Match the LIST-I with LIST-II

List-I (Thermodynamic Process)	List-II (Magnitude in kJ)
A. Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2text dm^3$2\text{ dm}^{3}$ to 20text dm^3$20\text{ dm}^{3}$ at 300 K.	I. 4
B. Work done in irreversible isothermal expansion of 1 mol ideal gas from 1text m^3$1\text{ m}^{3}$ to 3text m^3$3\text{ m}^{3}$ at 300 K against a constant pressure of 3kPa.	II. 11.5
C. Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and overlineC_v = frac32R$\overline{C}_{v} = \frac{3}{2}R$	III. 6
D. Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and overlineC_p = frac52R$\overline{C}_{p} = \frac{5}{2}R$	IV. 7

Choose the correct answer from the option given below:

### Core Logic The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis. Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work. Graph 2: Isochoric drop (vertical line at V=22.4L$V=22.4L$). Area = 0, so work done is zero. Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape. Graph 4: Direct expansion from V=22.4$V=22.4$ to V=44.8$V=44.8$ at pressure P=1$P=1$ up to P=2$P=2$ (a rectangle combined with a triangle). The total area under the upper curve from V=22.4$V=22.4$ to V=44.8$V=44.8$ covers the entire shaded region under the path down to the V-axis. Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Core Logic Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.

A. Standard free energy change (Delta_r G^circ$\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Delta_r G^circ = G_B^circ - G_A^circ$\Delta_r G^{\circ} = G_B^{\circ} - G_A^{\circ}$. Since B is higher than A, Delta_r G^circ$\Delta_r G^{\circ}$ is positive, not -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$. Statement A is false. B. Even if Delta_r G^circ$\Delta_r G^{\circ}$ is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false. C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false. D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is 0.84text kJ mol^-1$0.84\text{ kJ mol}^{-1}$. Thus Delta G = -0.84text kJ mol^-1$\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true. E. The difference from pure products (point B, equivalent to 2 moles NO_2$NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is 5.40$5.40$, and A to E is 0.84$0.84$. So drop from B to E is - (5.40 + 0.84) = -6.24text kJ mol^-1$- (5.40 + 0.84) = -6.24\text{ kJ mol}^{-1}$. Statement E is true. ### Step 1: Final Conclusion Only statements D and E are correct. ### Pattern Recognition The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Related Formula Delta_r H^circ = sum Delta_f H^circ(textproducts) - sum Delta_f H^circ(textreactants)$$\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$$ Delta_r S^circ = sum S^circ(textproducts) - sum S^circ(textreactants)$$\Delta_r S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$$ Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$ Delta_r G^circ = -2.303 RT log K$$\Delta_r G^{\circ} = -2.303 RT \log K$$ ### Core Logic For the reaction A_2(g) + B_2(g) rightleftharpoons 2AB(g)$A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$: 1. Enthalpy change (Delta_r H^circ$\Delta_r H^{\circ}$): Delta_r H^circ = [2 times Delta_f H^circ(AB)] - [Delta_f H^circ(A_2) + Delta_f H^circ(B_2)]$$\Delta_r H^{\circ} = [2 \times \Delta_f H^{\circ}(AB)] - [\Delta_f H^{\circ}(A_2) + \Delta_f H^{\circ}(B_2)]$$ Delta_r H^circ = (2 times 32) - (6 + x) = (64 - 6 - x) = (58 - x)text kJ mol^-1$$\Delta_r H^{\circ} = (2 \times 32) - (6 + x) = (64 - 6 - x) = (58 - x)\text{ kJ mol}^{-1}$$ 2. Entropy change (Delta_r S^circ$\Delta_r S^{\circ}$): Delta_r S^circ = [2 times S^circ(AB)] - [S^circ(A_2) + S^circ(B_2)]$$\Delta_r S^{\circ} = [2 \times S^{\circ}(AB)] - [S^{\circ}(A_2) + S^{\circ}(B_2)]$$ Delta_r S^circ = (2 times 222) - (146 + 280) = 444 - 426 = 18text J K^-1text mol^-1$$\Delta_r S^{\circ} = (2 \times 222) - (146 + 280) = 444 - 426 = 18\text{ J K}^{-1}\text{ mol}^{-1}$$ 3. Standard Gibbs Free Energy (Delta_r G^circ$\Delta_r G^{\circ}$) via Equilibrium Constant: Delta_r G^circ = -2.303 times R times T log K$$\Delta_r G^{\circ} = -2.303 \times R \times T \log K$$ Delta_r G^circ = -2.303 times 8.314 times 500 times 2.2 = -21063.8text J mol^-1 = -21.06text kJ mol^-1$$\Delta_r G^{\circ} = -2.303 \times 8.314 \times 500 \times 2.2 = -21063.8\text{ J mol}^{-1} = -21.06\text{ kJ mol}^{-1}$$ 4. Using Gibbs Equation: Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$ -21.06 = (58 - x) - 500 times left(frac181000right)$$-21.06 = (58 - x) - 500 \times \left(\frac{18}{1000}\right)$$ -21.06 = 58 - x - 9$$-21.06 = 58 - x - 9$$ -21.06 = 49 - x$-21.06 = 49 - x$ x = 49 + 21.06 = 70.06text kJ mol^-1$$x = 49 + 21.06 = 70.06\text{ kJ mol}^{-1}$$ Rounding off to nearest integer, x = 70$x = 70$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Core Logic Reaction for formation of methane: textC(texts) + 2textH_2(textg) rightarrow textCH_4(textg)$$\text{C}(\text{s}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g})$$ -X = (Delta H_textsub text of carbon) + 2 times (textB.E. of H-H) - 4 times (textB.E. of C-H)$-X = (\Delta H_{\text{sub}} \text{ of carbon}) + 2 \times (\text{B.E. of H-H}) - 4 \times (\text{B.E. of C-H})$ -X = Y + 2Z - 4(textB.E. of C-H)$-X = Y + 2Z - 4(\text{B.E. of C-H})$ ### Step 1: Rearranging for Bond Enthalpy Rearranging the equation for C-H bond enthalpy: textB.E. of C-H = fracX + Y + 2Z4$$\text{B.E. of C-H} = \frac{X + Y + 2Z}{4}$$ ### Pattern Recognition Sees: bond enthalpy derivation from heat of formation, sublimation, and dissociation. Trap: Sign convention errors when substituting formation enthalpies. ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics