A sample of n-octane (1.14mathrm~g$1.14\mathrm{~g}$) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5mathrm~kJ~K^-1$5\mathrm{~kJ~K^{-1}}$. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ________ mathrmkJ~mol^-1$\mathrm{kJ~mol^{-1}}$. (nearest integer)

### Related Formula Q = n_textreacted cdot Delta H_textneutralization$$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$$ Delta T = fracQm cdot c$$\Delta T = \frac{Q}{m \cdot c}$$ ### Core Logic The temperature rise depends directly on the total heat released (Q$Q$) normalized by the total heat capacity of the resulting mixed volume (m cdot c$m \cdot c$). Let's evaluate the millimoles of mathrmH^+$\mathrm{H}^+$ and mathrmOH^-$\mathrm{OH}^-$ that react in each mixture: 1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol$\sim -57.3\text{ kJ/mol}$). Total volume = 60text mL$60\text{ mL}$. 2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1. 3. **Option 3:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = mathrmNaOH = 20text mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only 20text mmol$20\text{ mmol}$ reacts. Total volume = 70text mL$70\text{ mL}$. 4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = 25text mmol$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile. Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol$30\text{ mmol}$ vs 20text mmol$20\text{ mmol}$) into a smaller volume (60text mL$60\text{ mL}$ vs 70text mL$70\text{ mL}$), yielding the largest increase in temperature Delta T$\Delta T$. ### Pattern Recognition To maximize Delta T$\Delta T$, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Related Formula Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps: Delta Htextnet = sum Delta Htextproducts - sum Delta Htextreactants$$\Delta H{\text{net}} = \sum \Delta H{\text{products}} - \sum \Delta H{\text{reactants}}$$ ### Core Logic The heat of formation of mathrmSO_2(g)$\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation: textTarget: mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g) quad Delta H_f = ?$$\text{Target: } \mathrm{S}(g) + \mathrm{O}{2}(g) ightarrow \mathrm{SO}{2}(g) \quad \Delta H_f = ?$$ Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so Delta H = -Q$\Delta H = -Q$): 1. mathrmS(g) + frac32mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_1 = -2xtext kcal$\mathrm{S}(g) + \frac{3}{2}\mathrm{O}_{2}(g) ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{1} = -2x\text{ kcal}$ 2. mathrmSO_2(g) + frac12mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_2 = -ytext kcal$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{2} = -y\text{ kcal}$ To isolate mathrmSO_2(g)$\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1): left[mathrmS(g) + frac32mathrmO2(g) ight] - left[mathrmSO2(g) + frac12mathrmO2(g) ight] ightarrow mathrmSO3(g) - mathrmSO3(g)$$\left[\mathrm{S}(g) + \frac{3}{2}\mathrm{O}{2}(g) ight] - \left[\mathrm{SO}{2}(g) + \frac{1}{2}\mathrm{O}{2}(g) ight] ightarrow \mathrm{SO}{3}(g) - \mathrm{SO}{3}(g)$$ mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g)$$\mathrm{S}(g) + \mathrm{O}{2}(g) ightarrow \mathrm{SO}{2}(g)$$ Now apply the same operation to the enthalpy values: Delta H_f = Delta H1 - Delta H_2 = -2x - (-y) = y - 2xtext kcal$$\Delta H_f = \Delta H{1} - \Delta H_{2} = -2x - (-y) = y - 2x\text{ kcal}$$ This matches Option (2). ### Pattern Recognition To isolate your target species on the desired side of the equation, use Hess's Law to add or subtract the given elemental equations. Make sure to invert the sign of the enthalpy change if you reverse a reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: 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

### 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