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 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

### Related Formula Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ w = -n R T lnleft(fracV_2V_1right)$$w = -n R T \ln\left(\frac{V_2}{V_1}\right)$$ Delta U = q + w$$\Delta U = q + w$$ ### Core Logic Since the expansion step is strictly **isothermal** (Delta T = 0$\Delta T = 0$): Delta U = 0$\Delta U = 0$ Now compute the work command parameter w$w$: w = -n R T lnleft(fracV_2V_1right) = -1 cdot 8.3 cdot 300 cdot lnleft(frac2010 ight)$$w = -n R T \ln\left(\frac{V_2}{V_1}\right) = -1 \cdot 8.3 \cdot 300 \cdot \ln\left(\frac{20}{10} ight)$$ w = -2490 cdot ln(2) = -2490 cdot (2.3 cdot log 2)$$w = -2490 \cdot \ln(2) = -2490 \cdot (2.3 \cdot \log 2)$$ w = -2490 cdot (2.3 cdot 0.30) = -2490 cdot 0.69 = -1718.1mathrm~J = -1.718mathrm~kJ$$w = -2490 \cdot (2.3 \cdot 0.30) = -2490 \cdot 0.69 = -1718.1\mathrm{~J} = -1.718\mathrm{~kJ}$$ Applying the first law equation constraint: q = -w = +1.718mathrm~kJ$$q = -w = +1.718\mathrm{~kJ}$$ Hence, Delta U = 0$\Delta U = 0$, q = 1.718mathrm~kJ$q = 1.718\mathrm{~kJ}$, w = -1.718mathrm~kJ$w = -1.718\mathrm{~kJ}$. ### Pattern Recognition Isothermal expansion of an ideal gas ALWAYS yields Delta U = 0$\Delta U = 0$. Work is negative (done by system) and heat exchange q$q$ matches work magnitude inversely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### 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