### Related Formula Reaction enthalpy based on atomization processes: Delta_r H = sum Delta_f H(textproducts) - sum Delta_f H(textreactants)$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$ ### Step 1: Map the Dissociation Reaction Consider the dissociation of gas phase water molecules into constituent gaseous atoms: mathrmH_2mathrmO_mathrm(g) rightarrow 2mathrmH_mathrm(g) + mathrmO_mathrm(g)$\mathrm{H}_2\mathrm{O}_{\mathrm{(g)}} \rightarrow 2\mathrm{H}_{\mathrm{(g)}} + \mathrm{O}_{\mathrm{(g)}}$ The total energy required corresponds to breaking exactly two mathrmO-mathrmH$\mathrm{O}-\mathrm{H}$ bonds: Delta_r H = 2 times textB.E.(mathrmO-mathrmH)$\Delta_r H = 2 \times \text{B.E.}(\mathrm{O}-\mathrm{H})$ ### Step 2: Calculate Delta_r H$\Delta_r H$ Using the enthalpies of formation: Delta_r H = [2 times Delta_f H(mathrmH_mathrm(g)) + Delta_f H(mathrmO_mathrm(g))] - Delta_f H(mathrmH_2mathrmO_mathrm(g))$\Delta_r H = [2 \times \Delta_f H(\mathrm{H}_{\mathrm{(g)}}) + \Delta_f H(\mathrm{O}_{\mathrm{(g)}})] - \Delta_f H(\mathrm{H}_2\mathrm{O}_{\mathrm{(g)}})$ Delta_r H = [2 times 220.0 + 250.0] - (-242.0)$\Delta_r H = [2 \times 220.0 + 250.0] - (-242.0)$ Delta_r H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0\,mathrmkJ\,mol^-1$\Delta_r H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0\,\mathrm{kJ\,mol}^{-1}$ ### Step 3: Solve for Single Bond Enthalpy 2 times textB.E.(mathrmO-mathrmH) = 932.0$2 \times \text{B.E.}(\mathrm{O}-\mathrm{H}) = 932.0$ textB.E.(mathrmO-mathrmH) = frac932.02 = 466\,mathrmkJ\,mol^-1$\text{B.E.}(\mathrm{O}-\mathrm{H}) = \frac{932.0}{2} = 466\,\mathrm{kJ\,mol}^{-1}$ ### Pattern Recognition Sees: Atomization state values used to evaluate single bond metrics. Shortcut: Remember textTotal Dissociation Energy = sum Delta_f H(textatoms) - Delta_f H(textmolecule)$\text{Total Dissociation Energy} = \sum \Delta_f H(\text{atoms}) - \Delta_f H(\text{molecule})$. Halving the result gives the average bond enthalpy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Core Logic Let us compute the concentration of both solutions: textConcentration of Solution 1 = frac10text mol10text L = 1text mol/L $ \text{Concentration of Solution 1} = \frac{10\text{ mol}}{10\text{ L}} = 1\text{ mol/L} $ textConcentration of Solution 2 = frac1text mol1text L = 1text mol/L $ \text{Concentration of Solution 2} = \frac{1\text{ mol}}{1\text{ L}} = 1\text{ mol/L} $ Since concentration is identical, both systems share matching compositions. Consequently, all **intensive properties** (independent of mass/size) like concentration, density, and molar heat capacity remain exactly equal. ### Step 1: Identifying the Variable Gibbs free energy (G$G$) is an **extensive property** that scales directly with the amount of matter in the system. Because Solution 1 contains a larger total mass and volume than Solution 2, its overall Gibbs free energy value is different. ### Pattern Recognition Shortcut: Look for the only extensive property in the options. Density, concentration, and molar parameters are always intensive. Gibbs free energy (G$G$) scales with total matter quantity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula The enthalpy of formation reaction can be evaluated by balancing atomization and bond cleavage details: Delta H_f^circ = sum Delta Htextatomization (reactants) - sum Delta Htextbonds broken/formed (products) $\Delta H_f^{\circ} = \sum \Delta H{\text{atomization (reactants)}} - \sum \Delta H{\text{bonds broken/formed (products)}} $ ### Core Logic The target formation reaction for ethylene (textC_2textH_4$\text{C}_2\text{H}_4$) from standard elemental states is: 2C(textgraphite) + 2H_2(g) ightarrow CH_2=CH_2(g) $ 2C(\text{graphite}) + 2H_2(g) ightarrow CH_2=CH_2(g) $ To construct this pathway: 1. Sublime 2 moles of solid graphite: 2 times Delta H_textsub^circ[C]$2 \times \Delta H_{\text{sub}}^{\circ}[C]$ 2. Dissociate 2 moles of gaseous textH-textH$\text{H}-\text{H}$ bonds: 2 times Delta_H-HH^circ$2 \times \Delta_{H-H}H^{\circ}$ 3. Form 1 mole of textC=textC$\text{C}=\text{C}$ double bonds: -1 times Delta_C=CH^circ$-1 \times \Delta_{C=C}H^{\circ}$ 4. Form 4 moles of textC-textH$\text{C}-\text{H}$ single bonds: -4 times Delta_C-HH^circ$-4 \times \Delta_{C-H}H^{\circ}$ ### Step 1: Arithmetic Calculation Delta H_f^circ = [2 times 710] + [2 times 436] - 611 - [4 times 414] $ Delta H_f^{\circ} = [2 \times 710] + [2 \times 436] - 611 - [4 \times 414] $ Delta H_f^circ = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25text kJ mol^-1 $ Delta H_f^{\circ} = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25\text{ kJ mol}^{-1} $ ### Pattern Recognition Shortcut: Group energy components systematically. Reactant state atomization costs +2292text kJ$+2292\text{ kJ}$. Exothermic structural bond formation releases -2267text kJ$-2267\text{ kJ}$. Net difference yields a small endothermic value of +25text kJ/mol$+25\text{ kJ/mol}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula Delta H_textdilution = Delta H_2 - Delta H_1$\Delta H_{\text{dilution}} = \Delta H_2 - \Delta H_1$ ### Core Logic Analyzing the thermodynamic statements: - Delta H$\Delta H$ values are negative, so the dissolution of HCl(g)$HCl(g)$ is clearly exothermic, eliminating option (1). - Since the enthalpy release changes when the moles of water solvent shift from 10 to 40 (-69.01$-69.01$ vs -72.79$-72.79$), the **heat of solution depends explicitly on the amount of solvent** (Statement 2 is true). - Let's check Statement 3: By subtracting equation (a) from (b): mathrmHClcdot10H_2O + 30mathrmH_2mathrmO rightarrow mathrmHClcdot40H_2O$\mathrm{HCl\cdot10H_2O} + 30\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl\cdot40H_2O}$ Delta H = -72.79 - (-69.01) = -3.78 mathrm~kJcdot mol^-1$\Delta H = -72.79 - (-69.01) = -3.78 \mathrm{~kJ\cdot mol^{-1}}$ The value is negative, indicating an exothermic process, so calling it +3.78$+3.78$ makes option (3) incorrect. ### Pattern Recognition The standard integral enthalpy of solution varies with solvent concentration until infinite dilution is achieved. Thus, concentration dependence is a core property of partial molar solution variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula Delta G = Delta H - TDelta S$\Delta G = \Delta H - T\Delta S$ ### Core Logic For a reaction to be spontaneous, the change in Gibbs free energy must be negative: Delta G < 0 implies Delta H - TDelta S < 0$\Delta G < 0 \implies \Delta H - T\Delta S < 0$ Given that both Delta H > 0$\Delta H > 0$ and Delta S > 0$\Delta S > 0$: Delta H < TDelta S implies T > fracDelta HDelta S$\Delta H < T\Delta S \implies T > \frac{\Delta H}{\Delta S}$ At the equilibrium temperature T_e$T_e$, Delta G = 0$\Delta G = 0$, which gives: T_e = fracDelta HDelta S$T_e = \frac{\Delta H}{\Delta S}$ Substituting this back into the inequality reveals that the reaction is spontaneous when: T > T_e$T > T_e$ ### Pattern Recognition When both Delta H$\Delta H$ and Delta S$\Delta S$ are positive, the reaction is entropy-driven and becomes spontaneous only at higher temperatures (T > T_e$T > T_e$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics