The formation enthalpies, Delta mathrmH_mathrmf^ominus$\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for mathrmH_(mathrmg)$\mathrm{H}_{(\mathrm{g})}$ and mathrmO_(mathrmg)$\mathrm{O}_{(\mathrm{g})}$ are 220.0 and 250.0~mathrmkJ~mol^-1$250.0~\mathrm{kJ~mol}^{-1}$ , respectively, at 298.15mathrmK$298.15\mathrm{K}$ , and Delta mathrmH_mathrmf^-$\Delta \mathrm{H}_{\mathrm{f}}^{-}$ for mathrmH_2mathrmO_(mathrmg)$\mathrm{H}_2\mathrm{O}_{(\mathrm{g})}$ is -242.0mathrmkJ\,mol^-1$-242.0\mathrm{kJ\,mol}^{-1}$ at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15mathrmK$298.15\mathrm{K}$ is \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (nearest integer).$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (nearest integer).$

### Core Logic According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Delta G$\Delta G$) dictates the spontaneity of a process. - If Delta G lt 0$\Delta G \lt 0$ (negative), the process is spontaneous. - If Delta G gt 0$\Delta G \gt 0$ (positive), the process is non-spontaneous. - If Delta G = 0$\Delta G = 0$, the system is in equilibrium (reversible process). Therefore, the statement "Delta G$\Delta G$ is positive for a spontaneous reaction" is factually incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided. ### Core Logic Let the given reactions be: (1) 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), quad Delta H_1 = -822 \, mathrmkJ/mol$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \quad \Delta H_1 = -822 \, \mathrm{kJ/mol}$ (2) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \quad \Delta H_2 = -110 \, \mathrm{kJ/mol}$ Target Reaction (3): 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ?$$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}, \quad \Delta H_3 = ?$$ To construct the target reaction: - We need 3 mathrmCO_(g)$3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3. - We need mathrmFe_2mathrmO_3(s)$\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and 2 mathrmFe_(s)$2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1). ### Step 1: Calculate Net Enthalpy Target Reaction (3) = 3 times (2) - (1)$3 \times (2) - (1)$ Delta H_3 = 3 times Delta H_2 - Delta H_1$$\Delta H_3 = 3 \times \Delta H_2 - \Delta H_1$$ Delta H_3 = 3(-110) - (-822)$$\Delta H_3 = 3(-110) - (-822)$$ Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol$$\Delta H_3 = -330 + 822 = 492 \, \mathrm{kJ/mol}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula W_textcyclic = textArea enclosed in P-V graph$$W_{\text{cyclic}} = \text{Area enclosed in P-V graph}$$ ### Core Logic The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume (V$V$) on the y-axis versus Pressure (P$P$) on the x-axis. The path A rightarrow B rightarrow C rightarrow A$A \rightarrow B \rightarrow C \rightarrow A$ is traced in a clockwise direction on the V$V$-P$P$ graph. Clockwise on a V$V$-P$P$ graph corresponds to anti-clockwise on a standard P$P$-V$V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for). ### Step 1: Calculating Area The enclosed region is a right-angled triangle. Base of triangle on P-axis = 30 - 10 = 20 text kPa$30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = 30 - 10 = 20 text dm^3$30 - 10 = 20 \text{ dm}^3$ textArea = frac12 times textbase times textheight$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3$$\text{Area} = \frac{1}{2} \times 20 \times 20 = 200 \text{ kPa}\cdot\text{dm}^3$$ ### Step 2: Unit conversion 1 text kPa = 10^3 text Pa$1 \text{ kPa} = 10^3 \text{ Pa}$ 1 text dm^3 = 1 text Litre = 10^-3 text m^3$1 \text{ dm}^3 = 1 \text{ Litre} = 10^{-3} \text{ m}^3$ W = 200 times 10^3 text Pa times 10^-3 text m^3$$W = 200 \times 10^3 \text{ Pa} \times 10^{-3} \text{ m}^3$$ W = 200 text J$$W = 200 \text{ J}$$ ### Pattern Recognition 1 text kPa cdot 1 text L = 1 text Joule$1 \text{ kPa} \cdot 1 \text{ L} = 1 \text{ Joule}$. This direct conversion saves time without converting explicitly to standard SI units (Pa$Pa$ and m^3$m^3$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula W = -2.303 \, nRT log left( fracV_2V_1 right)$$W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right)$$ ### Core Logic For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5text moles$n = 5\text{ moles}$ R = 8.314text J K^-1textmol^-1$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$ T = 300text K$T = 300\text{ K}$ V_1 = 10text L$V_1 = 10\text{ L}$ V_2 = 100text L$V_2 = 100\text{ L}$ ### Step 1: Calculating Work Done W = -2.303 times 5 times 8.314 times 300 times logleft( frac10010 right)$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log\left( \frac{100}{10} \right)$$ W = -2.303 times 5 times 8.314 times 300 times log(10)$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$$ W = -2.303 times 12471 times 1$$W = -2.303 \times 12471 \times 1$$ W = -28720.713text J$$W = -28720.713\text{ J}$$ ### Step 2: Final Formatting The question asks for work w = -xtext J$w = -x\text{ J}$. So x = 28720.713$x = 28720.713$, which rounds to 28721$28721$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula Delta_rG^ominus = -RT ln K_p$$\Delta_rG^{\ominus} = -RT \ln K_p$$ ### Step 1: Calculation Delta_rG^ominus = -8.314 times 10^-3 text kJ K^-1 mol^-1 times 298 text K times ln(2.47 times 10^-29)$$\Delta_rG^{\ominus} = -8.314 \times 10^{-3} \text{ kJ } K^{-1} mol^{-1} \times 298 \text{ K} \times \ln(2.47 \times 10^{-29})$$ = -8.314 times 10^-3 times 298 times (-65.87)$$= -8.314 \times 10^{-3} \times 298 \times (-65.87)$$ = 163.19 text kJ$$= 163.19 \text{ kJ}$$ ### Step 2: Nearest Integer Rounding 163.19$163.19$ to the nearest integer gives 163$163$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics