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

(C) The heat of formation is defined as the enthalpy change when $1 \ mole$ of a substance is formed from its constituent elements in their standard states.
For the given reaction: $H_{2(g)} + I_{2(g)} \to 2HI_{(g)}$,the enthalpy change is $\Delta H = -12.40 \ kcal$ for the formation of $2 \ moles$ of $HI$.
Therefore,the heat of formation of $1 \ mole$ of $HI$ is $\Delta H_f^o(HI) = \frac{-12.40 \ kcal}{2} = -6.20 \ kcal$.

(C) The enthalpy of formation $(\Delta H_f)$ is defined as the change in enthalpy when $1 \ mole$ of a substance is formed from its constituent elements in their standard states.
Since the enthalpy of elements in their standard state is defined as zero,the enthalpy of formation of a compound depends on the energy released or absorbed during the bond formation process.
Exothermic reactions result in a negative $\Delta H_f$,while endothermic reactions result in a positive $\Delta H_f$. Therefore,the heat of formation of compounds can be either negative or positive.

(C) The combustion reactions are given as follows:
$I. CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}, \Delta H_1 = -20 \ kcal$
$II. C_{(graphite)} + O_{2(g)} \to CO_{2(g)}, \Delta H_2 = -40 \ kcal$
$III. H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}, \Delta H_3 = -10 \ kcal$
We need to find the heat of formation of methane,which corresponds to the reaction:
$C_{(graphite)} + 2H_{2(g)} \to CH_{4(g)}, \Delta H_f = ?$
Using Hess's Law,we perform the operation: $II + 2 \times III - I$
$\Delta H_f = \Delta H_2 + 2(\Delta H_3) - \Delta H_1$
$\Delta H_f = -40 + 2(-10) - (-20)$
$\Delta H_f = -40 - 20 + 20 = -40 \ kcal$
Note: The provided options do not contain the correct result $-40 \ kcal$. Based on the calculation,the correct value is $-40 \ kcal$.

(B) The reaction $N_{2(g)} + O_{2(g)} \to 2NO_{(g)}$ is an endothermic reaction because heat is absorbed during the process.
All other reactions listed,such as combustion and neutralization,are exothermic reactions where heat is released.

(C) The heat of combustion is defined as the enthalpy change when $1 \ mol$ of a substance is completely burnt in the presence of oxygen.
Since combustion reactions involve the release of energy,the enthalpy change $(\Delta H_{combustion})$ is always negative.
Therefore,it is always an exothermic reaction.

(B) The target reaction for the heat of formation of $KOH_{(s)}$ is:
$K_{(s)} + \frac{1}{2} O_{2(g)} + \frac{1}{2} H_{2(g)} \to KOH_{(s)}$
Given equations:
$(I)$ $H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H_1 = -68.39 \ kcal$
$(II)$ $K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H_2 = -48 \ kcal$
$(III)$ $KOH + \text{water} \to KOH_{(aq)}; \Delta H_3 = -14 \ kcal$
To obtain the target reaction,we perform: $(II)$ + $(I)$ - $(III)$:
$K + H_2O + \frac{1}{2} O_2 + H_2 - KOH \to KOH_{(aq)} + \frac{1}{2} H_2 + H_2O - KOH_{(aq)}$
$K + \frac{1}{2} O_2 + \frac{1}{2} H_2 \to KOH$
Therefore,$\Delta H_f = \Delta H_2 + \Delta H_1 - \Delta H_3$
$\Delta H_f = -48 + (-68.39) - (-14) = -68.39 - 48 + 14 \ kcal$.

(A) The standard enthalpy of formation $(\Delta_fH^{\circ})$ is defined as the enthalpy change when $1 \text{ mole}$ of a substance is formed from its constituent elements in their most stable standard states.
For $CO_{2_{(g)}}$,the constituent elements are carbon (in its stable form,graphite) and oxygen (as $O_{2_{(g)}}$).
The reaction $C(\text{graphite}) + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$ represents the formation of $1 \text{ mole}$ of $CO_{2_{(g)}}$ from its elements in their standard states.
Therefore,option $A$ is correct.

(C) The reaction is $C_{\text{diamond}} \to C_{\text{graphite}}$,$\Delta H = -453.5 \ cal$.
Since the enthalpy change $(\Delta H)$ is negative,the reaction is exothermic.
This indicates that the product,graphite,has lower energy than the reactant,diamond.
Lower energy states are more stable,therefore,graphite is more stable than diamond.

