Chemical Thermodynamics MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Converting reaction internal energy into enthalpy:}\) The supplied reaction produces gaseous water. For an ideal-gas reaction at a common temperature, \[ \Delta_rH=\Delta_rU+\Delta n_gRT \] \(\textbf{Counting gaseous stoichiometric moles:}\) The reactant side contains two moles of hydrogen and one mole of oxygen. The product side contains two moles of water vapour. Therefore, \[ \Delta n_g=2-(2+1) \] \[ \Delta n_g=-1 \] \(\textbf{Evaluating the gaseous mole correction:}\) \[ \Delta n_gRT=(-1)(8.314)(300) \] \[ \Delta n_gRT=-2494.2\,\mathrm{J} \] \[ \Delta n_gRT=-2.4942\,\mathrm{kJ} \] \(\textbf{Reaction enthalpy for gaseous water formation:}\) The supplied internal-energy change is \[ \Delta_rU=-481.506\,\mathrm{kJ} \] Therefore, \[ \Delta_rH=-481.506-2.4942 \] \[ \Delta_rH=-484.0002\,\mathrm{kJ} \] Within the stated precision, \[ \Delta_rH=-484.0\,\mathrm{kJ} \] \(\textbf{Normalising to one mole of gaseous water:}\) The balanced reaction produces two moles of gaseous water. Therefore, \[ \Delta_fH[\mathrm{H_2O(g)}]=\frac{-484.0}{2} \] \[ \Delta_fH[\mathrm{H_2O(g)}]=-242.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Correcting the physical state of water:}\) The target formation enthalpy refers to liquid water. The supplied vaporisation process is \[ \mathrm{H_2O(l)\rightarrow H_2O(g)} \] with \[ \Delta_{\mathrm{vap}}H=+44.0\,\mathrm{kJ\,mol^{-1}} \] The reverse condensation process has enthalpy \[ \Delta_{\mathrm{cond}}H=-44.0\,\mathrm{kJ\,mol^{-1}} \] Therefore, the formation of liquid water can be obtained by forming gaseous water and then condensing it. \[ \Delta_fH[\mathrm{H_2O(l)}]=\Delta_fH[\mathrm{H_2O(g)}]+\Delta_{\mathrm{cond}}H \] Substituting, \[ \Delta_fH[\mathrm{H_2O(l)}]=-242.0-44.0 \] \[ \Delta_fH[\mathrm{H_2O(l)}]=-286.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Thermochemical verification:}\) The internal-energy-to-enthalpy correction must be applied to the balanced gas-phase reaction before changing the water product state. The reaction enthalpy must then be divided by two because a standard molar formation enthalpy refers to one mole of product. Condensation provides the final physical-state correction. The negative result indicates that formation of liquid water from hydrogen and oxygen in their elemental reference states is exothermic. \(\textbf{Final answer:}\) \[ \boxed{\Delta_fH^\circ[\mathrm{H_2O(l)}]=-286.0\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Using the common gaseous endpoint:}\) Both allotropes form the same gaseous molecular species X. Their sublimation enthalpies can therefore be compared using Hess's law. The mixture contains 0.50 mol of solid material in total. Let x be the initial amount of allotrope P. The amount of Q is consequently \[ n_Q=0.50-x \] \(\textbf{Constructing the total sublimation heat:}\) Sublimation of P requires 90 kJ per mole. Sublimation of Q requires 80 kJ per mole. Therefore, \[ q_P=90x \] \[ q_Q=80(0.50-x) \] The combined heat absorption is 43.0 kJ. Thus, \[ 90x+80(0.50-x)=43.0 \] Expanding, \[ 90x+40.0-80x=43.0 \] \[ 10x=3.0 \] \[ x=0.30\,\mathrm{mol} \] The remaining amount is \[ n_Q=0.50-0.30 \] \[ n_Q=0.20\,\mathrm{mol} \] \(\textbf{Determining the allotrope-conversion enthalpy:}\) To convert P into Q, first sublime P. \[ \mathrm{P(s)\rightarrow X(g)} \] This contributes \[ \Delta H_1=+90\,\mathrm{kJ\,mol^{-1}} \] Next, reverse the sublimation of Q. \[ \mathrm{X(g)\rightarrow Q(s)} \] The reversed enthalpy is \[ \Delta H_2=-80\,\mathrm{kJ\,mol^{-1}} \] Adding the steps gives \[ \mathrm{P(s)\rightarrow Q(s)} \] Therefore, \[ \Delta H_{\mathrm{P\rightarrow Q}}=90-80 \] \[ \Delta H_{\mathrm{P\rightarrow Q}}=+10\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Independent mixture check:}\) The heat required for the P fraction is \[ q_P=(0.30)(90)=27.0\,\mathrm{kJ} \] For the Q fraction, \[ q_Q=(0.20)(80)=16.0\,\mathrm{kJ} \] Their sum is \[ 27.0+16.0=43.0\,\mathrm{kJ} \] This reproduces the observed heat absorption. The positive allotrope-conversion enthalpy shows that Q has higher molar enthalpy than P at the specified conditions. A smaller sublimation enthalpy does not necessarily mean that the starting solid has lower enthalpy when both solids form the same gaseous endpoint. \(\textbf{Final answer:}\) \[ \boxed{\left(n_P,\Delta H_{\mathrm{P\rightarrow Q}}\right)=\left(0.30\,\mathrm{mol},+10\,\mathrm{kJ\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Identifying the required molecular rearrangement:}\) The target process changes gaseous tetraphosphorus molecules into gaseous diphosphorus molecules. Both reactant and product remain molecular rather than atomic. Therefore, the supplied complete atomisation enthalpies must be combined through a common gaseous-atom endpoint. \(\textbf{Atomising tetraphosphorus:}\) The first supplied equation is \[ \mathrm{P_4(g)\rightarrow4P(g)} \] Its enthalpy change is \[ \Delta H_1=+1100\,\mathrm{kJ} \] This produces four moles of gaseous phosphorus atoms from one mole of gaseous \(\mathrm{P_4}\). \(\textbf{Constructing two moles of diphosphorus:}\) The second supplied equation dissociates one mole of \(\mathrm{P_2}\). To obtain two moles of \(\mathrm{P_2}\) from four moles of phosphorus atoms, reverse the equation and multiply it by two. \[ \mathrm{4P(g)\rightarrow2P_2(g)} \] The enthalpy change becomes \[ \Delta H'_2=-2(490) \] \[ \Delta H'_2=-980\,\mathrm{kJ} \] \(\textbf{Combining the atomisation pathways:}\) Adding the two transformed reactions cancels the gaseous phosphorus atoms. The remaining equation is \[ \mathrm{P_4(g)\rightarrow2P_2(g)} \] By Hess's law, \[ \Delta_rH=1100-980 \] \[ \Delta_rH=+120\,\mathrm{kJ} \] This value corresponds to conversion of one mole of \(\mathrm{P_4}\). \(\textbf{Scaling to the experimental amount:}\) The amount converted is \[ n_{\mathrm{P_4}}=0.150\,\mathrm{mol} \] Therefore, \[ q_P=n_{\mathrm{P_4}}\Delta_rH \] \[ q_P=(0.150)(120) \] \[ q_P=+18.0\,\mathrm{kJ} \] \(\textbf{Stoichiometric verification:}\) One mole of \(\mathrm{P_4}\) contains four moles of phosphorus atoms. Two moles of \(\mathrm{P_2}\) contain the same number of phosphorus atoms. The transformation is therefore atomically balanced. Using only one reversed \(\mathrm{P_2}\) dissociation equation would leave two gaseous phosphorus atoms uncancelled. The positive result indicates that the specified molecular rearrangement absorbs enthalpy. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_rH,q_P\right)=\left(+120\,\mathrm{kJ\,mol^{-1}},+18.0\,\mathrm{kJ}\right)} \]

Explanation: \(\textbf{Counting bonds in hydrogen peroxide formation:}\) The formation equation consumes one H₂ molecule and one O₂ molecule. Therefore, one H–H bond and one O=O bond are broken. The product contains two O–H bonds and one O–O single bond. Let the unknown O–O bond enthalpy be x. \(\textbf{Energy required for reactant bond breaking:}\) \[ E_{\mathrm{broken}}=D(\mathrm{H-H})+D(\mathrm{O=O}) \] \[ E_{\mathrm{broken}}=436+498 \] \[ E_{\mathrm{broken}}=934\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Energy released by product bond formation:}\) \[ E_{\mathrm{formed}}=2D(\mathrm{O-H})+D(\mathrm{O-O}) \] \[ E_{\mathrm{formed}}=2(463)+x \] \[ E_{\mathrm{formed}}=926+x \] \(\textbf{Determining the O–O bond enthalpy:}\) The bond-enthalpy approximation gives \[ \Delta_fH\approx E_{\mathrm{broken}}-E_{\mathrm{formed}} \] Using the supplied formation enthalpy, \[ -135=934-(926+x) \] \[ -135=8-x \] \[ x=143\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Decomposition producing gaseous water:}\) First consider \[ \mathrm{H_2O_2(g)\rightarrow H_2O(g)+\frac12O_2(g)} \] The two O–H bonds are present on both sides and cancel in the net bond accounting. One O–O single bond is broken, and half a mole of O=O bonds is formed. Therefore, \[ \Delta H_{\mathrm{gas}}\approx D(\mathrm{O-O})-\frac12D(\mathrm{O=O}) \] \[ \Delta H_{\mathrm{gas}}\approx143-\frac12(498) \] \[ \Delta H_{\mathrm{gas}}\approx143-249 \] \[ \Delta H_{\mathrm{gas}}\approx-106\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Correcting the water product state:}\) The target reaction produces liquid water rather than gaseous water. Condensing one mole of water vapour contributes \[ \Delta H_{\mathrm{cond}}=-44\,\mathrm{kJ\,mol^{-1}} \] Therefore, \[ \Delta H_{\mathrm{target}}\approx\Delta H_{\mathrm{gas}}+\Delta H_{\mathrm{cond}} \] \[ \Delta H_{\mathrm{target}}\approx-106-44 \] \[ \Delta H_{\mathrm{target}}\approx-150\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Chemical verification:}\) The decomposition conserves two hydrogen atoms and two oxygen atoms. The O–H bonds cancel because the gaseous reactant and gaseous-water intermediate contain the same number of those bonds. The liquid-water phase correction is a separate thermochemical contribution, not an additional covalent bond. The negative result indicates that decomposition releases enthalpy under the stated approximation. \(\textbf{Final answer:}\) \[ \boxed{\left(D(\mathrm{O-O}),\Delta H_{\mathrm{target}}\right)=\left(143,-150\right)\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Separating the initial solid into gaseous ions:}\) The reaction begins with MX(s). The supplied lattice dissociation enthalpy is \[ \Delta H_1=+700\,\mathrm{kJ\,mol^{-1}} \] The corresponding process is \[ \mathrm{MX(s)\rightarrow M^+(g)+X^-(g)} \] \(\textbf{Removing hydration from the aqueous Y anion:}\) The initial Y⁻ ion is aqueous. To use the gaseous-ion lattice-formation pathway, reverse its hydration. The reverse of \[ \mathrm{Y^-(g)\rightarrow Y^-(aq)} \] has enthalpy \[ \Delta H_2=+400\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Forming the new ionic solid:}\) The desired product MY(s) forms from gaseous M⁺ and Y⁻. Its lattice formation enthalpy is the reverse of lattice dissociation. Therefore, \[ \Delta H_3=-650\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Hydrating the displaced X anion:}\) The final X⁻ ion is aqueous. Its hydration contributes \[ \Delta H_4=-330\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Summing the ion-exchange pathway:}\) When the four reactions are added, the gaseous ions cancel. The net equation is the specified ion-exchange reaction. Thus, \[ \Delta_rH=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4 \] \[ \Delta_rH=700+400-650-330 \] \[ \Delta_rH=+120\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Heat required for the actual extent:}\) The available amount of aqueous Y⁻ is 0.050 mol. MX is supplied in excess. Therefore, \[ \xi=0.050\,\mathrm{mol} \] The reaction enthalpy for this extent is \[ q_{\mathrm{reaction}}=(0.050)(120) \] \[ q_{\mathrm{reaction}}=+6.0\,\mathrm{kJ} \] \(\textbf{Temperature response of the surroundings:}\) The reaction absorbs 6.0 kJ from the calorimetric surroundings. Therefore, \[ q_{\mathrm{cal}}=-6000\,\mathrm{J} \] Using \[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \] we obtain \[ \Delta T=\frac{-6000}{1500} \] \[ \Delta T=-4.0\,\mathrm{K} \] \(\textbf{Energy and state check:}\) The starting Y⁻ ion is hydrated, whereas the final X⁻ ion is hydrated. These aqueous states must not be treated as gaseous ions when constructing the energy cycle. The enthalpy of hydration of the common cation does not enter the final expression because M⁺ begins and ends as part of a crystalline solid. The positive ion-exchange enthalpy explains the cooling of the calorimetric surroundings. The result assumes that the specified reaction proceeds to completion; the enthalpy data alone would not establish its equilibrium extent. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_rH,\Delta T\right)=\left(+120\,\mathrm{kJ\,mol^{-1}},-4.0\,\mathrm{K}\right)} \]

Explanation: \(\textbf{Constructing the MCl formation cycle:}\) MCl contains one \(\mathrm{M^+}\) ion and one chloride ion. The metal must be atomised and then ionised once. Half a mole of chlorine molecules must be converted into one mole of gaseous chlorine atoms. Finally, chlorine gains an electron before the gaseous ions form the lattice. \(\textbf{Energy before MCl lattice formation:}\) \[ \Delta H_{\mathrm{pre-lattice,MCl}}=150+500+120-350 \] \[ \Delta H_{\mathrm{pre-lattice,MCl}}=420\,\mathrm{kJ} \] Let the lattice formation enthalpy of MCl be L. The observed formation enthalpy gives \[ -400=420+L \] \[ L=-820\,\mathrm{kJ\,mol^{-1}} \] Lattice dissociation is the reverse process. Therefore, \[ \Delta H_{\mathrm{lattice,diss}}(\mathrm{MCl})=+820\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Identifying the additional ionisation required for MCl₂:}\) MCl₂ contains a divalent metal cation. Producing \(\mathrm{M^{2+}(g)}\) from neutral gaseous M requires two successive ionisations. Let the second ionisation enthalpy be x. The first ionisation contributes 500 kJ per mole, and the second contributes x. \(\textbf{Accounting for two chloride ions:}\) The formation equation for MCl₂ is \[ \mathrm{M(s)+Cl_2(g)\rightarrow MCl_2(s)} \] A full mole of chlorine molecules must dissociate. Therefore, \[ \Delta H_{\mathrm{Cl\ atoms}}=2(120) \] \[ \Delta H_{\mathrm{Cl\ atoms}}=240\,\mathrm{kJ} \] Two chloride ions form through two electron-gain steps. \[ \Delta H_{\mathrm{electron\ gain}}=2(-350) \] \[ \Delta H_{\mathrm{electron\ gain}}=-700\,\mathrm{kJ} \] \(\textbf{Using the MCl₂ formation enthalpy:}\) The complete cycle includes lattice formation of \(\mathrm{MCl_2}\). Therefore, \[ -600=150+500+x+240-700-2300 \] Combining the known terms, \[ -600=x-2110 \] Hence, \[ x=1510\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Independent MCl₂ check:}\) The energy before lattice formation is \[ 150+500+1510+240-700 \] \[ =1700\,\mathrm{kJ} \] Adding lattice formation gives \[ 1700-2300=-600\,\mathrm{kJ} \] This reproduces the supplied formation enthalpy. \(\textbf{Chemical interpretation:}\) The second ionisation cannot be replaced by a second copy of the first ionisation enthalpy. It removes an electron from an already positively charged gaseous ion. The two chloride ions require two electron-gain steps. The lattice formation enthalpy is negative, whereas the lattice dissociation enthalpy is positive. Both charge balancing and correct lattice-sign conventions are essential. \(\textbf{Final answer:}\) \[ \boxed{\left(IE_2,\Delta H_{\mathrm{lattice,diss}}(\mathrm{MCl})\right)=\left(1510,820\right)\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Mandatory first ionisation of the entire sample:}\) Every gaseous M atom loses at least one electron. Therefore, the complete 0.100 mol sample undergoes first ionisation. The enthalpy required is \[ \Delta H_1=nIE_1 \] \[ \Delta H_1=(0.100)(600) \] \[ \Delta H_1=60.0\,\mathrm{kJ} \] This step produces 0.100 mol of \(\mathrm{M^+}\) ions and 0.100 mol of electrons. \(\textbf{Energy available for second ionisation:}\) The measured total ionisation enthalpy is 132 kJ. The additional enthalpy beyond complete first ionisation is \[ \Delta H_2=132-60.0 \] \[ \Delta H_2=72.0\,\mathrm{kJ} \] This additional energy is associated with removal of a second electron from a fraction of the singly charged ions. \(\textbf{Amount undergoing second ionisation:}\) Let x be the amount of \(\mathrm{M^+}\) converted into \(\mathrm{M^{2+}}\). The second ionisation enthalpy is 1200 kJ per mole. Therefore, \[ xIE_2=72.0 \] \[ 1200x=72.0 \] \[ x=\frac{72.0}{1200} \] \[ x=0.060\,\mathrm{mol} \] Thus, \[ n_{\mathrm{M^{2+}}}=0.060\,\mathrm{mol} \] The remaining singly charged ions are \[ n_{\mathrm{M^+}}=0.100-0.060 \] \[ n_{\mathrm{M^+}}=0.040\,\mathrm{mol} \] \(\textbf{Counting all released electrons:}\) First ionisation releases one electron from each of the 0.100 mol of atoms. Second ionisation releases one additional electron from each of the 0.060 mol of doubly ionised atoms. Therefore, \[ n_{e^-}=0.100+0.060 \] \[ n_{e^-}=0.160\,\mathrm{mol} \] \(\textbf{Charge-balance verification:}\) The positive charge carried by the ions, in moles of elementary charge, is \[ (0.040)(1)+(0.060)(2) \] \[ =0.040+0.120 \] \[ =0.160 \] This equals the amount of negative charge carried by the released electrons. \(\textbf{Energy verification:}\) The first ionisation consumes 60.0 kJ. The second ionisation consumes \[ (0.060)(1200)=72.0\,\mathrm{kJ} \] The total is \[ 60.0+72.0=132\,\mathrm{kJ} \] Both the energy and charge balances are satisfied. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{M^{2+}}},n_{e^-}\right)=\left(0.060,0.160\right)\,\mathrm{mol}} \]

Explanation: \(\textbf{Recognising the physical state of iodine:}\) The target reaction uses solid iodine. However, the supplied I–I bond dissociation enthalpy refers to gaseous molecular iodine. Therefore, solid iodine must first be sublimed before its covalent bond is broken. \(\textbf{Subliming iodine:}\) The required phase transition is \[ \mathrm{I_2(s)\rightarrow I_2(g)} \] Its enthalpy contribution is \[ \Delta H_1=+62\,\mathrm{kJ} \] \(\textbf{Dissociating the gaseous reactant molecules:}\) Hydrogen dissociation is \[ \mathrm{H_2(g)\rightarrow2H(g)} \] with \[ \Delta H_2=+436\,\mathrm{kJ} \] Iodine dissociation is \[ \mathrm{I_2(g)\rightarrow2I(g)} \] with \[ \Delta H_3=+151\,\mathrm{kJ} \] \(\textbf{Total energy required before bond formation:}\) \[ E_{\mathrm{required}}=62+436+151 \] \[ E_{\mathrm{required}}=649\,\mathrm{kJ} \] \(\textbf{Forming gaseous hydrogen iodide:}\) The target reaction produces two moles of HI. Therefore, two moles of H–I bonds form. The corresponding enthalpy contribution is \[ \Delta H_4=-2D(\mathrm{H-I}) \] \[ \Delta H_4=-2(299) \] \[ \Delta H_4=-598\,\mathrm{kJ} \] \(\textbf{Overall reaction enthalpy:}\) By Hess's law, \[ \Delta_rH=649-598 \] \[ \Delta_rH=+51\,\mathrm{kJ} \] \(\textbf{Normalising to one mole of HI:}\) The balanced target reaction produces two moles of HI. The corresponding molar formation equation is \[ \mathrm{\frac12H_2(g)+\frac12I_2(s)\rightarrow HI(g)} \] Therefore, \[ \Delta_fH=\frac{+51}{2} \] \[ \Delta_fH=+25.5\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Thermochemical interpretation:}\) The target reaction requires both molecular bond dissociation and sublimation of solid iodine. Omitting sublimation would incorrectly treat solid iodine as gaseous iodine. The reaction enthalpy for producing two moles of HI differs numerically from the molar formation enthalpy for producing one mole. The final positive sign indicates enthalpy absorption for formation of HI under the stated reference conditions. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_rH,\Delta_fH\right)=\left(+51\,\mathrm{kJ},+25.5\,\mathrm{kJ\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Identifying the complete atomisation process:}\) Complete atomisation of gaseous methane produces one gaseous carbon atom and four gaseous hydrogen atoms per molecule. The thermochemical equation is \[ \mathrm{CH_4(g)\rightarrow C(g)+4H(g)} \] The supplied formation and elemental atomisation data can be combined to determine its enthalpy. \(\textbf{Producing gaseous carbon atoms:}\) One mole of graphite atoms is converted into gaseous carbon atoms with enthalpy \[ \Delta H_C=+716\,\mathrm{kJ} \] \(\textbf{Producing gaseous hydrogen atoms:}\) The formation equation for methane consumes two moles of H₂. Complete dissociation produces four moles of gaseous hydrogen atoms. Therefore, \[ \Delta H_H=2(436) \] \[ \Delta H_H=872\,\mathrm{kJ} \] \(\textbf{Enthalpy of gaseous atoms relative to the elemental states:}\) The combined atomisation enthalpy of the elemental reactants is \[ \Delta H_{\mathrm{elements\rightarrow atoms}}=716+872 \] \[ \Delta H_{\mathrm{elements\rightarrow atoms}}=1588\,\mathrm{kJ} \] \(\textbf{Recovering methane atomisation enthalpy:}\) The elemental formation reaction produces methane with enthalpy \(-75\,\mathrm{kJ}\). Reversing this formation reaction gives \[ \mathrm{CH_4(g)\rightarrow C(graphite)+2H_2(g)} \] with \[ \Delta H_{\mathrm{reverse}}=+75\,\mathrm{kJ} \] Following this reversal by elemental atomisation produces the target gaseous atoms. Therefore, \[ \Delta_{\mathrm{at}}H(\mathrm{CH_4})=75+1588 \] \[ \Delta_{\mathrm{at}}H(\mathrm{CH_4})=1663\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Separating the first bond dissociation:}\) The first C–H bond dissociation requires \[ D_1=440\,\mathrm{kJ\,mol^{-1}} \] The combined enthalpy of the remaining three successive dissociations is \[ D_{\mathrm{remaining,total}}=1663-440 \] \[ D_{\mathrm{remaining,total}}=1223\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Mean enthalpy of the remaining dissociations:}\) \[ \overline D_{\mathrm{remaining}}=\frac{1223}{3} \] \[ \overline D_{\mathrm{remaining}}=407.666\ldots\,\mathrm{kJ\,mol^{-1}} \] \[ \overline D_{\mathrm{remaining}}\approx407.7\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Chemical interpretation:}\) The overall atomisation enthalpy is not obtained by dividing the elemental atomisation energy directly among the methane bonds. The methane formation enthalpy must first be included because methane and the separated elemental reactants have different enthalpies. The first C–H dissociation occurs in methane, while subsequent dissociations occur in different gaseous fragments. Therefore, the remaining three bond dissociation enthalpies need not individually equal the first value. The calculated mean concerns the three remaining dissociation steps, not the mean over all four C–H bonds. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_{\mathrm{at}}H,\overline D_{\mathrm{remaining}}\right)=\left(1663,407.7\right)\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct because each formula unit of M₂X contains two monovalent metal cations. Producing two moles of \(\mathrm{M^+(g)}\) from two moles of gaseous neutral M atoms requires two first-ionisation events. A second ionisation of the same metal atom is unnecessary because the cation remains singly charged. Thus, the metal contribution involves two first-ionisation enthalpies rather than one first and one second ionisation enthalpy. Statement II is correct because the divalent anion must accept two electrons. The first electron-gain step is \[ \mathrm{X(g)+e^-\rightarrow X^-(g)} \] The second step is \[ \mathrm{X^-(g)+e^-\rightarrow X^{2-}(g)} \] These are distinct thermochemical transformations. The energy associated with adding an electron to a neutral atom need not have the same sign as that for adding an electron to an already negative ion. Statement III is correct because the second electron experiences electrostatic repulsion from the negatively charged gaseous ion. Consequently, the second electron-gain step can be endothermic even when the first electron-gain step is exothermic. The complete thermochemical cycle must retain both signed contributions. Statement IV is incorrect because lattice dissociation and lattice formation are reverse transformations. Lattice dissociation separates the ionic solid into gaseous ions and generally requires positive enthalpy. Lattice formation combines gaseous ions into the crystalline solid and has the opposite enthalpy sign. If the positive lattice dissociation enthalpy is denoted by L, the corresponding lattice formation contribution is \(-L\). The overall formation cycle must balance atoms, charges and electrons as well as thermochemical signs. Two gaseous metal cations collectively provide a charge of \(+2\), which balances the charge of one gaseous divalent anion. Therefore, Statements I, II and III are correct.

Explanation: \(\textbf{Interpreting the enthalpy-temperature slope:}\) For compatible thermochemical states and a constant reaction heat-capacity difference, \[ \Delta_rH(T_2)-\Delta_rH(T_1)=\Delta_rC_p(T_2-T_1) \] The graph therefore has slope \[ \frac{\Delta(\Delta_rH)}{\Delta T}=\Delta_rC_p \] \(\textbf{Calculating the graph slope:}\) At 300 K, the reaction enthalpy is \(+15.0\,\mathrm{kJ}\). At 500 K, it is \(+5.0\,\mathrm{kJ}\). Therefore, \[ \Delta(\Delta_rH)=5.0-15.0 \] \[ \Delta(\Delta_rH)=-10.0\,\mathrm{kJ} \] The temperature change is \[ \Delta T=500-300 \] \[ \Delta T=200\,\mathrm{K} \] Hence, \[ \Delta_rC_p=\frac{-10.0}{200} \] \[ \Delta_rC_p=-0.050\,\mathrm{kJ\,K^{-1}} \] Converting units, \[ \Delta_rC_p=-50\,\mathrm{J\,K^{-1}} \] \(\textbf{Locating the zero-enthalpy temperature:}\) The enthalpy at temperature T can be written as \[ \Delta_rH(T)=15.0-0.050(T-300) \] The zero-enthalpy temperature satisfies \[ 0=15.0-0.050(T-300) \] Rearranging, \[ 0.050(T-300)=15.0 \] \[ T-300=300 \] \[ T=600\,\mathrm{K} \] \(\textbf{Determining the higher-temperature sign:}\) The graph slopes downward because the reaction heat-capacity difference is negative. At temperatures above 600 K, the linear relation gives negative reaction enthalpy. Therefore, under the stated constant-property model, the reaction becomes exothermic above 600 K. \(\textbf{Physical qualification:}\) The calculation concerns the sign of reaction enthalpy, not the thermodynamic spontaneity of the reaction. A negative enthalpy change alone does not guarantee spontaneity. The extrapolation also depends on the explicit assumptions that heat-capacity differences remain constant and physical states do not change. The graph cannot be extended across an unaccounted phase transition without additional enthalpy information. Thus, option D correctly gives both the slope and its consequence.

Explanation: \(\textbf{Identifying the temperature correction:}\) The supplied sublimation enthalpy refers to 280 K. The requested sublimation occurs at 330 K. The two transitions have the same chemical species and corresponding physical states but different temperatures. A Hess-law cycle through heating and cooling steps can therefore relate their enthalpies. \(\textbf{Heating solid Z from 280 K to 330 K:}\) The temperature change is \[ \Delta T=330-280=50\,\mathrm{K} \] The solid's molar enthalpy increase is \[ \Delta H_s=C_{p,m}(s)\Delta T \] \[ \Delta H_s=40(50) \] \[ \Delta H_s=2000\,\mathrm{J\,mol^{-1}} \] \[ \Delta H_s=2.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Heating gaseous Z through the same interval:}\) The gaseous heat capacity is greater than the solid heat capacity. Its corresponding molar enthalpy increase is \[ \Delta H_g=C_{p,m}(g)\Delta T \] \[ \Delta H_g=60(50) \] \[ \Delta H_g=3000\,\mathrm{J\,mol^{-1}} \] \[ \Delta H_g=3.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Constructing a cycle for sublimation at 330 K:}\) Begin with solid Z at 330 K. First cool the solid to 280 K. This reverses the solid-heating step and contributes \[ \Delta H_1=-2.0\,\mathrm{kJ\,mol^{-1}} \] Next, sublime Z at 280 K. \[ \Delta H_2=+75.0\,\mathrm{kJ\,mol^{-1}} \] Finally, heat gaseous Z from 280 K to 330 K. \[ \Delta H_3=+3.0\,\mathrm{kJ\,mol^{-1}} \] The net process is sublimation at 330 K. Therefore, \[ \Delta_{\mathrm{sub}}H(330)=-2.0+75.0+3.0 \] \[ \Delta_{\mathrm{sub}}H(330)=+76.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Scaling to the actual sample:}\) The sample contains \[ n=0.200\,\mathrm{mol} \] At constant pressure, with only pressure-volume work, \[ q_P=n\Delta_{\mathrm{sub}}H(330) \] \[ q_P=(0.200)(76.0) \] \[ q_P=+15.2\,\mathrm{kJ} \] \(\textbf{Consistency check:}\) The gas gains more enthalpy than the solid for the same temperature rise because its heat capacity is greater. Consequently, the sublimation enthalpy increases by 1.0 kJ per mole over the stated temperature interval. The result would have the opposite temperature trend if the gas heat capacity were smaller than the solid heat capacity. The calculation remains conditional on the explicitly supplied phase-stability and constant-heat-capacity assumptions. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_{\mathrm{sub}}H(330\,\mathrm{K}),q_P\right)=\left(+76.0\,\mathrm{kJ\,mol^{-1}},+15.2\,\mathrm{kJ}\right)} \]

Explanation: \(\textbf{Step 1 — Enthalpy of dissolution:}\) For the specified ionic dissolution cycle, \[ \Delta_{\mathrm{sol}}H=\Delta_{\mathrm{lattice,diss}}H+\Delta_{\mathrm{hyd}}H(\mathrm{cation})+\Delta_{\mathrm{hyd}}H(\mathrm{anion}) \] Lattice dissociation requires energy and therefore contributes positively. Hydration of gaseous ions releases energy and contributes negatively. \(\textbf{Step 2 — Dissolution enthalpy of AX:}\) Using the supplied values, \[ \Delta_{\mathrm{sol}}H(\mathrm{AX})=720-410-300 \] \[ \Delta_{\mathrm{sol}}H(\mathrm{AX})=+10\,\mathrm{kJ\,mol^{-1}} \] The positive sign indicates that dissolution of AX is endothermic. For \(0.200\,\mathrm{mol}\) of AX, the heat absorbed by the dissolving system is \[ q_{\mathrm{AX}}=n\Delta_{\mathrm{sol}}H \] \[ q_{\mathrm{AX}}=(0.200)(+10) \] \[ q_{\mathrm{AX}}=+2.0\,\mathrm{kJ} \] Under the stated insulated conditions, this energy comes from the calorimetric contents. Therefore, \[ q_{\mathrm{cal,AX}}=-2.0\,\mathrm{kJ} \] \(\textbf{Step 3 — Predicted temperature change for AX:}\) The calorimetric heat relation is \[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \] Hence, \[ \Delta T_{\mathrm{AX}}=\frac{q_{\mathrm{cal,AX}}}{C_{\mathrm{cal}}} \] Converting kilojoules into joules, \[ \Delta T_{\mathrm{AX}}=\frac{-2000}{500} \] \[ \Delta T_{\mathrm{AX}}=-4.0\,\mathrm{K} \] Thus, AX would produce a temperature decrease of \(4.0\,\mathrm{K}\), not the observed increase of \(2.0\,\mathrm{K}\). \(\textbf{Step 4 — Dissolution enthalpy of AY:}\) For AY, \[ \Delta_{\mathrm{sol}}H(\mathrm{AY})=680-410-310 \] \[ \Delta_{\mathrm{sol}}H(\mathrm{AY})=-40\,\mathrm{kJ\,mol^{-1}} \] The negative sign indicates that dissolution of AY is exothermic. For \(0.200\,\mathrm{mol}\), \[ q_{\mathrm{AY}}=(0.200)(-40) \] \[ q_{\mathrm{AY}}=-8.0\,\mathrm{kJ} \] The dissolving system releases \(8.0\,\mathrm{kJ}\), which is absorbed by the calorimetric contents. Therefore, \[ q_{\mathrm{cal,AY}}=+8.0\,\mathrm{kJ} \] \(\textbf{Step 5 — Predicted temperature change for AY:}\) \[ \Delta T_{\mathrm{AY}}=\frac{q_{\mathrm{cal,AY}}}{C_{\mathrm{cal}}} \] \[ \Delta T_{\mathrm{AY}}=\frac{8000}{500} \] \[ \Delta T_{\mathrm{AY}}=+16.0\,\mathrm{K} \] Thus, AY predicts a temperature increase of \(16.0\,\mathrm{K}\). Although this agrees with the direction of the observed temperature change, it does not agree with its magnitude. \(\textbf{Step 6 — Independent calorimetric compatibility check:}\) The experimentally observed temperature increase is \[ \Delta T_{\mathrm{obs}}=+2.0\,\mathrm{K} \] Therefore, \[ q_{\mathrm{cal,obs}}=500(2.0) \] \[ q_{\mathrm{cal,obs}}=+1000\,\mathrm{J} \] \[ q_{\mathrm{cal,obs}}=+1.0\,\mathrm{kJ} \] The dissolving system must consequently release \(1.0\,\mathrm{kJ}\). For \(0.200\,\mathrm{mol}\), the molar dissolution enthalpy required by the observation is \[ \Delta_{\mathrm{sol}}H_{\mathrm{required}}=\frac{-1.0}{0.200} \] \[ \Delta_{\mathrm{sol}}H_{\mathrm{required}}=-5.0\,\mathrm{kJ\,mol^{-1}} \] Neither salt has this dissolution enthalpy. AX has \(+10\,\mathrm{kJ\,mol^{-1}}\), whereas AY has \(-40\,\mathrm{kJ\,mol^{-1}}\). \(\textbf{Final interpretation:}\) AX predicts cooling by \(4.0\,\mathrm{K}\), while AY predicts warming by \(16.0\,\mathrm{K}\). Neither prediction matches the observed warming of \(2.0\,\mathrm{K}\). Identifying an exothermic salt is insufficient to establish agreement with calorimetric measurements; the predicted magnitude must also match. \(\textbf{Final answer:}\) \[ \boxed{\Delta T_{\mathrm{AX}}=-4.0\,\mathrm{K},\qquad\Delta T_{\mathrm{AY}}=+16.0\,\mathrm{K}} \] Neither salt is consistent with the measured temperature change under the stated assumptions.

Explanation: \(\textbf{Writing the complete atomisation reaction:}\) The molecular formula of acetylene is C₂H₂. Complete atomisation separates all its constituent atoms into the gaseous atomic state. Therefore, \[ \mathrm{C_2H_2(g)\rightarrow2C(g)+2H(g)} \] The required enthalpy can be calculated through the standard elemental reference states. \(\textbf{Reversing the formation of acetylene:}\) The standard formation reaction is \[ \mathrm{2C(graphite)+H_2(g)\rightarrow C_2H_2(g)} \] Its enthalpy change is \[ \Delta_fH^\circ=+227\,\mathrm{kJ} \] Reversing this reaction gives \[ \mathrm{C_2H_2(g)\rightarrow2C(graphite)+H_2(g)} \] The enthalpy change becomes \[ \Delta H_1=-227\,\mathrm{kJ} \] \(\textbf{Atomising the carbon:}\) The reversed formation equation produces two moles of graphite carbon. Therefore, \[ \Delta H_C=2(716) \] \[ \Delta H_C=1432\,\mathrm{kJ} \] \(\textbf{Atomising the hydrogen:}\) One mole of molecular hydrogen is produced by the reversed formation reaction. Dissociation gives \[ \mathrm{H_2(g)\rightarrow2H(g)} \] with \[ \Delta H_H=+436\,\mathrm{kJ} \] \(\textbf{Total molar atomisation enthalpy:}\) By Hess's law, \[ \Delta_{\mathrm{at}}H=-227+1432+436 \] \[ \Delta_{\mathrm{at}}H=+1641\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Scaling to the actual acetylene amount:}\) The sample contains \[ n_{\mathrm{C_2H_2}}=0.050\,\mathrm{mol} \] Therefore, \[ q=(0.050)(1641) \] \[ q=+82.05\,\mathrm{kJ} \] \(\textbf{Counting the gaseous hydrogen atoms:}\) Each mole of acetylene produces two moles of gaseous hydrogen atoms. Thus, \[ n_{\mathrm{H(g)}}=2(0.050) \] \[ n_{\mathrm{H(g)}}=0.100\,\mathrm{mol} \] \(\textbf{Stoichiometric interpretation:}\) The positive formation enthalpy of acetylene means its gaseous state lies above the corresponding standard elemental reactants in enthalpy. Reversing formation therefore releases 227 kJ per mole under the supplied reference convention. The elemental carbon and hydrogen then require energy for atomisation. The complete atomisation process remains strongly endothermic. The final gaseous hydrogen amount is determined by the atomic composition of acetylene, not by the number of hydrogen molecules in its formation equation. \(\textbf{Final answer:}\) \[ \boxed{\left(q,n_{\mathrm{H(g)}}\right)=\left(+82.05\,\mathrm{kJ},0.100\,\mathrm{mol}\right)} \]

Explanation: The Assertion is true because the two pathways begin with the same solid X and end with the same gaseous species X. Their total enthalpy changes can therefore be compared using Hess's law. For Pathway I, the supplied enthalpy change is \[ \Delta H_I=+80\,\mathrm{kJ\,mol^{-1}} \] For Pathway II, the two consecutive enthalpy changes add algebraically. \[ \Delta H_{II}=+35+45 \] \[ \Delta H_{II}=+80\,\mathrm{kJ\,mol^{-1}} \] The two values agree. Thus, the thermochemical data are consistent with the state-function nature of enthalpy. However, the first step of Pathway II converts X(s) into Y(g). Although the initial state is solid and the final state is gaseous, the chemical species changes from X to Y. Consequently, that step is not a direct sublimation of X. Sublimation means conversion of a substance from the solid state into the gaseous state without changing its chemical identity. The second step of Pathway II converts gaseous Y into gaseous X. It is a chemical transformation rather than a phase transition. Together, the two steps constitute an indirect thermochemical route connecting solid X with gaseous X. The Reason correctly explains both the equality of the overall enthalpies and the distinction between direct sublimation and a chemically transforming indirect pathway. Both A and R are therefore true, and R correctly explains A.

