Heat of reaction, Bond energy and Hess law Questions in English

(A) The reaction is $OF_{2(g)} \to O_{(g)} + 2F_{(g)}$.
In this reaction,two $O-F$ bonds are broken.
The enthalpy of reaction $(\Delta_{rxn}H)$ is equal to the sum of the bond dissociation energies of the bonds broken.
$\Delta_{rxn}H = 2 \times \text{Bond Energy}(O-F)$.
Given $\Delta_{rxn}H = 368 \ kJ$.
Therefore,$368 = 2 \times \text{Bond Energy}(O-F)$.
Average $O-F$ bond energy $= \frac{368}{2} = 184 \ kJ/mol$.

(C) The enthalpy of neutralization of a strong acid with a strong base is always $-57.1 \, kJ/mol$ (or $-55.84 \, kJ/mol$ as given).
For a weak base like $NH_4OH$,some energy is consumed in the ionization of the weak base.
The enthalpy of neutralization of a weak base is given by:
$\Delta H_{\text{neutralization}}(NH_4OH) = \Delta H_{\text{neutralization}}(\text{strong acid-strong base}) + \Delta H_{\text{ionization}}(NH_4OH)$
Given:
$\Delta H_{\text{neutralization}}(NaOH) = -55.84 \, kJ/mol$
$\Delta H_{\text{neutralization}}(NH_4OH) = -51.34 \, kJ/mol$
Therefore,
$\Delta H_{\text{ionization}}(NH_4OH) = \Delta H_{\text{neutralization}}(NH_4OH) - \Delta H_{\text{neutralization}}(NaOH)$
$\Delta H_{\text{ionization}}(NH_4OH) = -51.34 - (-55.84) = 4.5 \, kJ/mol$
Thus,the enthalpy of ionization of $NH_4OH$ is $4.5 \, kJ/mol$.

(D) The enthalpy of reaction $\Delta_r H$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$.
For the reaction $2N_2O_{(g)} + O_{2(g)} \to 4NO_{(g)}$,the enthalpy change is:
$\Delta_r H = [4 \times \Delta_f H(NO)] - [2 \times \Delta_f H(N_2O) + 1 \times \Delta_f H(O_2)]$.
Given $\Delta_f H(N_2O) = 28 \ kJ \ mol^{-1}$,$\Delta_f H(NO) = 90 \ kJ \ mol^{-1}$,and $\Delta_f H(O_2) = 0 \ kJ \ mol^{-1}$ (standard state).
$\Delta_r H = [4 \times 90] - [2 \times 28 + 0]$
$\Delta_r H = 360 - 56 = 304 \ kJ$.

(D) The standard molar enthalpy of formation of $CO_2$ corresponds to the reaction:
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$
By definition,the enthalpy change for the combustion of one mole of a substance in its standard state is its standard molar enthalpy of combustion.
Since $C_{(graphite)}$ is the standard state of carbon,the enthalpy change for this reaction is equal to the standard molar enthalpy of combustion of carbon (graphite).

(B) The enthalpy of formation of $HCl(g)$ is given by the reaction: $\frac{1}{2} H_{2}(g) + \frac{1}{2} Cl_{2}(g) \to HCl(g)$,where $\Delta H_{f} = -92.3 \ kJ/mol$.
The enthalpy of solution is: $HCl(g) + aq \to H^{+}(aq) + Cl^{-}(aq)$,where $\Delta H_{sol} = -75.14 \ kJ/mol$.
Adding these two equations gives the formation reaction of the ions: $\frac{1}{2} H_{2}(g) + \frac{1}{2} Cl_{2}(g) + aq \to H^{+}(aq) + Cl^{-}(aq)$.
The total enthalpy change is $\Delta H_{f}(H^{+}) + \Delta H_{f}(Cl^{-}) = -92.3 + (-75.14) = -167.44 \ kJ/mol$.
Since $\Delta H_{f}(H^{+}) = 0 \ kJ/mol$,we have $\Delta H_{f}(Cl^{-}) = -167.44 \ kJ/mol$.

(B) The bond enthalpy of a specific bond in a polyatomic molecule is defined as the average of the bond dissociation enthalpies of that bond in the molecule.
Given the dissociation enthalpy of the first $O-H$ bond: $\Delta H_1 = 498 \ kJ \ mol^{-1}$.
Given the dissociation enthalpy of the second $O-H$ bond: $\Delta H_2 = 428 \ kJ \ mol^{-1}$.
Therefore,the average bond enthalpy of the $O-H$ bond is calculated as:
$\text{Bond Enthalpy} = \frac{\Delta H_1 + \Delta H_2}{2} = \frac{498 + 428}{2} = \frac{926}{2} = 463 \ kJ \ mol^{-1}$.

(D) The standard enthalpy of formation $(\Delta_fH^\circ)$ of an element in its most stable state at $298 \ K$ and $1 \ \text{bar}$ pressure is defined as zero.
$Hg\,(l)$ is the most stable state of mercury at standard conditions.
$Xe\,(g)$ is the most stable state of xenon at standard conditions.
$Br_2\,(l)$ is the most stable state of bromine at standard conditions.
$S_{\text{monoclinic}}$ is an allotrope of sulfur,but it is not the most stable form at $298 \ K$; the most stable form is $S_{\text{rhombic}}$.
Since the question asks for which has a standard enthalpy of formation equal to zero,all options $A$,$B$,and $C$ are technically correct as they represent elements in their standard states. However,in many textbook contexts,this question is often presented as a multiple-choice question where one might be expected to identify the exception. Given the options,$S_{\text{monoclinic}}$ is the only one that is definitely $NOT$ zero.

(C) We are given the following thermochemical equations:
$(1) HA \to A^{-} + H^{+} , \Delta H = Z$
$(2) H^{+} + OH^{-} \to H_2O , \Delta H = X$
$(3) HA + OH^{-} \to A^{-} + H_2O , \Delta H = Y$
According to Hess's Law,we can obtain equation $(3)$ by adding equation $(1)$ and equation $(2)$:
$(HA \to A^{-} + H^{+}) + (H^{+} + OH^{-} \to H_2O) \to HA + OH^{-} \to A^{-} + H_2O$
Therefore,the enthalpy change for equation $(3)$ is the sum of the enthalpy changes of equations $(1)$ and $(2)$:
$Y = Z + X$
Rearranging to solve for $Z$:
$Z = Y - X$

(B) The enthalpy change of a reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products:
$\Delta H = \sum (\text{Bond Energy})_{Reactants} - \sum (\text{Bond Energy})_{Products}$
For the reaction $H_2 + F_2 \longrightarrow 2HF$:
$\Delta H = [BE(H-H) + BE(F-F)] - [2 \times BE(H-F)]$
Substituting the given values:
$\Delta H = [434 + 158] - [2 \times 565]$
$\Delta H = 592 - 1130$
$\Delta H = -538 \ kJ/mol$

(B) reaction with a positive $\Delta H$ is an endothermic reaction.
$1$. The formation of ammonia $(N_2 + 3H_2 \to 2NH_3)$ is an exothermic process $(\Delta H < 0)$.
$2$. The formation of nitric oxide $(N_2 + O_2 \to 2NO)$ requires high temperatures to proceed,as it is an endothermic reaction $(\Delta H > 0)$.
$3$. The combustion of methane $(CH_4 + 2O_2 \to CO_2 + 2H_2O)$ is a highly exothermic process $(\Delta H < 0)$.
$4$. The formation of water $(H_2 + 1/2O_2 \to H_2O)$ is an exothermic process $(\Delta H < 0)$.
Therefore,the correct option is $B$.

(D) The given reaction is: $2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$
The enthalpy change of the reaction is calculated as: $\Delta_r H^{\circ} = \Sigma \Delta_f H^{\circ}(\text{products}) - \Sigma \Delta_f H^{\circ}(\text{reactants})$
$\Delta_r H^{\circ} = [\Delta_f H^{\circ}(Al_2O_3) + 2 \times \Delta_f H^{\circ}(Cr)] - [2 \times \Delta_f H^{\circ}(Al) + \Delta_f H^{\circ}(Cr_2O_3)]$
Since $Al$ and $Cr$ are in their standard elemental states,their enthalpy of formation is $0$.
$\Delta_r H^{\circ} = [-1596 + 2(0)] - [2(0) + (-1134)]$
$\Delta_r H^{\circ} = -1596 + 1134 = -462 \, kJ$

(A) The molar mass of propane $(C_3H_8)$ is $(3 \times 12.01) + (8 \times 1.008) = 44.11 \, g/mol$,which we approximate to $44 \, g/mol$.
The balanced chemical equation is: ${C_3H_8} + 5{O_2} \to 3{CO_2} + 4{H_2O}; \Delta{H^o} = -2219 \, kJ/mol$.
This means the combustion of $1 \, mol$ $(44 \, g)$ of propane releases $2219 \, kJ$ of heat.
For $15.5 \, g$ of propane,the heat released is calculated as:
$\text{Heat} = \frac{15.5 \, g}{44 \, g/mol} \times 2219 \, kJ/mol = 0.35227 \, mol \times 2219 \, kJ/mol \approx 781.7 \, kJ$.

