Thermodynamics practice · Mixed difficulty

Option B (Kerosene) is correct because kerosene has a higher calorific value compared to charcoal, wood, and dung.

Option C ($\mathrm{C}$ graphite $+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})$) is correct because the number of moles of gas increases (positive $\Delta S^\circ$) and increasing temperature favors the formation of products, thus decreasing $\Delta G^\circ$.

Option C (${5^0}C$) is correct because doubling the volume of both acid and base while still achieving complete neutralization results in the same temperature change.

Option A (III > I > II) is correct because the more negative the $\Delta_r G^0$ value, the larger the equilibrium constant K. Therefore, a smaller K corresponds to a less negative, or more positive, $\Delta_r G^0$.

Option C (-252 kJ) is correct because the enthalpy change for the overall reaction can be found by reversing reaction (a) and adding it to reaction (b), then summing the enthalpy changes.

Option D (d ln PdT = \fracΔ H_υ RT²$\frac{{\mathrm{d}\,\mathrm{ln}\,\mathrm{P}}}{{\mathrm{dT}}}\,\mathrm{ = }\,\frac{{\mathrm{\Delta }{\mathrm{H}_{\mathrm{\upsilon }}}}{{\mathrm{R}{\mathrm{T}^{\mathrm{2}}}}}$) is correct because it represents the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization.

Option B ($2.2 \times 10^{-1} \mathrm{~kJ}$) is correct because it represents the change in internal energy calculated using the first law of thermodynamics.

Option B ($\Delta H = - Ve$) is correct because as temperature increases, Kc decreases, indicating an exothermic reaction according to Van't Hoff equation.

Option D ($< \Delta E^\circ$) is correct because for combustion reactions with an increase in the number of moles of gas, $\Delta H^\circ$ is less than $\Delta E^\circ$.

Option D ($-2 R T$) is correct because the change in the number of moles of gas is -2, and $\Delta H - \Delta U = \Delta n_g RT$.

Option B (19.59 kJ) is correct because the Gibbs free energy is calculated using ∆G = ∆H - T∆S.

Option C is correct because an adiabatic process is defined as one where there is no heat exchange between the system and its surroundings, implying perfect heat insulation.

Option A (∆H>0 and ∆S>0) is correct because the reaction is endothermic (bond breaking) and increases disorder (more gas molecules).

Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the assertion and reason are true, but the reason doesn't directly explain why the assertion is true.

Option B (26.0 cal mol⁻¹ K⁻¹) is correct because the entropy of vaporization is calculated by dividing the molar heat of vaporization by the temperature in Kelvin.

Option C (101.325 J) is correct because 1 L atm is equal to 101.325 Joules.

Option D ($A$ and $R$ are true, $R$ explains $A$) is correct because at low temperatures, the -TΔS term is smaller, making ΔG positive unless -TΔS is greater than ΔH.

Option B is correct. $\Delta H = \Delta E + \Delta n_g RT$. $\Delta H = \Delta E$ when $\Delta n_g = 0$. In reaction II, $\Delta n_g = (2+2) - (1+3) = 0$. In reaction I, $\Delta n_g = 1 - (1+2) = -2$.

The value of $\Delta S_{\text {vaporization}}$ for chloroform is 87.3 J/mol-K, calculated using the formula $\Delta S = \frac{\Delta H}{T}$.

Option A (Increases) is correct because the negative enthalpy of solution indicates that the dissolution process is exothermic, releasing heat and increasing the water's temperature.

Option A (400 K) is correct because the temperature at equilibrium is calculated using T = ∆H/∆S.

The correct answer is C ((i) is exothermic and (ii) is endothermic) because reaction (i) has a negative ∆H (exothermic), while reaction (ii) has a positive ∆H (endothermic).

Option C (Isobaric) is correct because an isobaric process is defined as a thermodynamic process that occurs at constant pressure.

Option D (-1412 kJ/mol) is correct because the enthalpy of combustion is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products, considering stoichiometric coefficients.

Option A ($\frac{3 y-4 x}{3}$) is correct because by calculating the bond energies and using the given information, the P-P bond energy can be derived as (3y-4x)/3.

Option C (${\mathrm{H}_2}\mathrm{O}\left( l \right) \to {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)$) is correct because the conversion of liquid water to gaseous water (steam) increases disorder, resulting in a positive change in entropy.

Option B ($C(graphite) + 2H_2(g) \to CH_4(g)$) is correct because the standard heat of formation uses elements in their standard states (graphite for carbon and diatomic hydrogen gas).

Option B is correct because the enthalpy of a compound in its standard state is defined as its heat of formation.

Option A (62 kJ) is correct because it is calculated by integrating the heat capacity equation over the given temperature range and multiplying by the number of moles.

Option A (${\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } = {\mathrm{H}_2}\mathrm{O}$) is correct because the neutralization of a strong acid and strong base involves the combination of H+ and OH- ions to form water.