Explanation: \(\textbf{Recovering the lattice dissociation enthalpy of MX:}\) For the specified dissolution process, \[ \Delta_{\mathrm{sol}}H=\Delta_{\mathrm{lattice,diss}}H+\Delta_{\mathrm{hyd}}H \] The solution enthalpy of MX is \[ \Delta_{\mathrm{sol}}H(\mathrm{MX})=+12.0\,\mathrm{kJ\,mol^{-1}} \] Its total hydration enthalpy is \[ \Delta_{\mathrm{hyd}}H(\mathrm{MX})=-700\,\mathrm{kJ\,mol^{-1}} \] Let its lattice dissociation enthalpy be L. Then, \[ 12=L-700 \] Therefore, \[ L=+712\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Determining the solution enthalpy of MY:}\) For MY, the lattice dissociation enthalpy is \[ \Delta_{\mathrm{lattice,diss}}H(\mathrm{MY})=+650\,\mathrm{kJ\,mol^{-1}} \] Its total hydration enthalpy is \[ \Delta_{\mathrm{hyd}}H(\mathrm{MY})=-675\,\mathrm{kJ\,mol^{-1}} \] Therefore, \[ \Delta_{\mathrm{sol}}H(\mathrm{MY})=650-675 \] \[ \Delta_{\mathrm{sol}}H(\mathrm{MY})=-25\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Constructing the ion-exchange reaction:}\) The target process is \[ \mathrm{MX(s)+Y^-(aq)\rightarrow MY(s)+X^-(aq)} \] It can be constructed by dissolving MX and then precipitating MY. The dissolution step is \[ \mathrm{MX(s)\rightarrow M^+(aq)+X^-(aq)} \] with enthalpy \[ \Delta H_1=+12\,\mathrm{kJ} \] The precipitation of MY is the reverse of its dissolution: \[ \mathrm{M^+(aq)+Y^-(aq)\rightarrow MY(s)} \] Because MY dissolution has enthalpy \(-25\,\mathrm{kJ\,mol^{-1}}\), the reverse process has \[ \Delta H_2=+25\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Net ion-exchange enthalpy:}\) Adding the two steps cancels the common aqueous M⁺ ion. Therefore, \[ \Delta_rH=\Delta H_1+\Delta H_2 \] \[ \Delta_rH=12+25 \] \[ \Delta_rH=+37\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Thermochemical interpretation:}\) The process is endothermic under the specified conditions. Dissolution of MX absorbs enthalpy, while precipitation of MY also absorbs enthalpy because dissolution of MY is exothermic. The common cation cancels from the complete reaction, although its hydration was included in the separately supplied total hydration values. The positive reaction enthalpy does not by itself determine whether the ion-exchange reaction proceeds spontaneously or completely. The calculation establishes only the enthalpy difference between the specified initial and final states. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta_{\mathrm{lattice,diss}}H(\mathrm{MX}),\Delta_rH\right)=\left(+712,+37\right)\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Determining the maximum heat available from the metal:}\) The metal initially is hotter than the ice. If the final temperature reaches \(0^\circ\mathrm{C}\), the metal can cool through 120 K. Its heat-loss magnitude is \[ q_{\mathrm{metal}}=mc\Delta T \] Substituting, \[ q_{\mathrm{metal}}=(150)(0.50)(120) \] \[ q_{\mathrm{metal}}=9000\,\mathrm{J} \] \[ q_{\mathrm{metal}}=9.00\,\mathrm{kJ} \] \(\textbf{Heat required for complete melting:}\) The initial ice mass is 40 g. Its amount is \[ n_{\mathrm{ice}}=\frac{40}{18} \] \[ n_{\mathrm{ice}}=\frac{20}{9}\,\mathrm{mol} \] The heat required to melt all the ice is \[ q_{\mathrm{complete\ fusion}}=n_{\mathrm{ice}}\Delta_{\mathrm{fus}}H \] \[ q_{\mathrm{complete\ fusion}}=\frac{20}{9}(5.40) \] \[ q_{\mathrm{complete\ fusion}}=12.0\,\mathrm{kJ} \] \(\textbf{Selecting the correct final-state regime:}\) The metal can release only 9.00 kJ before reaching \(0^\circ\mathrm{C}\). Complete melting would require 12.0 kJ. Therefore, the available heat is insufficient to melt all the ice. Liquid water and ice must coexist at the final equilibrium state. Under the stated conditions, their common temperature is \(0^\circ\mathrm{C}\). \(\textbf{Calculating the amount of ice melted:}\) The molar enthalpy of fusion is 5.40 kJ per mole. Therefore, \[ n_{\mathrm{melted}}=\frac{q_{\mathrm{metal}}}{\Delta_{\mathrm{fus}}H} \] \[ n_{\mathrm{melted}}=\frac{9.00}{5.40} \] \[ n_{\mathrm{melted}}=\frac53\,\mathrm{mol} \] The corresponding mass is \[ m_{\mathrm{melted}}=\frac53(18) \] \[ m_{\mathrm{melted}}=30\,\mathrm{g} \] \(\textbf{Determining the remaining ice:}\) \[ m_{\mathrm{remaining}}=40-30 \] \[ m_{\mathrm{remaining}}=10\,\mathrm{g} \] \(\textbf{Physical verification:}\) The energy released by cooling the metal is exactly the energy required to melt 30 g of ice. No heat remains available to warm the resulting water above \(0^\circ\mathrm{C}\). The final mixture contains 30 g of liquid water and 10 g of ice. The calculation illustrates why the possible phase-change energy must be checked before treating all supplied heat as sensible heating. \(\textbf{Final answer:}\) \[ \boxed{T_f=0^\circ\mathrm{C},\quad m_{\mathrm{ice,remaining}}=10\,\mathrm{g}} \]

Explanation: \(\textbf{Heat released before any condensation:}\) The combustion enthalpy producing gaseous water is \[ \Delta_cH_g=-802\,\mathrm{kJ\,mol^{-1}} \] The amount of methane combusted is 0.250 mol. Therefore, \[ q_{\mathrm{combustion}}=(0.250)(-802) \] \[ q_{\mathrm{combustion}}=-200.5\,\mathrm{kJ} \] \(\textbf{Separating the measured condensation contribution:}\) The measured total heat transfer is \[ q_{\mathrm{measured}}=-211.5\,\mathrm{kJ} \] The additional enthalpy change associated with condensation is \[ q_{\mathrm{cond}}=q_{\mathrm{measured}}-q_{\mathrm{combustion}} \] Substituting, \[ q_{\mathrm{cond}}=-211.5-(-200.5) \] \[ q_{\mathrm{cond}}=-11.0\,\mathrm{kJ} \] \(\textbf{Determining the amount of water condensed:}\) One mole of condensing water vapour releases 44 kJ. Therefore, \[ n_{\mathrm{condensed}}=\frac{11.0}{44} \] \[ n_{\mathrm{condensed}}=0.250\,\mathrm{mol} \] \(\textbf{Calculating the total water produced:}\) The balanced combustion equation produces two moles of water per mole of methane. Thus, \[ n_{\mathrm{water,total}}=2(0.250) \] \[ n_{\mathrm{water,total}}=0.500\,\mathrm{mol} \] The