Thermodynamics practice · Mixed difficulty

Option D ($x = y + z$) is correct because adding the second and third reactions results in the first reaction, so their enthalpy changes must also add up.

Option B (–111.7 kJ/mol) is correct because it is calculated using Hess's law and accounting for the change in the number of moles of gas.

Option C ($H(g) + {e^ - } \to {H^ - }(g)$) is correct because the formation of $H^-$ from $H$ is exothermic due to high hydration energy.

Option A (113.8 kJ per mol) is correct because it is the result of calculating the Gibbs free energy change using the given enthalpy and entropy changes.

Option B ( - , - , + ) is correct because combustion reactions are exothermic (negative Δ H), spontaneous (negative Δ G), and increase disorder (positive Δ S).

Option A ($\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Cl}) < \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{F})$) is correct because the electron gain enthalpy of chlorine is more negative (higher electron affinity) than that of fluorine due to fluorine's small size and strong interelectronic repulsions.

Option D (More than that of carbohydrate and protein) is correct because fats have a higher calorific value due to their chemical structure and higher proportion of carbon-hydrogen bonds.

Option D is the closest. The heat of formation of H2O(g) will be less negative than H2O(l) because energy is required to vaporize the liquid into gas. Therefore, the heat of formation of H2O(g) will be closer to zero than -68.0 kcal, making -58.3 kcal the most reasonable answer.

Option A (Spontaneous) is correct because the Gibbs free energy change ($\Delta G$) is negative, indicating a spontaneous reaction. $\Delta G = \Delta H - T\Delta S$, and with the given values, $\Delta G$ is negative.

Option C (Isolated system) is correct because a thermos flask is designed to prevent the exchange of both heat and matter with the surroundings, approximating an isolated system.

Option B (${\left( {\frac{{\partial E}}{{\partial P}}} \right)_T}$= 0) is correct because for an ideal gas, internal energy (E) depends only on temperature (T), not on pressure (P).

Option D is correct because neither exothermic reactions are always spontaneous, nor reactions with increased entropy are always spontaneous; spontaneity depends on both enthalpy and entropy.

Option D ($\mathrm{NaOH}\,\mathrm{and}\,\mathrm{HCl}$) is correct because the reaction between a strong acid (HCl) and a strong base (NaOH) releases the maximum amount of heat during neutralization.

Option B ($R/M$) is correct because the difference between molar heat capacity at constant pressure and constant volume ($c_p - c_v$) for an ideal gas is equal to $R/M$, where R is the gas constant and M is the molar mass.

Option C (Negative) is correct because exothermic reactions release heat, resulting in a negative enthalpy change.

The correct answer is C (Electrolysis of water) because electrolysis requires an external energy source to drive the non-spontaneous decomposition of water.

Option A (43.4 J/mol) is correct because it represents the entropy change calculated using the heat exchanged and the temperature at the boiling point of water.

Option C is correct. Statement I is correct because the internal energy of an ideal gas depends only on temperature, and enthalpy is related to internal energy. Statement II is incorrect because the enthalpy of an ideal gas is a function of temperature only, not pressure.

Option D ($1200 \mathrm{~cal}$) is correct because it accurately calculates the work done in an adiabatic expansion using the formula $W = \frac{{nR(T_2 - T_1)}}{{1 - \gamma}}$.

Option C (${Xe}({g})+2 {F}_2({g}) \rightarrow {XeF}_4({g})$) is correct because the formation reaction involves elements in their standard states forming one mole of the product.

Option B (R) is correct because the expression represents the difference between heat capacity at constant pressure and constant volume, which equals the gas constant R for an ideal gas.

Option D (All of the above) is correct because all the given statements are thermodynamically correct and well-established relationships.

Option A (Decrease in Helmholtz energy (∆A)) is correct because the net work done by a system at constant temperature and volume is equal to the decrease in Helmholtz free energy.

Option C is correct because the heat of neutralization of a strong acid and a strong base is approximately -57.3 kJ/mol.

Option C (-333, -349, -325, -296) is correct because electron gain enthalpy generally becomes less negative down the group, but fluorine is an exception due to its small size and strong interelectronic repulsions.

Option D (For a cyclic process=Q=-W) is correct because in a cyclic process, the change in internal energy is zero, so Q = -W according to the first law of thermodynamics.

Option C (One in which Gibbs energy change is negative) is correct because a negative Gibbs free energy change is the direct criterion for spontaneity, regardless of whether the reaction is endothermic or exothermic.

Option C (Third law) is correct because the third law of thermodynamics states that the entropy of a perfect crystal at absolute zero is zero.

Option B ($q_1+q_2+2 q_3$) is correct because the overall heat of reaction is the sum of the heats of each step, considering the stoichiometry.

Option D (Lighting of an electric lamp) is correct because it involves electrical energy conversion, not a chemical reaction involving oxidation.