(D) The reaction for the dimerization of $NO_2$ is: $2NO_{2(g)} \rightarrow N_2O_{4(g)}$.
The enthalpy change of the reaction $(\Delta H_{reaction})$ is calculated using the standard heats of formation $(\Delta H_f^o)$:
$\Delta H_{reaction} = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
$\Delta H_{reaction} = \Delta H_f^o(N_2O_4) - 2 \times \Delta H_f^o(NO_2)$.
Substituting the given values: $\Delta H_{reaction} = 2.0 - 2 \times 8.0$.
$\Delta H_{reaction} = 2.0 - 16.0 = -14.0 \ kcal$.

(B) The goal is to find $\Delta H$ for the transition: $C(\text{graphite}) \to C(\text{diamond})$.
Using Hess's Law,we reverse the first equation and add it to the second:
$CO_{2(g)} \to C(\text{diamond}) + O_{2(g)}; \Delta H = +395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
Adding these two equations gives:
$C(\text{graphite}) \to C(\text{diamond}); \Delta H = 395 - 393.5 = +1.5 \text{ kJ}$.

(D) The stability of a compound is inversely proportional to its enthalpy of formation. $A$ positive value of heat of formation indicates that the compound is endothermic and less stable relative to its constituent elements. Among the given values,$+64.8 \ kcal$ is the highest positive value,indicating the least stable product.

(C) The standard heat of formation $(\Delta H_f^o)$ is defined as the enthalpy change when $1 \text{ mole}$ of a substance is formed from its constituent elements in their most stable standard states.
For methane $(CH_{4(g)})$,the constituent elements are carbon and hydrogen.
The most stable state of carbon at standard conditions is graphite.
The most stable state of hydrogen is diatomic gas $(H_{2(g)})$.
Therefore,the correct equation is: $C(\text{graphite}) + 2H_{2(g)} \rightarrow CH_{4(g)}$.

(B) The heat of formation is defined as the enthalpy change when $1 \ mole$ of a substance is formed from its constituent elements in their standard states.
For water $(H_2O)$,the reaction is: $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$.
Option $B$ represents the formation of exactly $1 \ mole$ of $H_2O$ from its elements $H_2$ and $O_2$ in their standard states.

(B) The reaction is $2Mg + SiO_2 \to 2MgO + Si$.
The heat of reaction $\Delta H_{rxn}$ is calculated as:
$\Delta H_{rxn} = [2 \times \Delta H_f(MgO) + 1 \times \Delta H_f(Si)] - [2 \times \Delta H_f(Mg) + 1 \times \Delta H_f(SiO_2)]$.
Since $Mg$ and $Si$ are in their standard states,their heat of formation $\Delta H_f$ is $0 \ kJ$.
$\Delta H_{rxn} = 2 \times (-34.7) - (-48.4)$.
$\Delta H_{rxn} = -69.4 + 48.4 = -21.0 \ kJ$.

(A) Let the given equations be:
$(i) H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H_1 = 131 \ kJ$
$(ii) CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H_2 = -282 \ kJ$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H_3 = -242 \ kJ$
Adding equations $(i)$,$(ii)$,and $(iii)$:
$(H_2O_{(g)} + C_{(s)}) + (CO_{(g)} + \frac{1}{2}O_{2(g)}) + (H_{2(g)} + \frac{1}{2}O_{2(g)}) \to (CO_{(g)} + H_{2(g)}) + CO_{2(g)} + H_2O_{(g)}$
Canceling common terms on both sides:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
Therefore,$X = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.

(D) The heat of transition is defined as the enthalpy change that occurs when one allotropic form of a substance is converted into another allotropic form.

(D) The given reaction is $C(\text{diamond}) \to C(\text{graphite})$ with $\Delta H = -1.89 \ kJ$.
This means diamond is at a higher energy level than graphite by $1.89 \ kJ/mol$.
When $1 \ mol$ $(12 \ g)$ of diamond burns,it releases $1.89 \ kJ$ more heat than $1 \ mol$ of graphite.
For $6 \ g$ of diamond,the amount of substance is $6/12 = 0.5 \ mol$.
Therefore,the extra heat liberated compared to graphite is $0.5 \times 1.89 \ kJ = 0.945 \ kJ$.
Thus,the heat liberated in the first case (diamond) is more than in the second case (graphite) by $0.945 \ kJ$.