(A) The enthalpy change of the reaction is given by: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$
Substituting the given values:
$-57.8 = \Delta H_f^o((CH_3)_3C-OCH_3) - [\Delta H_f^o((CH_3)_2C=CH_2) + \Delta H_f^o(CH_3OH)]$
$-57.8 = -313.6 - [\Delta H_f^o((CH_3)_2C=CH_2) + (-238.7)]$
$-57.8 = -313.6 - \Delta H_f^o((CH_3)_2C=CH_2) + 238.7$
$\Delta H_f^o((CH_3)_2C=CH_2) = -313.6 + 238.7 + 57.8$
$\Delta H_f^o((CH_3)_2C=CH_2) = -17.1 \ kJ \ mol^{-1}$

(B) Given reactions:
$(i) \ C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H_1 = -393.5 \ kJ$
$(ii) \ C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H_2 = -1410.9 \ kJ$
$(iii) \ H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H_3 = -285.8 \ kJ$
Target reaction: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)}$
To obtain the target reaction,perform: $2 \times (i) + 2 \times (iii) - (ii)$
$\Delta H = 2 \times (-393.5) + 2 \times (-285.8) - (-1410.9)$
$\Delta H = -787.0 - 571.6 + 1410.9$
$\Delta H = -1358.6 + 1410.9 = 52.3 \ kJ$

(A) The combustion reactions are:
$(i) \ C(s) + O_2(g) \longrightarrow CO_2(g) \quad \Delta H = x \, kJ \, mol^{-1}$
$(ii) \ S(s) + O_2(g) \longrightarrow SO_2(g) \quad \Delta H = y \, kJ \, mol^{-1}$
$(iii) \ CS_2(l) + 3O_2(g) \longrightarrow CO_2(g) + 2SO_2(g) \quad \Delta H = z \, kJ \, mol^{-1}$
We need the heat of formation of $CS_2$:
$(iv) \ C(s) + 2S(s) \longrightarrow CS_2(l) \quad \Delta H_f = ?$
To obtain equation $(iv)$,perform: $(i) + 2 \times (ii) - (iii)$
$\Delta H_f = x + 2y - z \, kJ \, mol^{-1}$

(A) The enthalpy of formation of $N_2O$ (experimental) is $\Delta_f H = 82 \, kJ \, mol^{-1}$.
The reaction is: $N_2 + \frac{1}{2} O_2 \longrightarrow N_2O$.
Using bond energies: $(N \equiv N) + \frac{1}{2}(O = O) \longrightarrow (N = N) + (N = O)$.
$\Delta_f H^{\circ} = \Sigma B.E._{\text{reactants}} - \Sigma B.E._{\text{products}}$.
$\Delta_f H^{\circ} = [946 + \frac{1}{2}(498)] - [418 + 607] = [946 + 249] - [1025] = 1195 - 1025 = 170 \, kJ \, mol^{-1}$.
Resonance energy = $\Delta_f H_{\text{experimental}} - \Delta_f H_{\text{calculated}}$.
Resonance energy = $82 - 170 = -88 \, kJ \, mol^{-1}$.

(B) The reaction is: $NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}$
The enthalpy of reaction is given by: $\Delta H^o = [\sum \Delta H_f^o (\text{products})] - [\sum \Delta H_f^o (\text{reactants})]$
$\Delta H^o = [\frac{1}{2} \Delta H_f^o (N_2) + \Delta H_f^o (CO_2)] - [\Delta H_f^o (NO) + \Delta H_f^o (CO)]$
Since $\Delta H_f^o (N_2) = 0$ (standard state of an element),we have:
$-372.2 = [0 + (-393.5)] - [\Delta H_f^o (NO) + (-110.5)]$
$-372.2 = -393.5 - \Delta H_f^o (NO) + 110.5$
$-372.2 = -283.0 - \Delta H_f^o (NO)$
$\Delta H_f^o (NO) = -283.0 + 372.2 = 89.2 \, kJ \, mol^{-1}$

(A) The calorific value is calculated by dividing the heat of combustion by the molar mass of the substance.
$Molar \, mass \, of \, CH_4 = 12 + (4 \times 1) = 16 \, g \, mol^{-1}$.
$\text{Calorific value} = \frac{\text{Heat of combustion}}{\text{Molar mass}} = \frac{809 \, kJ \, mol^{-1}}{16 \, g \, mol^{-1}} = 50.6 \, kJ \, g^{-1}$.

