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)

### Core Logic Let's treat the given parameters as terms for exothermic heats evolved on the product side: 1) C(textdiamond) ightarrow C(textgraphite)$C(\text{diamond}) ightarrow C(\text{graphite})$, Delta H = -X$Delta H = -X$ 2) C(textdiamond) + O_2(g) ightarrow CO_2(g)$C(\text{diamond}) + O_{2}(g) ightarrow CO_{2}(g)$, Delta H = -Y$Delta H = -Y$ 3) C(textgraphite) + O_2(g) ightarrow CO_2(g)$C(\text{graphite}) + O_{2}(g) ightarrow CO_{2}(g)$, Delta H = -Z$Delta H = -Z$ By subtracting Equation (3) from Equation (2): C(textdiamond) - C(textgraphite) ightarrow 0 implies C(textdiamond) ightarrow C(textgraphite)$$C(\text{diamond}) - C(\text{graphite}) ightarrow 0 implies C(\text{diamond}) ightarrow C(\text{graphite})$$ Delta H = (-Y) - (-Z) = Z - Y$$Delta H = (-Y) - (-Z) = Z - Y$$ Matching this to Equation (1): -X = Z - Y implies X = Y - Z$$-X = Z - Y implies X = Y - Z$$ ### Pattern Recognition Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Core Logic Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water: V_m(textice) > V_m(textwater)$$V_m(\text{ice}) > V_m(\text{water})$$ According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water:

If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15mathrmK$273.15\mathrm{K}$, the melting point decreases, causing the entire solid phase (ice) to disappear completely. ### Pattern Recognition Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### 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