### Related Formula Delta H_textsol = textLattice Energy (L.E.) + Delta H_texthyd(textCation) + Delta H_texthyd(textAnion) $\Delta H_{\text{sol}} = \text{Lattice Energy (L.E.)} + \Delta H_{\text{hyd}}(\text{Cation}) + \Delta H_{\text{hyd}}(\text{Anion}) $ ### Core Logic According to Hess's Law, the dissolution process can be mapped as follows:

Given parameters: - Lattice Energy of textKCl$\text{KCl}$ breaking into gaseous ions = -(-textz) = textztext kJ/mol$= -(-\text{z}) = \text{z}\text{ kJ/mol}$ (since lattice energy released on formation is given as -textz$-\text{z}$). - Hydration energy of textK^+ = -textxtext kJ/mol$\text{K}^+ = -\text{x}\text{ kJ/mol}$ - Hydration energy of textCl^- = -textytext kJ/mol$\text{Cl}^- = -\text{y}\text{ kJ/mol}$ ### Step 1: Computation Substituting the values into the governing formulation: Delta H_textsol = textz + (-textx) + (-texty) $\Delta H_{\text{sol}} = \text{z} + (-\text{x}) + (-\text{y}) $ Delta H_textsol = textz - textx - texty = textz - (textx + texty) $\Delta H_{\text{sol}} = \text{z} - \text{x} - \text{y} = \text{z} - (\text{x} + \text{y}) $ ### Pattern Recognition To dissolve an ionic crystal, energy equal to the lattice energy must be supplied (endothermic step, +textz$+\text{z}$), and hydration releases energy (exothermic steps, -textx$-\text{x}$ and -texty$-\text{y}$). Net heat of solution is simply the sum of these parts: textz - textx - texty$\text{z} - \text{x} - \text{y}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Related Formula DeltatextfH^theta = 0 quad textfor an element in its reference/most stable standard state $\Delta{\text{f}}H^{\theta} = 0 \quad \text{for an element in its reference/most stable standard state} $ ### Core Logic - Standard state conditions prescribe a pressure of 1text bar$1\text{ \bar}$. Temperature is not fixed by definition but is explicitly specified (often reference tables use 298.15text K$298.15\text{ K}$). - Oxygen naturally and stably exists as diatomic gas molecules (textO_2(g)$\text{O}_2(g)$) at standard thresholds. - The enthalpy of formation of an element in its reference elemental state is identically zero at any reference temperature: DeltatextfH_500^theta[textO2(g)] = 0 $\Delta{\text{f}}H_{500}^{\theta}[\text{O}2(g)] = 0 $ Conversely, atomic oxygen gas (textO(g)$\text{O}(g)$) is not the reference phase, so its formation enthalpy is non-zero. ### Step 1: Verification of Options Statement (4) accurately aligns with thermodynamic core definitions, while statement (1) and (2) mistakenly conflate standard ambient reference states with STP conditions (273.15text K, 1text atm$273.15\text{ K}, 1\text{ atm}$). ### Pattern Recognition Standard state definitions checklist: Pressure = 1text bar$= 1\text{ \bar}$. Temperature is variable/assigned independently. Elements in their most stable natural form take DeltatextfH^theta = 0$\Delta{\text{f}}H^{\theta} = 0$ at all thermal profiles. ### 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 An endothermic profile specifies that Delta H > 0$\Delta H > 0$. For the system to become spontaneous (Delta G < 0$\Delta G < 0$) specifically when shifting to higher temperatures (T$T$), the temperature-dependent entropic subtraction term (-TDelta S$-T\Delta S$) must outweigh the enthalpic barrier. This transition demands a positive structural entropy step, i.e., Delta S > 0$\Delta S > 0$. Hence, both Delta H$\Delta H$ and Delta S$\Delta S$ are positive. ### Pattern Recognition Spontaneity driven purely by elevated thermal thresholds mandates matching positive signs for enthalpy and entropy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula Delta S_textrxn^0 = sum S_textproducts^0 - sum S_textreactants^0 quad textand quad T = fracDelta H^0Delta S^0 quad textat equilibrium (Delta G^0 = 0text)$\Delta S_{\text{rxn}}^{0} = \sum S_{\text{products}}^{0} - \sum S_{\text{reactants}}^{0} \quad \text{and} \quad T = \frac{\Delta H^0}{\Delta S^0} \quad \text{at equilibrium (}\Delta G^0 = 0\text{)}$ ### Core Logic First, calculate the standard entropy change for the reaction system (Delta S_textrxn^0$\Delta S_{\text{rxn}}^{0}$): Delta S_textrxn^0 = S^0(XY_5) - left[ frac12S^0(X_2) + frac52S^0(Y_2) right]$\Delta S_{\text{rxn}}^{0} = S^{0}(XY_5) - \left[ \frac{1}{2}S^{0}(X_2) + \frac{5}{2}S^{0}(Y_2) \right]$ Delta S_textrxn^0 = 110 - left[ left(frac12 times 70right) + left(frac52 times 50right) right] = 110 - [35 + 125]$\Delta S_{\text{rxn}}^{0} = 110 - \left[ \left(\frac{1}{2} \times 70\right) + \left(\frac{5}{2} \times 50\right) \right] = 110 - [35 + 125]$ Delta S_textrxn^0 = 110 - 160 = -50text J K^-1text mol^-1$\Delta S_{\text{rxn}}^{0} = 110 - 160 = -50\text{ J K}^{-1}\text{ mol}^{-1}$ At thermodynamic equilibrium, the change in Gibbs free energy drops to zero (Delta G^0 = 0$\Delta G^0 = 0$): 0 = Delta H^0 - TDelta S^0 implies T = fracDelta H^0Delta S^0$0 = \Delta H^0 - T\Delta S^0 \implies T = \frac{\Delta H^0}{\Delta S^0}$ Convert the enthalpy value into Joules (Delta H^0 = -35 times 10^3text J/mol$\Delta H^0 = -35 \times 10^3\text{ J/mol}$) and substitute the parameters: T = frac-35000text J mol^-1-50text J K^-1text mol^-1 = 700text Kelvin$T = \frac{-35000\text{ J mol}^{-1}}{-50\text{ J K}^{-1}\text{ mol}^{-1}} = 700\text{ Kelvin}$ ### Pattern Recognition Ensure all variables use matching energy units (Joules vs. Kilojoules) before setting up your final division step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics