### 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

### Related Formula For any state function like Internal Energy (U$U$), the cyclic integral over a complete closed loop is identically zero: oint dU = 0 implies Delta U_textcyclic = 0$\oint dU = 0 \implies \Delta U_{\text{cyclic}} = 0$ ### Core Logic Internal energy (U$U$) depends only on the initial and final states of the thermodynamic system, not on the path followed. In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state A$A$. ### Step 1: Final Evaluation Since every transformation begins and ends at point A$A$: Delta U_textCase-I = 0$\Delta U_{\text{Case-I}} = 0$ Delta U_textCase-II = 0$\Delta U_{\text{Case-II}} = 0$ Delta U_textCase-III = 0$\Delta U_{\text{Case-III}} = 0$ Therefore, Delta Utext (Case-I) = Delta Utext (Case-II) = Delta Utext (Case-III)$\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$. ### Pattern Recognition Do not waste time calculating path areas or values if the question asks for a state function change (Delta U$\Delta U$, Delta H$\Delta H$, Delta S$\Delta S$, Delta G$\Delta G$) over a cyclic loop. The answer is instantly zero for all cases! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants)$\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{Products}) - \sum \Delta H_f(\text{Reactants})$ ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6$C_6H_6$): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l)$C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$ Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol$\Delta H_c = -3267.0\mathrm{\ kJ/mol}$ - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol$\Delta H_f[CO_2] = -393.5\mathrm{\ kJ/mol}$ - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol$\Delta H_f[H_2O] = -286.0\mathrm{\ kJ/mol}$ - Delta H_f[O_2] = 0mathrm\ kJ/mol$\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$ ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6]$-3267 = [6(-393.5) + 3(-286.0)] - \Delta H_f[C_6H_6]$ -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6]$-3267 = [-2361.0 - 858.0] - \Delta H_f[C_6H_6]$ -3267 = -3219.0 - Delta H_f[C_6H_6]$-3267 = -3219.0 - \Delta H_f[C_6H_6]$ Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol$\Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48\mathrm{\ kJ/mol}$ ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2$CO_2$ by 6$6$ and H_2O$H_2O$ by 3$3$) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula q_p = n cdot C_p cdot Delta T$q_p = n \cdot C_p \cdot \Delta T$ fracDelta HDelta U = fracC_pC_v$\frac{\Delta H}{\Delta U} = \frac{C_p}{C_v}$ ### Core Logic Argon is a monoatomic gas, hence: C_v = frac32R, quad C_p = frac52R$C_v = \frac{3}{2}R, \quad C_p = \frac{5}{2}R$ Given heat supply happens at constant atmospheric pressure parameter conditions (1.00text atm$1.00\text{ atm}$), so q = q_p = Delta H = 500text J$q = q_p = \Delta H = 500\text{ J}$ . Step 1: Compute Final Temperature 500 = 0.5 cdot left(frac52 cdot 8.3 ight) cdot (T_f - 298)$500 = 0.5 \cdot \left(\frac{5}{2} \cdot 8.3 ight) \cdot (T_f - 298)$ 500 = 1.25 cdot 8.3 cdot (T_f - 298) implies 500 = 10.375 cdot (T_f - 298)$500 = 1.25 \cdot 8.3 \cdot (T_f - 298) \implies 500 = 10.375 \cdot (T_f - 298)$ T_f - 298 = frac50010.375 simeq 48.2 implies T_f simeq 346.2mathrmK ightarrow 348mathrmK$T_f - 298 = \frac{500}{10.375} \simeq 48.2 \implies T_f \simeq 346.2\mathrm{K} ightarrow 348\mathrm{K}$ Step 2: Compute Change in Internal Energy Delta U = n cdot C_v cdot Delta T$\Delta U = n \cdot C_v \cdot \Delta T$ Alternatively, using the ratio : Delta U = fracC_vC_p cdot Delta H = frac35 cdot 500 = 300mathrmJ$\Delta U = \frac{C_v}{C_p} \cdot \Delta H = \frac{3}{5} \cdot 500 = 300\mathrm{J}$ This maps perfectly to option (1). ### Pattern Recognition For monoatomic ideal gases under constant external pressure systems, exactly 60\%$60\%$ of the net enthalpy transfer (\% Delta H$\% \Delta H$) goes towards internal kinetic velocity changes (Delta U$\Delta U$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Core Logic Free expansion means expansion against a vacuum (P_ext = 0$P_{ext} = 0$). Work done: w = -P_ext Delta V$w = -P_{ext} \Delta V$. Since P_ext = 0$P_{ext} = 0$, w = 0$w = 0$. Adiabatic condition means there is no heat exchange with the surroundings. Heat transfer: q = 0$q = 0$. According to the First Law of Thermodynamics, Delta U = q + w$\Delta U = q + w$. Since q = 0$q = 0$ and w = 0$w = 0$, the change in internal energy Delta U = 0$\Delta U = 0$. For an ideal gas, internal energy is a function of temperature only (Delta U = nC_vDelta T$\Delta U = nC_v\Delta T$). If Delta U = 0$\Delta U = 0$, then Delta T = 0$\Delta T = 0$. ### Step 1: Final Parameter Check Evaluating all parameters simultaneously: q = 0$q = 0$ w = 0$w = 0$ Delta T = 0$\Delta T = 0$ ### Pattern Recognition Adiabatic + Free Expansion of IDEAL gas rightarrow$\rightarrow$ Nothing changes thermodynamically except volume and pressure. q = 0, w = 0, Delta U = 0, Delta T = 0, Delta H = 0$q = 0, w = 0, \Delta U = 0, \Delta T = 0, \Delta H = 0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics