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

(B) Let the bond dissociation energies be $E_{XY} = a$,$E_{X_2} = a$,and $E_{Y_2} = 0.5a$.
The reactions are:
$1$) $XY(g) \to X(g) + Y(g)$; $\Delta H_1 = a$
$2$) $X_2(g) \to 2X(g)$; $\Delta H_2 = a$
$3$) $Y_2(g) \to 2Y(g)$; $\Delta H_3 = 0.5a$
The formation reaction of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \to XY(g)$.
Using Hess's Law: $\Delta_f H = [\frac{1}{2} \Delta H_2 + \frac{1}{2} \Delta H_3] - \Delta H_1$
$-200 = [\frac{1}{2}a + \frac{1}{2}(0.5a)] - a$
$-200 = 0.5a + 0.25a - a$
$-200 = -0.25a$
$a = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.

(C) The statement provided is the definition of $Hess's$ Law of Constant Heat Summation. $Hess's$ Law states that the total enthalpy change for a chemical reaction is the same,whether the reaction occurs in one step or in several steps. Therefore,the correct scientist is $Hess$.

(D) The given reaction is $2CO_{(g)} + O_{2(g)} \rightarrow 2CO_{2(g)}$.
This reaction involves the burning of carbon monoxide $(CO)$ in the presence of oxygen $(O_2)$ to produce carbon dioxide $(CO_2)$.
Since the substance is reacting with oxygen to release energy,this is a combustion reaction.
Therefore,the enthalpy change associated with this reaction is known as the enthalpy of combustion.

(A) The target reaction is: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$.
We can obtain this by manipulating the given equations:
$(1) 2Zn + O_2 \rightarrow 2ZnO, \Delta G^o = -616 \, J$
$(2) 2ZnS \rightarrow 2Zn + S_2, \Delta G^o = +293 \, J$ (Reversing the second given equation)
$(3) S_2 + 2O_2 \rightarrow 2SO_2, \Delta G^o = -408 \, J$
Adding these three equations:
$(2Zn + O_2) + (2ZnS) + (S_2 + 2O_2) \rightarrow (2ZnO) + (2Zn + S_2) + (2SO_2)$
Canceling common terms ($2Zn$ and $S_2$):
$2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$
The total $\Delta G^o = (-616) + (+293) + (-408) = -731 \, J$.

(C) The reaction for the formation of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$.
The enthalpy of formation is given by: $\Delta_f H = [\frac{1}{2} BE(H-H) + \frac{1}{2} BE(Cl-Cl)] - BE(H-Cl)$.
Substituting the given values: $-90 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 240] - BE(H-Cl)$.
$-90 = [215 + 120] - BE(H-Cl)$.
$-90 = 335 - BE(H-Cl)$.
$BE(H-Cl) = 335 + 90 = 425 \, kJ/mol$.

(C) The reaction is: $C_{2}H_{4(g)} + H_{2(g)} \rightarrow C_{2}H_{6(g)}$.
Enthalpy change $\Delta H$ is calculated as: $\Delta H = \sum (B.E.)_{Reactants} - \sum (B.E.)_{Products}$.
Reactants: $1 \times (C=C) + 4 \times (C-H) + 1 \times (H-H) = 145 + 4(99) + 103 = 145 + 396 + 103 = 644 \, Kcal \, mol^{-1}$.
Products: $1 \times (C-C) + 6 \times (C-H) = 80 + 6(99) = 80 + 594 = 674 \, Kcal \, mol^{-1}$.
$\Delta H = 644 - 674 = -30 \, Kcal \, mol^{-1}$.

(B) The heat of neutralization for a strong acid and a strong base is constant,which is approximately $-13.7 \, kcal \, eq^{-1}$.
For $1 \, M \, HCl$ (a monobasic acid),$1 \, L$ contains $1 \, mole$ of $H^+$ ions. Thus,heat liberated $x = 1 \times 13.7 = 13.7 \, kcal$.
For $1 \, M \, H_2SO_4$ (a dibasic acid),$1 \, L$ contains $2 \, moles$ of $H^+$ ions. Thus,heat liberated $y = 2 \times 13.7 = 27.4 \, kcal$.
Comparing $x$ and $y$,we get $y = 2x$ or $x = 0.5y$.

(A) The enthalpy change of a reaction is calculated using bond energies as: $\Delta H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$.
For the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$:
$\Delta H = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$.
Substituting the given values:
$\Delta H = [435 + 192] - [2 \times 364]$.
$\Delta H = 627 - 728$.
$\Delta H = -101 \, kJ \, mol^{-1}$.

(A) The enthalpy of vaporization is the energy required to convert liquid water to gaseous water: $H_2O(l) \rightarrow H_2O(g)$.
We can obtain this reaction by subtracting the first equation from the second equation:
$(H_2(g) + 1/2 O_2(g)$ $\rightarrow H_2O(g)) - (H_2(g) + 1/2 O_2(g)$ $\rightarrow H_2O(l))$
$\Delta H_{vap} = \Delta H_2 - \Delta H_1$
$\Delta H_{vap} = (-241.84 \, KJ \, mol^{-1}) - (-285.77 \, KJ \, mol^{-1})$
$\Delta H_{vap} = -241.84 + 285.77 = +43.93 \, KJ \, mol^{-1}$.

(B) The given combustion reactions are:
$(i)$ $C_{(s)} + O_{2(g)} \to CO_{2(g)} \, \Delta H_1 = -394 \, kJ \, mol^{-1}$
(ii) $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \, \Delta H_2 = -285 \, kJ \, mol^{-1}$
We need to find the heat of formation of $CO$:
(iii) $C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)} \, \Delta H_f = ?$
Using Hess's Law:
$(iii) = (i) - (ii)$
$\Delta H_f = \Delta H_1 - \Delta H_2$
$\Delta H_f = -394 - (-285) = -394 + 285 = -109 \, kJ \, mol^{-1}$

(C) Given:
$\Delta H_f(NH_3) = -46.0 \ kJ \ mol^{-1}$
$\Delta H_f(H_2O) = -286.0 \ kJ \ mol^{-1}$
$\Delta H_f(N_2) = 0 \ kJ \ mol^{-1}$ (element in standard state)
$\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$ (element in standard state)
The enthalpy change for the reaction is given by:
$\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$\Delta H_{rxn} = [2 \times \Delta H_f(N_2) + 6 \times \Delta H_f(H_2O)] - [4 \times \Delta H_f(NH_3) + 3 \times \Delta H_f(O_2)]$
Substituting the values:
$\Delta H_{rxn} = [2(0) + 6(-286.0)] - [4(-46.0) + 3(0)]$
$\Delta H_{rxn} = [-1716] - [-184]$
$\Delta H_{rxn} = -1716 + 184 = -1532 \ kJ$

(A) The given equations are:
$(1) \ S + \frac{3}{2} O_2 \to SO_3, \ \Delta H_1 = -2x \ \text{kcal}$
$(2) \ SO_2 + \frac{1}{2} O_2 \to SO_3, \ \Delta H_2 = -y \ \text{kcal}$
To find the heat of formation of $SO_2$,we need the reaction: $S + O_2 \to SO_2$.
Subtract equation $(2)$ from equation $(1)$:
$(S + \frac{3}{2} O_2) - (SO_2 + \frac{1}{2} O_2) \to SO_3 - SO_3$
$S + O_2 - SO_2 \to 0$
$S + O_2 \to SO_2$
The enthalpy change is $\Delta H = \Delta H_1 - \Delta H_2 = (-2x) - (-y) = y - 2x \ \text{kcal}$.

(A) To find the enthalpy change for the reaction $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$,we can manipulate the given equations:
$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H_1 = -8.4 \, Kcal$
$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H_2 = -17.3 \, Kcal$
$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H_3 = -12.5 \, Kcal$
$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H_4 = +3.9 \, Kcal$
We need the target equation: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$.
Adding equations $(i)$,$(ii)$,and $(iii)$ gives:
$NH_3(g) + HCl(g) + 2aq \rightarrow NH_4Cl(aq)$,$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -8.4 - 17.3 - 12.5 = -38.2 \, Kcal$.
Now,subtract equation $(iv)$ from this result (or add the reverse of equation $(iv)$):
$NH_4Cl(aq) \rightarrow NH_4Cl(s) + aq$,$\Delta H = -3.9 \, Kcal$.
Adding the equations:
$NH_3(g) + HCl(g) + 2aq + NH_4Cl(aq) \rightarrow NH_4Cl(aq) + NH_4Cl(s) + aq$
$NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$
$\Delta H_{total} = -38.2 + (-3.9) = -42.1 \, Kcal$.

(A) According to Hess's Law,the reaction $(iv)$ is the sum of reactions $(i)$,$(ii)$,and $(iii)$.
$\Delta H_{(iv)} = \Delta H_{(i)} + \Delta H_{(ii)} + \Delta H_{(iii)}$
$-x = 131 + (-282) + (-242)$
$-x = 131 - 524$
$-x = -393$
$x = 393 \ kJ$

(B) The standard molar enthalpy of formation $(\Delta_f H^\circ)$ of an element in its most stable state at $298 \ K$ and $1 \ bar$ pressure is defined as zero.
Among the given options,$Cl_{2(g)}$ is the most stable form of chlorine at $298 \ K$,so its enthalpy of formation is zero.
$Br_{2(l)}$ is the most stable form of bromine at $298 \ K$,but the option provided was $Br_{2(l)}$ (liquid),which is correct; however,$Cl_{2(g)}$ is also a standard element in its stable state.
$H_2O_{(g)}$ and $CH_{4(g)}$ are compounds,so their enthalpy of formation is not zero.

(B) The combustion reaction of ethanol is:
$C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)$
The change in the number of gaseous moles is $\Delta n_g = n_p(g) - n_r(g) = 2 - 3 = -1$.
The heat of combustion in a bomb calorimeter is the change in internal energy,$\Delta U = -670.48 \, kcal \, mol^{-1}$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
Given $R = 1.987 \times 10^{-3} \, kcal \, K^{-1} \, mol^{-1} \approx 2 \times 10^{-3} \, kcal \, K^{-1} \, mol^{-1}$ and $T = 298 \, K$.
$\Delta H = -670.48 + (-1) \times (2 \times 10^{-3}) \times 298$
$\Delta H = -670.48 - 0.596 \approx -671.08 \, kcal \, mol^{-1}$.

(C) The reaction for the formation of $ICl_{(g)}$ from its atoms is: $I_{(g)} + Cl_{(g)} \to ICl_{(g)}$
The enthalpy change for this reaction is given by the difference between the sum of bond energies of reactants and products: $\Delta H = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$
Given that $\Delta H_f^o(ICl) = 17.57 \, J \, mol^{-1}$,$\Delta H_f^o(Cl) = 121.34 \, J \, mol^{-1}$,and $\Delta H_f^o(I) = 106.96 \, J \, mol^{-1}$.
For the reaction $I_{(g)} + Cl_{(g)} \to ICl_{(g)}$,the enthalpy change is: $\Delta H = \Delta H_f^o(ICl) - [\Delta H_f^o(I) + \Delta H_f^o(Cl)]$
$\Delta H = 17.57 - (106.96 + 121.34) = 17.57 - 228.3 = -210.73 \, J \, mol^{-1}$
Since bond dissociation energy $(B.E.)$ is the energy required to break the bond,$B.E.(I-Cl) = -\Delta H = 210.73 \, J \, mol^{-1}$.

(B) The enthalpy of combustion of carbon is defined as the enthalpy change when $1 \ mol$ of carbon is completely burnt in oxygen to form $CO_2$.
The reaction is: $C(s) + O_2(g) \to CO_2(g)$.
The enthalpy of formation of $CO_2$ is given as $-94.3 \ kcal/mol$.
Since the enthalpy of formation of $CO_2$ is equivalent to the enthalpy of combustion of carbon,the value is $-94.3 \ kcal$.

(A) The reaction $Cl_{2(g)} \rightarrow 2Cl_{(g)}$ involves the breaking of a chemical bond between two chlorine atoms.
Bond dissociation is an endothermic process because energy must be supplied to break the bond.
Therefore,the enthalpy change $\Delta H$ for this reaction is positive $(> 0)$.

(A) The enthalpy change for the reaction between a strong acid and a strong base,such as $NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$,is known as the heat of neutralization.

(B) The Kirchhoff equation relates the change in the enthalpy of a reaction to the change in temperature. The equation is given by: $\Delta H_2 = \Delta H_1 + \int_{T_1}^{T_2} \Delta C_p \ dT$. Therefore,temperature is the factor that affects the heat of reaction.

(B) The heat of neutralization for a reaction between a strong acid and a strong base is approximately $-57.1 \ kJ \ mol^{-1}$ (or $-13.7 \ kcal \ mol^{-1}$),which is often rounded to $56 \ kJ \ mol^{-1}$.
$HNO_3$ is a strong acid and $LiOH$ is a strong base.
Therefore,the reaction between $HNO_3$ and $LiOH$ will have a heat of neutralization of $56 \ kJ \ mol^{-1}$.

(B) The heat of formation of $KOH_{(s)}$ corresponds to the reaction: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow KOH_{(s)}$.
To obtain this,we manipulate the given equations:
$(i) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H_1 = -68.39 \, kcal$
$(ii) \ K_{(s)} + H_2O_{(l)} \rightarrow KOH_{(aq)} + \frac{1}{2}H_{2(g)} ; \Delta H_2 = -48.0 \, kcal$
$(iii) \ KOH_{(s)} \rightarrow KOH_{(aq)} ; \Delta H_3 = -14.0 \, kcal$
Adding $(i)$ and $(ii)$ gives: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)}$ $\rightarrow KOH_{(aq)} ; \Delta H = \Delta H_1 + \Delta H_2 = -68.39 - 48.0 = -116.39 \, kcal$.
Subtracting $(iii)$ from this result: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)}$ $\rightarrow KOH_{(s)} ; \Delta H = (-116.39) - (-14.0) = -68.39 - 48.0 + 14.0 \, kcal$.

(B) The reaction $3O_2 \rightarrow 2O_3$ has a positive enthalpy change $(\Delta H > 0)$,which means it is an endothermic process.
Since the formation of $O_3$ from $O_2$ requires energy,$O_3$ is at a higher energy state than $O_2$.
According to the principle of stability,substances at higher energy levels are less stable.
Therefore,$O_3$ is less stable than $O_2$,and it tends to decompose back into $O_2$ to achieve a more stable,lower-energy state.

(B) The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is: $H^+ + OH^- \rightarrow H_2O$,$\Delta H = -57.1 \ kJ/mol$ (often approximated as $56 \ kJ/mol$).
Since $0.2 \ mol$ of $KOH$ is the limiting reagent,it will neutralize only $0.2 \ mol$ of $HNO_3$.
The heat of neutralization is calculated as: $\text{Heat} = \text{moles of water formed} \times 56 \ kJ/mol$.
$\text{Heat} = 0.2 \ mol \times 56 \ kJ/mol = 11.2 \ kJ$.

(A) The formation reaction for $CS_2$ is: $C(s) + 2S(s) \rightarrow CS_2(l)$.
Given equations:
$(1) \ C + O_2 \rightarrow CO_2$ : $\Delta H_1 = -395 \ kJ$
$(2) \ S + O_2 \rightarrow SO_2$ : $\Delta H_2 = -295 \ kJ$
$(3) \ CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ : $\Delta H_3 = -1110 \ kJ$
To obtain the target equation,perform: $(1) + 2 \times (2) - (3)$:
$\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$
$\Delta H_f = -395 + 2(-295) - (-1110)$
$\Delta H_f = -395 - 590 + 1110$
$\Delta H_f = 1110 - 985 = 125 \ kJ/mol$.

(D) The enthalpy of neutralization of a strong acid and a strong base is $-57.4 \, kJ/mol$.
For a weak acid like $HCN$,the enthalpy of neutralization is given by the sum of the enthalpy of ionization of the weak acid and the enthalpy of neutralization of $H^+$ and $OH^-$.
Let $\Delta H_{ion}$ be the enthalpy of ionization of $HCN$.
$\Delta H_{neut} = \Delta H_{ion} + \Delta H_{H^+ + OH^-} \implies -12.13 = \Delta H_{ion} + (-57.4)$.
$\Delta H_{ion} = -12.13 + 57.4 = 45.27 \, kJ/mol$.
Rounding to the nearest provided option,the value is $45.19 \, kJ/mol$.

(A) Given reactions:
$(i) \, C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) \, ; \, \Delta H = -783 \, kcal$
$(ii) \, C(s) + O_2(g) \rightarrow CO_2(g) \, ; \, \Delta H = -97 \, kcal$
$(iii) \, H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \, ; \, \Delta H = -68 \, kcal$
We need to find $\Delta H_f$ for: $6C(s) + 3H_2(g) \rightarrow C_6H_6(l)$
Using Hess's Law:
$6 \times (ii)$ $\Rightarrow 6C(s) + 6O_2(g)$ $\rightarrow 6CO_2(g) \, ; \, \Delta H = 6 \times (-97) = -582 \, kcal$
$3 \times (iii)$ $\Rightarrow 3H_2(g) + \frac{3}{2} O_2(g)$ $\rightarrow 3H_2O(l) \, ; \, \Delta H = 3 \times (-68) = -204 \, kcal$
Reverse of $(i)$ $\Rightarrow 6CO_2(g) + 3H_2O(l)$ $\rightarrow C_6H_6(l) + \frac{15}{2} O_2(g) \, ; \, \Delta H = +783 \, kcal$
Adding these equations:
$6C(s) + 3H_2(g) \rightarrow C_6H_6(l)$
$\Delta H_f = -582 - 204 + 783 = -3 \, kcal$

(B) The stability of a compound is inversely proportional to its heat of formation (enthalpy of formation).
Compounds with more negative (lower) values of enthalpy of formation are more stable.
Since the heat of formation of $Y$ $(-156 \ kJ)$ is more negative than that of $X$ $(-84 \ kJ)$,$Y$ is more stable than $X$.
Therefore,$X$ is less stable than $Y$.

(C) The combustion reaction for methane is:
$CH_{4_{(g)}} + 2O_{2_{(g)}} \to CO_{2_{(g)}} + 2H_2O_{(l)}$; $\Delta H = -210.8 \, kcal$
The enthalpy of reaction is given by:
$\Delta H = [\Delta H_f(CO_{2_{(g)}}) + 2 \times \Delta H_f(H_2O_{(l)})] - [\Delta H_f(CH_{4_{(g)}}) + 2 \times \Delta H_f(O_{2_{(g)}})]$
Substituting the given values (where $\Delta H_f(O_{2_{(g)}}) = 0$):
$-210.8 = [-94.2 + 2 \times (-68.3)] - \Delta H_f(CH_{4_{(g)}})$
$-210.8 = -94.2 - 136.6 - \Delta H_f(CH_{4_{(g)}})$
$-210.8 = -230.8 - \Delta H_f(CH_{4_{(g)}})$
$\Delta H_f(CH_{4_{(g)}}) = -230.8 + 210.8 = -20.0 \, kcal$

(C) The chemical equation for the formation of $HCl$ is: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \to HCl_{(g)}$
The enthalpy of reaction is calculated as: $\Delta_fH = \Sigma \text{Bond energies of reactants} - \Sigma \text{Bond energies of products}$
$\Delta_fH = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - [BE(H-Cl)]$
$\Delta_fH = [\frac{1}{2}(434) + \frac{1}{2}(242)] - 431$
$\Delta_fH = [217 + 121] - 431 = 338 - 431 = -93 \ kJ \ mol^{-1}$

(B) The heat of neutralization of a strong acid with a strong base is defined as the heat released when $1 \ gram-equivalent$ of $H^+$ ions reacts with $1 \ gram-equivalent$ of $OH^-$ ions to form $1 \ mol$ of $H_2O$.
This value is constant at $-13.7 \ Kcal$ per equivalent.
$H_2SO_4$ is a dibasic acid,meaning $1 \ mol$ of $H_2SO_4$ provides $2 \ moles$ of $H^+$ ions (i.e.,$2 \ gram-equivalents$).
Therefore,the heat of neutralization for $1 \ mol$ of $H_2SO_4$ is $2 \times (-13.7 \ Kcal) = -27.4 \ Kcal$.

(B) The enthalpy of solution of an anhydrous salt is given by the relation: $\Delta H_{\text{sol(anhydrous)}} = \Delta H_{\text{hydration}} + \Delta H_{\text{sol(hydrated)}}$.
Given: $\Delta H_{\text{sol(anhydrous)}} = -21.0 \, J \, \text{mol}^{-1}$ (since heat is released,it is exothermic).
Given: $\Delta H_{\text{hydration}} = -29.4 \, J \, \text{mol}^{-1}$.
Substituting the values: $-21.0 = -29.4 + \Delta H_{\text{sol(hydrated)}}$.
$\Delta H_{\text{sol(hydrated)}} = -21.0 + 29.4 = +8.4 \, J \, \text{mol}^{-1}$.

(A) The total enthalpy change $\Delta H$ for the process $\frac{1}{2} Cl_{2(g)} \to Cl^-_{(aq)}$ is the sum of the enthalpy changes of the individual steps:
$\Delta H = \frac{1}{2} \Delta_{diss} H^\Theta + \Delta_{eg} H^\Theta + \Delta_{hyd} H^\Theta$
Substituting the given values:
$\Delta H = \frac{1}{2} (240) + (-349) + (-381)$
$\Delta H = 120 - 349 - 381$
$\Delta H = -610 \ kJ \ mol^{-1}$

(C) The enthalpy of reaction is given by the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
Given: $\Delta H^o = -352.8 \ kJ$,$\Delta H_f^o(HF) = -268.3 \ kJ \ mol^{-1}$,and $\Delta H_f^o(F_2) = 0 \ kJ \ mol^{-1}$ (standard state),$\Delta H_f^o(Cl_2) = 0 \ kJ \ mol^{-1}$ (standard state).
Substituting the values into the equation:
$-352.8 = [2 \times \Delta H_f^o(HF) + 1 \times \Delta H_f^o(Cl_2)] - [1 \times \Delta H_f^o(F_2) + 2 \times \Delta H_f^o(HCl)]$
$-352.8 = [2 \times (-268.3) + 0] - [0 + 2 \times \Delta H_f^o(HCl)]$
$-352.8 = -536.6 - 2 \times \Delta H_f^o(HCl)$
$2 \times \Delta H_f^o(HCl) = -536.6 + 352.8$
$2 \times \Delta H_f^o(HCl) = -183.8$
$\Delta H_f^o(HCl) = -91.9 \ kJ \ mol^{-1}$.

(B) The enthalpy of atomization of $PH_3$ corresponds to the breaking of $3$ $P-H$ bonds: $3 \times BE(P-H) = 228 \, kcal \, mol^{-1}$.
Thus,$BE(P-H) = 228 / 3 = 76 \, kcal \, mol^{-1}$.
For $P_2H_4$,the atomization involves breaking $1$ $P-P$ bond and $4$ $P-H$ bonds: $BE(P-P) + 4 \times BE(P-H) = 355 \, kcal \, mol^{-1}$.
Substituting the value of $BE(P-H)$: $BE(P-P) + 4 \times 76 = 355$.
$BE(P-P) + 304 = 355$.
$BE(P-P) = 355 - 304 = 51 \, kcal \, mol^{-1}$.

(A) By adding the first three equations,we obtain the target equation $C_{(s)} + O_{2(g)} \to CO_{2(g)}$.
Summing the enthalpies: $\Delta H = 131 + (-282) + (-242) \ \text{kJ}$.
$x = 131 - 282 - 242 = -393 \ \text{kJ}$.

(A) The combustion reaction of $CS_2$ is: $CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$.
According to Hess's Law,the enthalpy of combustion $\Delta H_c^o$ is given by: $\Delta H_c^o = [\Delta H_f^o(CO_2) + 2 \times \Delta H_f^o(SO_2)] - [\Delta H_f^o(CS_2) + 3 \times \Delta H_f^o(O_2)]$.
Given: $\Delta H_f^o(CO_2) = -393 \, kJ \, mol^{-1}$,$\Delta H_f^o(SO_2) = -297 \, kJ \, mol^{-1}$,$\Delta H_f^o(CS_2) = +117 \, kJ \, mol^{-1}$,and $\Delta H_f^o(O_2) = 0$.
Substituting the values: $\Delta H_c^o = [-393 + 2(-297)] - [117 + 3(0)]$.
$\Delta H_c^o = [-393 - 594] - 117$.
$\Delta H_c^o = -987 - 117 = -1104 \, kJ \, mol^{-1}$.