(B) The combustion reaction for $C_2H_{4(g)}$ is: $C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)}$
$\Delta H_{combustion} = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$
$\Delta H_{combustion} = [2 \times \Delta H_f^o(CO_2) + 2 \times \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_2H_4) + 3 \times \Delta H_f^o(O_2)]$
Given: $\Delta H_f^o(C_2H_4) = 52 \ kJ \ mol^{-1}$,$\Delta H_f^o(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta H_f^o(H_2O) = -286 \ kJ \ mol^{-1}$,and $\Delta H_f^o(O_2) = 0 \ kJ \ mol^{-1}$.
$\Delta H_{combustion} = [2(-394) + 2(-286)] - [52 + 3(0)]$
$\Delta H_{combustion} = [-788 - 572] - 52 = -1360 - 52 = -1412 \ kJ \ mol^{-1}$.

(B) The heat of formation of $CO$ corresponds to the reaction: $C(s) + \frac{1}{2}O_2(g) \to CO(g)$.
Given:
$(i) \ C(s) + O_2(g) \to CO_2(g)$; $\Delta H_1 = -394 \ kJ \ mol^{-1}$
$(ii) \ 2CO(g) + O_2(g) \to 2CO_2(g)$; $\Delta H_2 = -569 \ kJ \ mol^{-1}$
Divide equation $(ii)$ by $2$:
$(iii) \ CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)$; $\Delta H_3 = \frac{-569}{2} = -284.5 \ kJ \ mol^{-1}$
Subtract equation $(iii)$ from equation $(i)$:
$(C + O_2) - (CO + \frac{1}{2}O_2) \to CO_2 - CO_2$
$C(s) + \frac{1}{2}O_2(g) \to CO(g)$
$\Delta H_f = \Delta H_1 - \Delta H_3 = -394 - (-284.5) = -109.5 \ kJ \ mol^{-1}$.

(C) The standard heat of formation $(\Delta H_f)$ is defined as the enthalpy change when $1 \ mole$ of a compound is formed from its constituent elements in their standard states.
For the given reaction $H_2 + Cl_2 \to 2HCl$,the enthalpy change is $\Delta H = -44 \ kcal$ (since heat is released,the reaction is exothermic).
Since $2 \ moles$ of $HCl$ are formed in this reaction,the heat of formation for $1 \ mole$ of $HCl$ is calculated as $\Delta H_f = \frac{-44 \ kcal}{2} = -22 \ kcal \ mol^{-1}$.

(B) In an exothermic reaction,heat is released into the surroundings.
Therefore,the enthalpy change,$\Delta H$,is negative $(\Delta H < 0)$.
Thus,the correct option is $(B)$.

(D) The reaction $C_{(s)} + 2S_{(s)} \to CS_{2(l)}$ represents the formation of one mole of $CS_{2}$ from its constituent elements in their standard states.
By definition,the enthalpy change accompanying the formation of one mole of a substance from its constituent elements in their standard states is called the standard heat of formation.

(A) The molar mass of acetaldehyde $(CH_3CHO)$ is $12 \times 2 + 1 \times 4 + 16 = 44 \ g \ mol^{-1}$.
Given that $2.2016 \ g$ of $CH_3CHO$ releases $13.95 \ kcal$ of heat.
Therefore,the heat of combustion per mole is calculated as:
$\Delta H = \frac{13.95 \ kcal}{2.2016 \ g} \times 44 \ g \ mol^{-1} = 278.7 \ kcal \ mol^{-1}$.
Rounding to the nearest integer,we get $279 \ kcal \ mol^{-1}$.

(D) The heat of formation of $CO_2$ corresponds to the reaction: $C + O_2 \to CO_2$.
By adding the two given equations:
$(i) C + \frac{1}{2} O_2 \to CO; \Delta H_1 = -42 \ kJ$
$(ii) CO + \frac{1}{2} O_2 \to CO_2; \Delta H_2 = -24 \ kJ$
Adding $(i)$ and $(ii)$ gives:
$C + \frac{1}{2} O_2 + CO + \frac{1}{2} O_2 \to CO + CO_2$
$C + O_2 \to CO_2$
According to Hess's Law,the total enthalpy change is the sum of the individual enthalpy changes:
$\Delta H_{total} = \Delta H_1 + \Delta H_2 = -42 \ kJ + (-24 \ kJ) = -66 \ kJ$.

(D) The standard molar enthalpy of formation of a compound is defined as the enthalpy change when $1 \ mol$ of the compound is formed from its constituent elements in their most stable states at $298 \ K$ and $1 \ bar$ pressure.
For $CO_2$,the reaction is: $C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)$.
This reaction also represents the combustion of carbon (graphite) in oxygen to form $CO_2$.
Therefore,the standard molar enthalpy of formation of $CO_2$ is equal to the standard molar enthalpy of combustion of carbon (graphite).