(B) The enthalpy of neutralization of a strong acid with a strong base is $-57.3 \, kJ/equiv$.
For a weak base like $NH_4OH$, the enthalpy of neutralization is given by: $\Delta H_{neut} = \Delta H_{ionization} + \Delta H_{neutralization(strong)}$.
Given $\Delta H_{neut} = -51.40 \, kJ/equiv$ and $\Delta H_{neutralization(strong)} = -57.3 \, kJ/equiv$.
Therefore, $\Delta H_{ionization} = \Delta H_{neut} - \Delta H_{neutralization(strong)}$.
$\Delta H_{ionization} = -51.40 - (-57.3) = 5.9 \, kJ$.

(B) The given reactions are:
$(i) S_R + O_{2(g)} \to SO_{2(g)}; \Delta H_1 = -296.90 \ kJ$
$(ii) S_M + O_{2(g)} \to SO_{2(g)}; \Delta H_2 = -299.40 \ kJ$
We need to find the enthalpy change for the transition: $S_R \to S_M$.
Subtracting equation $(ii)$ from equation $(i)$:
$(S_R + O_{2(g)}) - (S_M + O_{2(g)}) \to SO_{2(g)} - SO_{2(g)}$
$S_R - S_M \to 0$
$S_R \to S_M$
Therefore,$\Delta H_{transition} = \Delta H_1 - \Delta H_2 = -296.90 - (-299.40) = 2.5 \ kJ$.

(C) For $NH_3$: $NH_{3(g)} \rightarrow N_{(g)} + 3H_{(g)}$,$\Delta H = 150 \ kJ \ mol^{-1}$.
Since there are $3$ $N-H$ bonds,the bond enthalpy of $N-H$ is $\frac{150}{3} = 50 \ kJ \ mol^{-1}$.
For $N_2H_4$: $N_2H_{4(g)} \rightarrow 2N_{(g)} + 4H_{(g)}$,$\Delta H = 310 \ kJ \ mol^{-1}$.
The atomization enthalpy is the sum of bond enthalpies: $\Delta H_{atom} = 1 \times \Delta H(N-N) + 4 \times \Delta H(N-H)$.
Substituting the values: $310 = \Delta H(N-N) + 4(50)$.
$310 = \Delta H(N-N) + 200$.
$\Delta H(N-N) = 310 - 200 = 110 \ kJ \ mol^{-1}$.

(C) The enthalpy of combustion of carbon is given as $\Delta_{c}H = -393 \, kJ \, mol^{-1}$.
This means that the combustion of $1 \, mol$ $(12 \, g)$ of carbon releases $393 \, kJ$ of heat.
Therefore,the heat produced by the combustion of $1 \, g$ of carbon is calculated as:
$\text{Heat} = \frac{393 \, kJ}{12 \, g} = 32.75 \, kJ \, g^{-1}$.
Thus,the heat produced by $1 \, g$ of carbon is $32.75 \, kJ$.

(B) The given thermochemical equation is: $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$.
From the stoichiometry of the reaction,the formation of $2 \ mol$ of $HCl$ releases $184.6 \ kJ$ of heat.
Therefore,the enthalpy change for the formation of $1 \ mol$ of $HCl$ is $\Delta H = \frac{-184.6 \ kJ}{2} = -92.3 \ kJ/mol$.
For the formation of $5.67 \ mol$ of $HCl$,the enthalpy change is:
$\Delta H_{total} = 5.67 \ mol \times (-92.3 \ kJ/mol) \approx -523.34 \ kJ$.
Rounding to the nearest integer,the value is $-523 \ kJ$.