(B) We need to find the enthalpy of formation of $H_2O_2(l)$,which corresponds to the reaction: $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$.
From the given equations:
$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$
$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$
$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$
Perform the operation: $\frac{1}{2} [ (ii) + 2(iii) - (i) ]$:
$\frac{1}{2} [ (N_2H_4 + O_2 - N_2H_4 - 2H_2O_2 + 2H_2 + O_2) \rightarrow (N_2 + 2H_2O - N_2 - 4H_2O + 2H_2O) ]$
$\frac{1}{2} [ 2H_2 + 2O_2 - 2H_2O_2 \rightarrow 0 ]$
$H_2 + O_2 \rightarrow H_2O_2$
$\Delta_f H^circ (H_2O_2) = \frac{1}{2} [ \Delta_r H_2^\circ + 2 \Delta_r H_3^\circ - \Delta_r H_1^\circ ]$
$= \frac{1}{2} [ -622 + 2(-285) - (-818) ]$
$= \frac{1}{2} [ -622 - 570 + 818 ]$
$= \frac{1}{2} [ -374 ] = -187 \, kJ/mol$.

(C) The combustion of carbon is an exothermic process,meaning it releases heat energy.
For the reaction $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,the standard enthalpy of combustion is approximately $-94 \, kcal/mol$.
Option $C$ correctly represents the physical states of the reactants and products along with the correct sign for the enthalpy change ($\Delta H < 0$ for exothermic reactions).

(C) For any chemical reaction,the enthalpy change is given by $\Delta H = H_P - H_R$.
In an exothermic reaction,heat is released,which means the enthalpy of the products is less than the enthalpy of the reactants $(\Delta H < 0)$.
Therefore,$H_P < H_R$,which implies $H_R > H_P$.

(A) The hydrogenation reaction of cyclohexene is:
$C_6H_{10} + H_2 \rightarrow C_6H_{12}$
The heat of reaction $(\Delta H_{hydro})$ is given by the difference between the sum of enthalpies of combustion of reactants and products:
$\Delta H_{hydro} = \sum \Delta H_{c}(\text{reactants}) - \sum \Delta H_{c}(\text{products})$
$\Delta H_{hydro} = [\Delta H_{c}(C_6H_{10}) + \Delta H_{c}(H_2)] - [\Delta H_{c}(C_6H_{12})]$
Substituting the given values:
$\Delta H_{hydro} = [-3800 + (-241)] - [-3920]$
$\Delta H_{hydro} = -4041 + 3920$
$\Delta H_{hydro} = -121 \, kJ \, mol^{-1}$

(B) In reaction $(i)$,$HCl$ is produced in the gaseous state.
In reaction $(ii)$,$HCl$ is produced in the liquid state.
The condensation of $HCl_{(g)}$ to $HCl_{(\ell)}$ is an exothermic process,which releases additional heat.
Therefore,the total heat released in reaction $(ii)$ $(y)$ will be greater than the heat released in reaction $(i)$ $(x)$.
Thus,$y > x$ or $x < y$.

(D) The combustion reactions are:
$S_{\text{rhombic}} + O_2 \to SO_2$; $\Delta H_1 = -70,960 \ cal$ (Note: Combustion is exothermic,so values are negative)
$S_{\text{monoclinic}} + O_2 \to SO_2$; $\Delta H_2 = -71,030 \ cal$
To find the transition enthalpy $(S_{\text{rhombic}} \to S_{\text{monoclinic}})$,subtract the second equation from the first:
$\Delta H_{\text{transition}} = \Delta H_1 - \Delta H_2$
$\Delta H_{\text{transition}} = (-70,960) - (-71,030)$
$\Delta H_{\text{transition}} = -70,960 + 71,030 = +70 \ cal$