(A) The heat of formation of $CO$ corresponds to the reaction: $C + \frac{1}{2} O_2 \to CO$.
Given equations:
$(1) C + O_2 \to CO_2; \Delta H = X$
$(2) CO + \frac{1}{2} O_2 \to CO_2; \Delta H = Y$
To obtain the target reaction,subtract equation $(2)$ from equation $(1)$:
$(C + O_2) - (CO + \frac{1}{2} O_2) = CO_2 - CO_2$
$C + \frac{1}{2} O_2 - CO = 0$
$C + \frac{1}{2} O_2 \to CO$
Therefore,the enthalpy change is $\Delta H = X - Y$.

(A) The heat of reaction $\Delta H_r$ is calculated using the enthalpies of formation of products and reactants:
$\Delta H_r = \Delta H_f(SO_3) - [\Delta H_f(SO_2) + \frac{1}{2} \Delta H_f(O_2)]$
Given:
$\Delta H_f(SO_2) = -298.2 \ kJ/mol$
$\Delta H_f(SO_3) = -395.2 \ kJ/mol$
$\Delta H_f(O_2) = 0 \ kJ/mol$ (standard state)
Substituting the values:
$\Delta H_r = -395.2 - (-298.2) = -395.2 + 298.2 = -97.0 \ kJ$

(C) The target reaction is: $C(s) + 2S(s) \to CS_2(l), \Delta H_f = ?$
Given reactions:
$(i) C(s) + O_2(g) \to CO_2(g), \Delta H_1 = -393.3 \text{ kJ mol}^{-1}$
$(ii) S(s) + O_2(g) \to SO_2(g), \Delta H_2 = -293.72 \text{ kJ mol}^{-1}$
$(iii) CS_2(l) + 3O_2(g) \to CO_2(g) + 2SO_2(g), \Delta H_3 = -1108.76 \text{ kJ mol}^{-1}$
Using Hess's Law: $\Delta H_f = \Delta H_1 + 2 \times \Delta H_2 - \Delta H_3$
$\Delta H_f = -393.3 + 2(-293.72) - (-1108.76)$
$\Delta H_f = -393.3 - 587.44 + 1108.76$
$\Delta H_f = +128.02 \text{ kJ mol}^{-1}$

(D) The given reactions are:
$1$) $BaCl_{2(s)} + aq \to BaCl_{2(aq)}$,$\Delta H_1 = -20.6 \, kJ \, mol^{-1}$
$2$) $BaCl_2 \cdot 2H_2O_{(s)} + aq \to BaCl_{2(aq)}$,$\Delta H_2 = 8.8 \, kJ \, mol^{-1}$
To find the enthalpy change for $BaCl_{2(s)} + 2H_2O_{(l)} \to BaCl_2 \cdot 2H_2O_{(s)}$,we subtract equation $(2)$ from equation $(1)$:
$\Delta H = \Delta H_1 - \Delta H_2$
$\Delta H = (-20.6) - (8.8) \, kJ \, mol^{-1} = -29.4 \, kJ \, mol^{-1}$.

(A) The heat of formation of $SO_2$ is the enthalpy change for the reaction: $S + O_2 \to SO_2$.
Given equations:
$(i) \ S + \frac{3}{2} O_2 \to SO_3 + 2x \ kcal$
$(ii) \ SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ kcal$
To obtain the target reaction,subtract equation $(ii)$ from equation $(i)$:
$(S + \frac{3}{2} O_2) - (SO_2 + \frac{1}{2} O_2) \to (SO_3 - SO_3) + (2x - y) \ kcal$
$S + O_2 - SO_2 \to 2x - y \ kcal$
$S + O_2 \to SO_2 + (2x - y) \ kcal$
Thus,the heat of formation of $SO_2$ is $(2x - y) \ kcal$.

(C) The combustion reactions are:
$(i)$ $C(s) + O_2(g) \to CO_2(g), \Delta H_1 = -395.5 \, kJ \, mol^{-1}$
(ii) $H_2(g) + \frac{1}{2} O_2(g) \to H_2O(l), \Delta H_2 = -285.8 \, kJ \, mol^{-1}$
(iii) $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l), \Delta H_3 = -890.4 \, kJ \, mol^{-1}$
We need the enthalpy of formation of methane: $C(s) + 2H_2(g) \to CH_4(g), \Delta H_f = ?$
Using Hess's Law: $\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$
$\Delta H_f = -395.5 + 2(-285.8) - (-890.4)$
$\Delta H_f = -395.5 - 571.6 + 890.4$
$\Delta H_f = -76.7 \, kJ \, mol^{-1}$.
Note: The closest option provided is $-74.7 \, kJ \, mol^{-1}$.