(C) The target reaction is the formation of $Fe_2O_3$ from its elements: $2Fe + 3/2 O_2 \to Fe_2O_3$.
Given equations:
$(i) \ Fe + 1/2 O_2 \to FeO, \Delta H_1 = -x \, kcal$
$(ii) \ 2FeO + 1/2 O_2 \to Fe_2O_3, \Delta H_2 = -y \, kcal$
To obtain the target reaction,multiply equation $(i)$ by $2$ and add it to equation $(ii)$:
$2 \times (Fe + 1/2 O_2 \to FeO) \implies 2Fe + O_2 \to 2FeO, \Delta H = -2x \, kcal$
Adding this to $(ii)$:
$(2Fe + O_2) + (2FeO + 1/2 O_2) \to 2FeO + Fe_2O_3$
$2Fe + 3/2 O_2 \to Fe_2O_3, \Delta H = -2x + (-y) = -(2x + y) \, kcal$
Thus,the heat of formation is $2x + y \, kcal$.

(A) The decomposition reaction of $HgO$ is:
$HgO \longrightarrow Hg + \frac{1}{2} O_2$
Since the formation enthalpy is $\Delta _f H^o = -90.5 \, kJ \, mol^{-1}$,the enthalpy of decomposition is $\Delta H = +90.5 \, kJ \, mol^{-1}$.
This means $90.5 \, kJ$ of heat is required to produce $1 \, mol$ of $Hg$.
The molar mass of $Hg$ is $200.59 \, g \, mol^{-1}$ (approximated as $200 \, g \, mol^{-1}$ for calculation).
So,$90.5 \, kJ$ of heat produces $200 \, g$ of $Hg$.
For $1600 \, kJ$ of heat,the amount of $Hg$ produced is:
$\text{Mass of } Hg = \frac{1600 \, kJ \times 200 \, g}{90.5 \, kJ} = 3535.9 \, g \approx 3.536 \, kg$.

(A) $i) \, C + O_2 \longrightarrow CO_2$
Heat for $12 \, g$ ($1$ mole) of $C = \frac{x}{0.8} \times 12 = 15x \, cal$.
$ii) \, C + \frac{1}{2} O_2 \longrightarrow CO$
Heat for $12 \, g$ ($1$ mole) of $C = \frac{y}{0.8} \times 12 = 15y \, cal$.
Subtracting $(ii)$ from $(i)$:
$CO + \frac{1}{2} O_2 \longrightarrow CO_2$
Heat of reaction for $1 \, mole$ $(28 \, g)$ of $CO = (15x - 15y) \, cal$.
Heat released for $1.86 \, g$ of $CO = \frac{15(x - y)}{28} \times 1.86 \approx 0.996(x - y) \approx (x - y) \, cal$.

(C) The reaction for the formation of water vapor is: $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(g)}; \Delta H_1$
The reaction for the formation of liquid water is: $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)}; \Delta H_2$
Subtracting the second equation from the first gives: $H_2O_{(l)} \to H_2O_{(g)}; \Delta H = \Delta H_1 - \Delta H_2$
Since the conversion of liquid water to water vapor (vaporization) is an endothermic process,$\Delta H > 0$.
Therefore,$\Delta H_1 - \Delta H_2 > 0$,which implies $\Delta H_1 > \Delta H_2$.

(C) The formation reaction of ethylene is: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)} \quad \Delta H_f = ?$
Given equations:
$(i) C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H_1 = -393.5\,kJ$
$(ii) H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)} \quad \Delta H_2 = -286.2\,kJ$
$(iii) C_2H_{4(g)} + 3 O_{2(g)} \to 2 CO_{2(g)} + 2 H_2O_{(l)} \quad \Delta H_3 = -1410.8\,kJ$
To obtain the formation reaction,perform: $2 \times (i) + 2 \times (ii) - (iii)$
$\Delta H_f = 2(\Delta H_1) + 2(\Delta H_2) - (\Delta H_3)$
$\Delta H_f = 2(-393.5) + 2(-286.2) - (-1410.8)$
$\Delta H_f = -787.0 - 572.4 + 1410.8$
$\Delta H_f = -1359.4 + 1410.8 = +51.4\,kJ\,mol^{-1}$

(D) The chemical equation for the formation of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \longrightarrow HCl(g)$
The enthalpy of formation is given by: ${\Delta _f}H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$
$-90 = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - BE(H-Cl)$
$-90 = [\frac{1}{2}(430) + \frac{1}{2}(240)] - BE(H-Cl)$
$-90 = [215 + 120] - BE(H-Cl)$
$-90 = 335 - BE(H-Cl)$
$BE(H-Cl) = 335 + 90 = 425 \, kJ \, mol^{-1}$

