Relationship Between ΔG, Reaction Quotient (Q), & Equilibrium Constant (K)
Physical Chemistry is heavily interconnected. One of the most important conceptual bridges you must cross for JEE and NEET is connecting Thermodynamics (Gibbs Free Energy) with Chemical Equilibrium.
How does the spontaneity of a reaction (ΔG) relate to its Equilibrium Constant (K)? Let's derive the master equation that examiners love to test year after year.
Video Tutorial: The Master Thermodynamic Equation
Watch Abhishek Sengar sir from CHEMCA expertly derive the relationship between ΔG° and K, and break down the three distinct cases that predict spontaneity.
1. The General Thermodynamic Equation
Before equilibrium is reached, any chemical reaction at a specific moment has a Gibbs Free Energy Change (ΔG). This value determines the immediate spontaneity of the reaction. It is linked to the Standard Gibbs Free Energy (ΔG°) and the Reaction Quotient (Q) by the following equation:
Where:
- ΔG = Change in Gibbs Free Energy (Variable)
- ΔG° = Standard Gibbs Free Energy Change (Constant for a specific reaction)
- R = Universal Gas Constant = 8.314 J K-1 mol-1
- T = Temperature in Kelvin
- Q = Reaction Quotient (Snapshot of the concentration ratio)
2. Arriving at Equilibrium
What happens when the reaction finally reaches equilibrium? Two crucial things occur mathematically:
- The driving force of the reaction vanishes, meaning ΔG = 0.
- The reaction snapshot (Q) finally hits its target, meaning Q = K (the Equilibrium Constant).
Let's substitute these two conditions back into our general equation:
0 = ΔG° + RT ln(K)
In chemistry numericals, dealing with natural log (ln, base e) is difficult. We always convert it to the common log (log10) by multiplying by 2.303.
The highly usable, final formula becomes:
ΔG° = - 2.303 RT log(K)
3. Predicting ΔG° from the value of K (The 3 Cases)
Because R and T (in Kelvin) are strictly positive values, the sign of ΔG° is completely controlled by the value of log(K). Let's analyze the math.
Fig: Visually tracking how the sign of the logarithm perfectly controls the thermodynamic outcome.
Practice Questions for JEE & NEET
Let's test your ability to apply these thermodynamic conditions to real exam questions!
Question 1: You are told that a specific chemical reaction has an equilibrium constant Kc = 108. Without doing any math, what can you definitively say about the Standard Gibbs Free Energy Change (ΔG°) of this reaction?
Answer: ΔG° is highly Negative (The reaction is highly spontaneous in the forward direction).
Reasoning:
We apply Case 1 from our rules. Because Kc = 108, it is massively larger than 1.
Therefore, log(108) = +8 (a positive number).
When we plug this into the formula ΔG° = -2.303 RT log(K), the negative sign at the front of the formula remains. A negative ΔG° indicates a reaction that naturally drives forward to completion!
Question 2 (The R-Value Trap): When using the formula ΔG° = -2.303 RT log(K) to calculate energy in Joules, why must we strictly use R = 8.314 instead of R = 0.0821?
Answer: Because of matching units of Energy.
Reasoning:
This is a classic trap that costs students a -1 mark.
- The value R = 0.0821 has the units L·atm / (mol·K). It is used in gas law calculations involving liters and atmospheres.
- Gibbs Free Energy (ΔG°) is a measure of Energy, typically expressed in Joules (J) or kiloJoules (kJ).
- Therefore, you must use the Universal Gas Constant that is formulated in energy units: R = 8.314 J / (mol·K).
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