Q. Use the following data:
| Substance |
ΔfH°(500 K) / kJ mol⁻¹ |
S°(500 K) / JK⁻¹ mol⁻¹ |
| AB(g) |
32 |
222 |
| A₂(g) |
6 |
146 |
| B₂(g) |
x |
280 |
One mole each of A₂(g) and B₂(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K.
\[ \text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g) \]
The value of x (in kJ mol⁻¹) is _____. (Nearest integer)
(Given : log K = 2.2, R = 8.3 J K⁻¹ mol⁻¹ )
Step-by-Step Explanation
Step 1: Calculate Standard Gibbs Free Energy change (\(\Delta G^\circ\)).
The relationship between \(\Delta G^\circ\) and Equilibrium Constant \(K\) is:
\( \Delta G^\circ = -2.303 \times R \times T \times \log K \)
Given: \( T = 500 \text{ K} \), \( \log K = 2.2 \), \( R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \)
\( \Delta G^\circ = -2.303 \times 8.3 \times 500 \times 2.2 \)
\( \Delta G^\circ = -21026.39 \text{ J mol}^{-1} \approx -21.026 \text{ kJ mol}^{-1} \)
Step 2: Calculate Standard Entropy change (\(\Delta S^\circ\)) for the reaction.
For the reaction: \( \text{A}_2 + \text{B}_2 \rightarrow 2\text{AB} \)
\( \Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}) \)
\( \Delta S^\circ = [2 \times S^\circ(\text{AB})] - [S^\circ(\text{A}_2) + S^\circ(\text{B}_2)] \)
\( \Delta S^\circ = [2 \times 222] - [146 + 280] \)
\( \Delta S^\circ = 444 - 426 = 18 \text{ J K}^{-1} \text{ mol}^{-1} = 0.018 \text{ kJ K}^{-1} \text{ mol}^{-1} \)
Step 3: Relate \(\Delta G^\circ, \Delta H^\circ,\) and \(\Delta S^\circ\).
Using the Gibbs-Helmholtz equation:
\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)
\( -21.026 = \Delta H^\circ - (500 \times 0.018) \)
\( -21.026 = \Delta H^\circ - 9 \)
\( \Delta H^\circ = -21.026 + 9 = -12.026 \text{ kJ mol}^{-1} \)
Step 4: Calculate Enthalpy of reaction using Enthalpies of Formation.
\( \Delta H^\circ = [2 \times \Delta_f H^\circ(\text{AB})] - [\Delta_f H^\circ(\text{A}_2) + \Delta_f H^\circ(\text{B}_2)] \)
\( -12.026 = [2 \times 32] - [6 + x] \)
\( -12.026 = 64 - 6 - x \)
\( -12.026 = 58 - x \)
\( x = 58 + 12.026 = 70.026 \text{ kJ mol}^{-1} \)
Step 5: Final Result.
Rounding to the nearest integer, we get:
x = 70
Related Theory
This problem is a classic example of how thermodynamics and chemical equilibrium are intertwined in the JEE Main syllabus. It requires the simultaneous application of the Gibbs-Helmholtz equation and the equilibrium constant relationship.
1. Gibbs Free Energy and Equilibrium (\(\Delta G^\circ\) vs K)
The standard Gibbs free energy change (\(\Delta G^\circ\)) tells us about the spontaneity of a reaction when reactants and products are in their standard states. The expression \(\Delta G^\circ = -RT \ln K\) (or \(-2.303 RT \log K\)) links the thermodynamics of the system to the extent of the reaction at equilibrium. A large positive \(K\) results in a negative \(\Delta G^\circ\), indicating the reaction is spontaneous in the forward direction.
2. The Gibbs-Helmholtz Equation
At constant temperature, the relationship is defined as:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
This equation is vital for understanding how enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) compete to determine the spontaneity of a reaction.
- Enthalpy (\(\Delta H\)) represents the heat energy change.
- Entropy (\(\Delta S\)) represents the change in the degree of disorder or randomness.
3. Enthalpy and Entropy of Reaction
For any generic reaction \(aA + bB \rightarrow cC + dD\):
- \(\Delta H^\circ_{rxn} = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})\)
- \(\Delta S^\circ_{rxn} = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})\)
Note: While \(\Delta_f H^\circ\) for elements in their standard state is zero at 298K, at other temperatures like 500K, it may have a non-zero value as seen for \(A_2\) in this problem.
4. Key Exam Strategies for JEE
- Unit Consistency: This is the biggest trap. Entropy (\(S\)) is almost always given in Joules (J), while Enthalpy (\(H\)) and Gibbs Free Energy (\(G\)) are in kiloJoules (kJ). You must convert them to the same unit before adding or subtracting.
- Temperature: Always use temperature in Kelvin (K).
- Stoichiometric Coefficients: Don't forget to multiply the molar values by the coefficients in the balanced chemical equation (e.g., \(2 \times \Delta_f H^\circ(\text{AB})\)).
5. Relation to Other Concepts
This topic bridges Chemical Thermodynamics and Chemical Equilibrium. Understanding these relationships is also essential for Electrochemistry, where \(\Delta G^\circ = -nFE^\circ_{cell}\), further connecting thermodynamics to cell potential.
Frequently Asked Questions (FAQs)
1. Why is ΔfH° of A₂(g) given as 6 kJ/mol instead of 0?
Enthalpy of formation for elements is zero only at 298 K (standard reference). At 500 K, the enthalpy changes due to heat capacity (\(C_p\)).
2. What is the difference between ΔG and ΔG°?
ΔG° is at standard conditions (1 bar, specific T), while ΔG is at any arbitrary condition. At equilibrium, ΔG = 0, but ΔG° is not necessarily zero.
3. How do I choose between 22.4 L and 22.7 L for molar volume?
Use 22.4 L for 1 atm (older STP) and 22.7 L for 1 bar (modern STP). Usually, JEE Main specifies the values or uses 22.4 L.
4. What does a negative ΔG° signify?
It means the equilibrium constant K > 1, and the products are favored at equilibrium.
5. Can Entropy be negative?
Absolute entropy (\(S^\circ\)) is always positive (Third Law of Thermodynamics), but the change in entropy (\(\Delta S^\circ\)) can be negative.