(A) The enthalpy of reaction is given by the sum of bond energies of reactants minus the sum of bond energies of products: $\Delta H = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$.
For the reaction $2H_2 + O_2 \to 2H_2O$,the bonds broken are $2(H-H)$ and $1(O=O)$,and the bonds formed are $4(O-H)$.
$\Delta H = [2 \times BE(H-H) + 1 \times BE(O=O)] - [4 \times BE(O-H)]$.
Substituting the given values: $-571 = [2 \times 435 + 498] - 4 \times BE(O-H)$.
$-571 = [870 + 498] - 4 \times BE(O-H)$.
$-571 = 1368 - 4 \times BE(O-H)$.
$4 \times BE(O-H) = 1368 + 571 = 1939$.
$BE(O-H) = \frac{1939}{4} = 484.75 \ kJ \ mol^{-1} \approx 484 \ kJ \ mol^{-1}$.

(A) Since the enthalpy of solution $\Delta H$ is negative,the process is exothermic.
Therefore,the heat released during the dissolution increases the temperature of the water.

(B) The given reaction is: $\frac{1}{2} X_2O_{(s)} \to X_{(s)} + \frac{1}{4} O_{2(g)}$,$\Delta H = 90 \ kJ$.
To find the enthalpy change for the reaction of $1 \ mol$ of $O_2$ with metal $X$,we first reverse the reaction:
$X_{(s)} + \frac{1}{4} O_{2(g)} \to \frac{1}{2} X_2O_{(s)}$,$\Delta H = -90 \ kJ$.
Now,to obtain the reaction for $1 \ mol$ of $O_2$,we multiply the entire equation by $4$:
$4X_{(s)} + O_{2(g)} \to 2X_2O_{(s)}$.
The new enthalpy change is $\Delta H = -90 \ kJ \times 4 = -360 \ kJ$.

(A) The reaction for the formation of one mole of water from gaseous atoms is: $2H_{(g)} + O_{(g)} \longrightarrow H_2O_{(g)}$.
In one mole of $H_2O$,there are two $O-H$ bonds.
The energy released during the formation of two $O-H$ bonds is equal to $2 \times \Delta H_{O-H}$.
Given $\Delta H_{O-H} = 109 \ kcal \ mol^{-1}$.
Therefore,energy released $= 2 \times 109 = 218 \ kcal \ mol^{-1}$.
Since energy is released,the enthalpy change is $\Delta H = -218 \ kcal \ mol^{-1}$.

(A) The reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)}$
Calculate $\Delta C_p$ for the reaction:
$\Delta C_p = C_{p(H_2O)} - [C_{p(H_2)} + \frac{1}{2} C_{p(O_2)}]$
$\Delta C_p = 33.58 - [28.84 + \frac{1}{2}(29.37)]$
$\Delta C_p = 33.58 - [28.84 + 14.685] = 33.58 - 43.525 = -9.945 \ J \ K^{-1} \ mol^{-1} = -0.009945 \ kJ \ K^{-1} \ mol^{-1}$
Using Kirchhoff's equation: $\Delta H_2 = \Delta H_1 + \Delta C_p (T_2 - T_1)$
$\Delta H_2 = -241.82 \ kJ \ mol^{-1} + (-0.009945 \ kJ \ K^{-1} \ mol^{-1}) \times (373 \ K - 298 \ K)$
$\Delta H_2 = -241.82 + (-0.009945 \times 75)$
$\Delta H_2 = -241.82 - 0.745875 \approx -242.57 \ kJ \ mol^{-1}$
Rounding to one decimal place,we get $-242.6 \ kJ \ mol^{-1}$.

(A) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.
Sublimation is the direct conversion of a solid to a vapor.
This process can be represented as a two-step process: Solid $\rightarrow$ Liquid (fusion) followed by Liquid $\rightarrow$ Vapor (vaporization).
Therefore,$\Delta H_{sublimation} = \Delta H_{fusion} + \Delta H_{vap}$.
Given $\Delta H_{fusion} = x$ and $\Delta H_{vap} = y$,we have $\Delta H_{sublimation} = x + y$.

