understanding the Reaction: AB(s) ⇌ A⁺(aq) + B⁻(aq)
First, let’s break down the reaction itself: **AB(s) ⇌ A⁺(aq) + B⁻(aq)**. This equation tells us that solid AB is in equilibrium with its ions, A⁺ and B⁻, in an aqueous solution. The double arrow (⇌) indicates that the reaction is reversible, meaning it can proceed in both directions. The solid AB can dissolve into ions (forward reaction), and the ions can combine to reform the solid (reverse reaction). The equilibrium constant, **K**, gives us a measure of the relative amounts of reactants and products at equilibrium. In this case, **K = 0.0005** at 298 K, which is a pretty small value. This tells us that at equilibrium, there are significantly more reactants (AB) than products (A⁺ and B⁻). To truly understand the thermodynamics of this reaction, we need to calculate the Gibbs free energy change (**ΔG**). The Gibbs free energy is a thermodynamic potential that can be used to predict the spontaneity of a reaction at a constant temperature and pressure. A negative **ΔG** indicates a spontaneous reaction, while a positive **ΔG** indicates a non-spontaneous reaction. The relationship between **ΔG**, the equilibrium constant **K**, and the temperature **T** is given by the equation: **ΔG° = -RTlnK**, where **R** is the ideal gas constant (8.314 J/(mol·K)). This equation gives us the standard Gibbs free energy change (**ΔG°**), which is the change in Gibbs free energy when all reactants and products are in their standard states (1 M for solutions, 1 atm for gases, and pure solids or liquids). However, our reaction is not under standard conditions since we have specific concentrations of A⁺ and B⁻. To calculate the Gibbs free energy change under non-standard conditions (**ΔG**), we need to use another equation: **ΔG = ΔG° + RTlnQ**, where **Q** is the reaction quotient. The reaction quotient **Q** is a measure of the relative amounts of products and reactants at any given time, not necessarily at equilibrium. For our reaction, **Q = [A⁺][B⁻]**. By calculating **Q** and using the given concentrations, we can determine the Gibbs free energy change under these specific conditions. Understanding these concepts is crucial for predicting the direction a reaction will proceed to reach equilibrium and for determining the feasibility of a reaction under different conditions.
Calculating ΔG°: Standard Gibbs Free Energy Change
Alright, let's crunch some numbers! First, we need to calculate the standard Gibbs free energy change (**ΔG°**) using the equation:
**ΔG° = -RTlnK**
Where:
- **R** is the ideal gas constant, which is 8.314 J/(mol·K)
- **T** is the temperature in Kelvin, which is given as 298 K
- **K** is the equilibrium constant, which is given as 0.0005
Let's plug these values into the equation:
**ΔG° = -(8.314 J/(mol·K)) × (298 K) × ln(0.0005)**
First, we calculate the natural logarithm of 0.0005:
**ln(0.0005) ≈ -7.6009**
Now, we substitute this back into the equation:
**ΔG° = -(8.314 J/(mol·K)) × (298 K) × (-7.6009)**
**ΔG° ≈ 18850.4 J/mol**
To convert this to kJ/mol, we divide by 1000:
**ΔG° ≈ 18.85 kJ/mol**
So, the standard Gibbs free energy change for this reaction is approximately **18.85 kJ/mol**. This positive value indicates that the reaction is non-spontaneous under standard conditions, meaning it requires energy input to proceed in the forward direction. However, standard conditions are not always what we have in real-world scenarios. To get the full picture, we need to consider the actual concentrations of the reactants and products in our specific system.
Determining Q: The Reaction Quotient
Now that we have **ΔG°**, we need to calculate the reaction quotient (**Q**) to find the Gibbs free energy change under the given conditions. Remember, the reaction quotient **Q** is a measure of the relative amounts of products and reactants at a particular time, and it helps us determine the direction the reaction will shift to reach equilibrium. For the reaction **AB(s) ⇌ A⁺(aq) + B⁻(aq)**, the reaction quotient is given by:
**Q = [A⁺][B⁻]**
We are given the concentrations of **A⁺** and **B⁻** as:
- [A⁺] = 0.330 M
- [B⁻] = 0.660 M
Plugging these values into the equation for **Q**:
**Q = (0.330 M) × (0.660 M)**
**Q = 0.2178**
So, the reaction quotient **Q** for this reaction under the given conditions is **0.2178**. This value is greater than the equilibrium constant **K** (0.0005), which tells us that there are relatively more products than there would be at equilibrium. The reaction will therefore shift towards the reactants (the reverse reaction) to reach equilibrium. Knowing **Q** is crucial because it allows us to calculate the Gibbs free energy change under non-standard conditions, giving us a more accurate picture of the reaction's spontaneity in our specific situation. Next, we'll use **Q** and **ΔG°** to calculate **ΔG**.
Calculating ΔG: Gibbs Free Energy Change Under Non-Standard Conditions
Okay, guys, we're in the home stretch now! We've calculated the standard Gibbs free energy change (**ΔG°**) and the reaction quotient (**Q**). Now, we can finally calculate the Gibbs free energy change (**ΔG**) under the given non-standard conditions using the equation:
**ΔG = ΔG° + RTlnQ**
We have:
- **ΔG° ≈ 18.85 kJ/mol** (which we converted from J/mol)
- **R = 8.314 J/(mol·K)**
- **T = 298 K**
- **Q = 0.2178**
First, let's plug these values into the equation:
**ΔG = (18.85 kJ/mol) + (8.314 J/(mol·K) × 298 K × ln(0.2178))**
We need to make sure our units are consistent. Since **ΔG°** is in kJ/mol, let's convert the **RTlnQ** term to kJ/mol as well. First, we calculate the natural logarithm of 0.2178:
**ln(0.2178) ≈ -1.524**
Now, we multiply this by **RT**:
**(8.314 J/(mol·K)) × (298 K) × (-1.524) ≈ -3785.3 J/mol**
Convert this to kJ/mol by dividing by 1000:
**-3785.3 J/mol ≈ -3.785 kJ/mol**
Now, we can substitute this back into the equation for **ΔG**:
**ΔG = (18.85 kJ/mol) + (-3.785 kJ/mol)**
**ΔG ≈ 15.065 kJ/mol**
Rounding to a reasonable number of significant figures, we get:
**ΔG ≈ 15.1 kJ/mol**
So, the Gibbs free energy change for the reaction under these specific conditions is approximately **15.1 kJ/mol**. This positive value indicates that the reaction is still non-spontaneous under these conditions, meaning it requires energy input to proceed in the forward direction. The fact that **ΔG** is smaller than **ΔG°** suggests that the reaction is closer to being spontaneous than it would be under standard conditions, but it still needs an external push.
Conclusion: The Significance of ΔG
Alright, guys, we made it! We successfully calculated the Gibbs free energy change (**ΔG**) for the reaction **AB(s) ⇌ A⁺(aq) + B⁻(aq)** under non-standard conditions. We found that **ΔG** is approximately **15.1 kJ/mol**, which is a positive value. This result is super important because it tells us that under the given conditions ([A⁺] = 0.330 M** and **[B⁻] = 0.660 M** at 298 K), the reaction is non-spontaneous. This means that the forward reaction, where solid **AB** dissociates into its ions, will not occur on its own without an external input of energy. To put it simply, if you mix **AB** in water under these conditions, it won't readily dissolve into **A⁺** and **B⁻** ions. The positive **ΔG** value also indicates that the reverse reaction, where **A⁺** and **B⁻** ions combine to form solid **AB**, is more favorable under these conditions. This doesn't mean the forward reaction can't happen, but it does mean it won't happen spontaneously. We might need to add energy, change the temperature, or alter the concentrations of the ions to make the forward reaction more favorable. By understanding the Gibbs free energy change, we can predict the behavior of chemical reactions and manipulate conditions to achieve desired outcomes. This is a fundamental concept in chemistry, and it has wide-ranging applications in various fields, from designing new chemical processes to understanding biological systems. So, next time you see a chemical reaction, remember **ΔG** and how it dictates the spontaneity of the process! Understanding the Gibbs free energy change is crucial for predicting the spontaneity and equilibrium of chemical reactions. By calculating **ΔG**, we can determine whether a reaction will proceed spontaneously under given conditions, providing valuable insights for various applications in chemistry and beyond.