fraction condensed is \[ f_{\mathrm{condensed}}=\frac{0.250}{0.500} \] \[ f_{\mathrm{condensed}}=0.50 \] Therefore, 50% of the water condenses. \(\textbf{Complete-condensation comparison:}\) If all 0.500 mol of product water condensed, the phase-change contribution would be \[ q_{\mathrm{cond,complete}}=(0.500)(-44) \] \[ q_{\mathrm{cond,complete}}=-22.0\,\mathrm{kJ} \] The corresponding total heat transfer would be \[ q_{\mathrm{complete}}=-200.5-22.0 \] \[ q_{\mathrm{complete}}=-222.5\,\mathrm{kJ} \] \(\textbf{Thermochemical interpretation:}\) The gaseous-water combustion enthalpy provides the common reference for both experiments. Partial condensation makes the measured heat more negative because condensation releases additional enthalpy. The measured heat does not imply incomplete methane combustion; methane is completely consumed in both cases. The difference arises entirely from the physical-state distribution of the product water. \(\textbf{Final answer:}\) \[ \boxed{\left(f_{\mathrm{condensed}},|q_{\mathrm{complete}}|\right)=(50\%,222.5\,\mathrm{kJ})} \]

Explanation: \(\textbf{Enthalpy released by the exothermic reaction:}\) The conversion of A into B releases 75 kJ per mole. The amount converted is \[ n_A=0.400\,\mathrm{mol} \] Therefore, \[ \Delta H_1=(0.400)(-75) \] \[ \Delta H_1=-30.0\,\mathrm{kJ} \] \(\textbf{Energy absorbed by the calorimetric assembly:}\) The assembly warms by 3.0 K. Its heat capacity is \(1000\,\mathrm{J\,K^{-1}}\). Thus, \[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \] \[ q_{\mathrm{cal}}=(1000)(3.0) \] \[ q_{\mathrm{cal}}=3000\,\mathrm{J} \] \[ q_{\mathrm{cal}}=+3.0\,\mathrm{kJ} \] \(\textbf{Determining the net reaction enthalpy:}\) The overall apparatus is insulated. Therefore, the total reaction enthalpy change is balanced by the energy gained by the calorimetric assembly under the specified conditions. \[ \Delta H_{\mathrm{reactions}}=-q_{\mathrm{cal}} \] \[ \Delta H_{\mathrm{reactions}}=-3.0\,\mathrm{kJ} \] \(\textbf{Recovering the carbonate decomposition contribution:}\) The reaction enthalpies are additive. \[ \Delta H_{\mathrm{reactions}}=\Delta H_1+\Delta H_2 \] Therefore, \[ -3.0=-30.0+\Delta H_2 \] \[ \Delta H_2=+27.0\,\mathrm{kJ} \] \(\textbf{Calculating the decomposition extent:}\) Calcium carbonate decomposition absorbs 180 kJ per mole. Therefore, \[ n_{\mathrm{decomposed}}=\frac{27.0}{180} \] \[ n_{\mathrm{decomposed}}=0.150\,\mathrm{mol} \] One mole of decomposed calcium carbonate produces one mole of carbon dioxide. Hence, \[ n_{\mathrm{CO_2}}=0.150\,\mathrm{mol} \] \(\textbf{Calculating the remaining carbonate:}\) \[ n_{\mathrm{remaining}}=0.200-0.150 \] \[ n_{\mathrm{remaining}}=0.050\,\mathrm{mol} \] \(\textbf{Energy verification:}\) The first reaction releases 30.0 kJ. Carbonate decomposition absorbs \[ (0.150)(180)=27.0\,\mathrm{kJ} \] The net energy released is 3.0 kJ, matching the calorimetric energy gain. Thus, the warmer calorimeter does not imply that the endothermic decomposition failed to occur. An exothermic process can supply energy simultaneously consumed by an endothermic process. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{CO_2}},n_{\mathrm{CaCO_3,remaining}}\right)=(0.150,0.050)\,\mathrm{mol}} \]

Explanation: \(\textbf{Interpreting the strong-acid reaction:}\) A strong monoprotic acid is effectively ionised under the stated aqueous conditions. Its neutralisation enthalpy corresponds primarily to formation of liquid water from aqueous hydrogen and hydroxide ions. The relevant net ionic equation is \[ \mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)} \] with \[ \Delta H_{\mathrm{neutralisation}}=-57.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Separating ionisation from weak-acid neutralisation:}\) A weak acid HA must first undergo ionisation: \[ \mathrm{HA(aq)\rightarrow H^+(aq)+A^-(aq)} \] The resulting hydrogen ions then undergo neutralisation by hydroxide ions. The overall process is \[ \mathrm{HA(aq)+OH^-(aq)\rightarrow A^-(aq)+H_2O(l)} \] \(\textbf{Recovering the weak-acid ionisation enthalpy:}\) By Hess's law, \[ \Delta H_{\mathrm{weak}}=\Delta H_{\mathrm{ionisation}}+\Delta H_{\mathrm{strong}} \] Substitution gives \[ -54.0=\Delta H_{\mathrm{ionisation}}-57.0 \] Therefore, \[ \Delta H_{\mathrm{ionisation}}=+3.0\,\mathrm{kJ\,mol^{-1}} \] The positive value indicates enthalpy absorption during the specified ionisation step. \(\textbf{Heat released in the strong-acid experiment:}\) \[ q_{\mathrm{strong}}=(0.200)(-57.0) \] \[ q_{\mathrm{strong}}=-11.40\,\mathrm{kJ} \] The calorimeter absorbs 11.40 kJ. Therefore, \[ \Delta T_{\mathrm{strong}}=\frac{11400}{1500} \] \[ \Delta T_{\mathrm{strong}}=7.60\,\mathrm{K} \] \(\textbf{Heat released in the weak-acid experiment:}\) \[ q_{\mathrm{weak}}=(0.200)(-54.0) \] \[ q_{\mathrm{weak}}=-10.80\,\mathrm{kJ} \] Thus, \[ \Delta T_{\mathrm{weak}}=\frac{10800}{1500} \] \[ \Delta T_{\mathrm{weak}}=7.20\,\mathrm{K} \] \(\textbf{Temperature-rise difference:}\) \[ \Delta T_{\mathrm{strong}}-\Delta T_{\mathrm{weak}}=7.60-7.20 \] \[ \Delta T_{\mathrm{strong}}-\Delta T_{\mathrm{weak}}=+0.40\,\mathrm{K} \] \(\textbf{Thermochemical interpretation:}\) Both neutralisation reactions release heat. However, the weak-acid experiment releases less heat because part of the neutralisation energy is offset by the ionisation enthalpy. The smaller temperature rise does not imply that the weak acid was incompletely neutralised. The two experiments are explicitly stated to consume equal acid amounts completely. The calculation depends on comparing compatible aqueous conditions and separating the ionisation step from the common ionic neutralisation step. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta H_{\mathrm{ionisation}},\Delta T_{\mathrm{strong}}-\Delta T_{\mathrm{weak}}\right)=\left(+3.0\,\mathrm{kJ\,mol^{-1}},+0.40\,\mathrm{K}\right)} \]