(D) The given thermochemical equations are:
$C(s) + O_2(g) \to CO_2(g) \quad \Delta H_1 = -x \ kJ \dots (i)$
$H_2(g) + \frac{1}{2} O_2(g) \to H_2O(l) \quad \Delta H_2 = -y \ kJ \dots (ii)$
$CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l) \quad \Delta H_3 = z \ kJ \dots (iii)$
Note: Heat of combustion is energy released,so $\Delta H$ is negative. Given $z$ is the heat of combustion,$\Delta H_3 = -z \ kJ$.
We need the heat of formation of methane:
$C(s) + 2H_2(g) \to CH_4(g) \quad \Delta H_f = ?$
To obtain this,we perform: $(i) + 2 \times (ii) - (iii)$
$\Delta H_f = (-x) + 2(-y) - (-z) = -x - 2y + z \ kJ$.

(B) The required reaction is $C_6H_{10} + H_2 \to C_6H_{12}$,$\Delta H_1 = ? \dots (1)$
Given combustion reactions:
$H_2 + \frac{1}{2}O_2 \to H_2O$,$\Delta H_2 = -241 \, kJ/mol \dots (2)$
$C_6H_{10} + \frac{17}{2}O_2 \to 6CO_2 + 5H_2O$,$\Delta H_3 = -3800 \, kJ/mol \dots (3)$
$C_6H_{12} + 9O_2 \to 6CO_2 + 6H_2O$,$\Delta H_4 = -3920 \, kJ/mol \dots (4)$
Applying Hess's Law,the required reaction $(1)$ is obtained by $(2) + (3) - (4)$:
$\Delta H_1 = (\Delta H_2 + \Delta H_3) - \Delta H_4$
$\Delta H_1 = (-241 - 3800) - (-3920)$
$\Delta H_1 = -4041 + 3920 = -121 \, kJ/mol$.

(D) Let the bond energy of $A_2$ be $x$. Then,the bond energy of $AB$ is $x$ and the bond energy of $B_2$ is $0.5x$.
The reaction for the formation of $AB$ is:
$\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$; $\Delta H_f = -100 \, kJ \, mol^{-1}$.
The enthalpy of reaction is given by the sum of bond energies of reactants minus the sum of bond energies of products:
$\Delta H = [\frac{1}{2} BE(A_2) + \frac{1}{2} BE(B_2)] - [BE(AB)]$
Substituting the values:
$-100 = [\frac{1}{2}(x) + \frac{1}{2}(0.5x)] - x$
$-100 = 0.5x + 0.25x - x$
$-100 = -0.25x$
$x = \frac{100}{0.25} = 400 \, kJ \, mol^{-1}$.

(B) The standard enthalpy change for the reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
For the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$,the enthalpy change is:
$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.
Given $\Delta H_f^o(H_{2(g)}) = 0 \ kJ \ mol^{-1}$ (standard state of an element).
$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$.
$\Delta H^o = -352.3 - (-393.5) = 41.2 \ kJ$.

(A) The reaction is: $2H_2O_{2(l)} \to 2H_2O_{(l)} + O_{2(g)}$
$\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$\Delta H = [2 \times \Delta H_f(H_2O_{(l)}) + \Delta H_f(O_{2(g)})] - [2 \times \Delta H_f(H_2O_{2(l)})]$
Given: $\Delta H_f(H_2O_{(l)}) = -286 \ kJ/mol$,$\Delta H_f(H_2O_{2(l)}) = -188 \ kJ/mol$,and $\Delta H_f(O_{2(g)}) = 0 \ kJ/mol$ (standard state).
$\Delta H = [2 \times (-286) + 0] - [2 \times (-188)]$
$\Delta H = [-572] - [-376]$
$\Delta H = -572 + 376 = -196 \ kJ/mol$

(A) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} \to ICl_{(g)}$.
We can calculate the enthalpy of reaction $(\Delta_r H)$ using the given bond dissociation and sublimation energies:
$\Delta_r H = [\frac{1}{2} \Delta H_{sub}(I_2) + \frac{1}{2} BE(I-I) + \frac{1}{2} BE(Cl-Cl)] - [BE(I-Cl)]$
Substituting the given values:
$\Delta_r H = [\frac{1}{2}(62.76) + \frac{1}{2}(151.0) + \frac{1}{2}(242.3)] - [211.3]$
$\Delta_r H = [31.38 + 75.5 + 121.15] - 211.3$
$\Delta_r H = 228.03 - 211.3 = 16.73 \ kJ \ mol^{-1}$
Thus,the standard enthalpy of formation $\Delta_f H^o(ICl) \approx +16.8 \ kJ \ mol^{-1}$.

(B) We are given the following combustion reactions:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(l)}$; $\Delta H = -3270 \ kJ \ mol^{-1} \dots (i)$
$C(gr) + O_{2(g)} \to CO_{2(g)}$; $\Delta H = -394 \ kJ \ mol^{-1} \dots (ii)$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H = -286 \ kJ \ mol^{-1} \dots (iii)$
We need to find the enthalpy of formation of $C_6H_{6(l)}$:
$6C(gr) + 3H_{2(g)} \to C_6H_{6(l)}$; $\Delta H_f = ? \dots (iv)$
By performing the operation $6 \times (ii) + 3 \times (iii) - (i)$,we get:
$\Delta H_f = [6 \times (-394) + 3 \times (-286)] - (-3270)$
$\Delta H_f = [-2364 - 858] + 3270$
$\Delta H_f = -3222 + 3270 = +48 \ kJ \ mol^{-1}$

(B) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
For the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$,the expression is:
$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.
Given that $\Delta H_f^o(H_{2(g)}) = 0 \ kJ/mol$ (standard state of an element):
$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$.
$\Delta H^o = -352.3 - (-393.5) = 41.2 \ kJ$.

(C) Let the bond dissociation energies ($B$.$E$.) of $X_2$,$Y_2$,and $XY$ be $x \ kJ \ mol^{-1}$,$0.5x \ kJ \ mol^{-1}$,and $x \ kJ \ mol^{-1}$ respectively.
The chemical equation for the formation of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g)$.
The enthalpy of reaction is given by: $\Delta H = \Sigma(B.E.)_{\text{reactants}} - \Sigma(B.E.)_{\text{products}}$.
Substituting the values: $-200 = [\frac{1}{2} \times (x) + \frac{1}{2} \times (0.5x)] - [1 \times (x)]$.
$-200 = [0.5x + 0.25x] - x$.
$-200 = 0.75x - x$.
$-200 = -0.25x$.
$x = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.
Therefore,the bond dissociation energy of $X_2$ is $800 \ kJ \ mol^{-1}$.

(D) The formation reaction of ethane is: $2 C(graphite) + 3 H_{2(g)} \longrightarrow C_{2}H_{6(g)}$.
Using the formula for heat of formation from heats of combustion: $\Delta_{f} H^{\circ} = \sum \Delta H_{comb}(reactants) - \sum \Delta H_{comb}(products)$.
$\Delta_{f} H^{\circ} = [2 \times \Delta H_{comb}(C) + 3 \times \Delta H_{comb}(H_{2})] - [\Delta H_{comb}(C_{2}H_{6})]$.
Substituting the given values: $\Delta_{f} H^{\circ} = [2 \times (-393.5) + 3 \times (-286)] - [-1560]$.
$\Delta_{f} H^{\circ} = [-787 - 858] + 1560$.
$\Delta_{f} H^{\circ} = -1645 + 1560 = -85 \; kJ/mol$.

(D) The enthalpy of atomisation $(\Delta H_{atom})$ is the energy required to convert one mole of a substance into its gaseous atoms.
For $Br_{2(l)}$,the process is:
$Br_{2(l)}$ $\xrightarrow{\Delta H_{vap}} Br_{2(g)}$ $\xrightarrow{\Delta H_{BE}} 2Br_{(g)}$
Therefore,$\Delta H_{atom} = \Delta H_{vap} + \Delta H_{BE}$.
Given $\Delta H_{atom} = x$ and $\Delta H_{BE} = y$,we have $x = \Delta H_{vap} + y$.
Since the enthalpy of vaporisation $(\Delta H_{vap})$ is always positive,it follows that $x > y$.

(A) The combustion reactions are:
$(i) CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_{2}O_{(l)}; \Delta_cH = -890.3 \, kJ \, mol^{-1}$
$(ii) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta_cH = -393.5 \, kJ \, mol^{-1}$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(l)}; \Delta_cH = -285.8 \, kJ \, mol^{-1}$
The formation reaction of $CH_{4(g)}$ is:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
Using Hess's Law:
$\Delta_fH_{CH_4} = \Delta_cH_{C(s)} + 2(\Delta_cH_{H_2(g)}) - \Delta_cH_{CH_4(g)}$
$\Delta_fH_{CH_4} = [-393.5 + 2(-285.8) - (-890.3)] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = [-393.5 - 571.6 + 890.3] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = -74.8 \, kJ \, mol^{-1}$
Thus,the enthalpy of formation is $-74.8 \, kJ \, mol^{-1}$,which corresponds to option $(i)$.