(D) The heat of neutralization of a strong acid and a strong base is constant at $-57.1 \ kJ \ mol^{-1}$.
Since methanoic acid $(HCOOH)$ is a weak acid,some of the heat evolved during the neutralization process is consumed in the dissociation of the weak acid.
Therefore,the net heat evolved is less than the standard heat of neutralization of a strong acid and strong base,which is equivalent to the heat of formation of water $(x)$.
Thus,the heat evolved is less than $x$.

(C) The reaction for the formation of sulphur dioxide is: $S(s) + O_2(g) \to SO_2(g)$.
Given that $0.5 \ g$ of sulphur releases $4.6 \ kJ$ of heat.
The molar mass of sulphur $(S)$ is $32 \ g \ mol^{-1}$.
Heat released for $1 \ mol$ $(32 \ g)$ of sulphur is calculated as:
$\Delta H_f = \frac{-4.6 \ kJ}{0.5 \ g} \times 32 \ g \ mol^{-1} = -294.4 \ kJ \ mol^{-1}$.

(B) The combustion reaction is $C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$,where $\Delta H = -72 \ kcal$ for $1 \ mol$ of glucose.
Since the molecular mass of glucose is $180 \ g/mol$,the energy released for $180 \ g$ of glucose is $72 \ kcal$.
Therefore,the energy required for the production of $1.6 \ g$ of glucose is calculated as:
$\text{Energy} = \frac{72 \ kcal}{180 \ g} \times 1.6 \ g = 0.4 \times 1.6 = 0.64 \ kcal$.

(A) The formation reaction of $CH_4$ is: $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$.
We are given:
$(i)$ $C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H_1 = -394 \ kJ$
$(ii)$ $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}, \Delta H_2 = -568 \ kJ$
$(iii)$ $CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}, \Delta H_3 = -892 \ kJ$
To obtain the formation reaction,perform $(i)$ + $(ii)$ - $(iii)$:
$\Delta H_f = \Delta H_1 + \Delta H_2 - \Delta H_3$
$\Delta H_f = -394 + (-568) - (-892)$
$\Delta H_f = -394 - 568 + 892$
$\Delta H_f = -962 + 892 = -70 \ kJ$.

(A) The formation reaction for $PCl_{5(s)}$ is: $P_{(s)} + 2.5Cl_{2(g)} \to PCl_{5(s)}$.
From the given equations:
$(1) \ 2P_{(s)} + 3Cl_{2(g)} \to 2PCl_{3(l)}; \Delta H_1 = -151.8 \ kcal$
$(2) \ PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)}; \Delta H_2 = -32.8 \ kcal$
Divide equation $(1)$ by $2$:
$P_{(s)} + 1.5Cl_{2(g)} \to PCl_{3(l)}; \Delta H_3 = \frac{-151.8}{2} = -75.9 \ kcal$
Add the modified equation $(1)$ and equation $(2)$:
$(P_{(s)} + 1.5Cl_{2(g)}) + (PCl_{3(l)} + Cl_{2(g)}) \to PCl_{3(l)} + PCl_{5(s)}$
$P_{(s)} + 2.5Cl_{2(g)} \to PCl_{5(s)}$
The total enthalpy change is $\Delta H = \Delta H_3 + \Delta H_2 = -75.9 \ kcal + (-32.8 \ kcal) = -108.7 \ kcal$.

(A) The heat of neutralization is an intensive property related to the amount of substance reacted.
When $50 \ cm^3$ of acid and $50 \ cm^3$ of base are mixed,the total volume is $100 \ cm^3$. The heat evolved $(q)$ is given by $q = mc\Delta T$.
Since the amount of acid and base is increased by a factor of $5$ (from $50 \ cm^3$ to $250 \ cm^3$),the total heat evolved $(q')$ becomes $5q$.
However,the total mass of the solution $(m')$ also increases by a factor of $5$ (from $100 \ cm^3$ to $500 \ cm^3$).
Using the formula $\Delta T' = \frac{q'}{m'c} = \frac{5q}{5mc} = \frac{q}{mc} = \Delta T$.
Therefore,the temperature rise remains $5 \ ^oC$.

(A) The given reactions are:
$1) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_1 = -285.8 \ kJ \ mol^{-1}$
$2) \ H_2O_{(l)} \to H_2O_{(g)}$; $\Delta H_2 = 44 \ kJ \ mol^{-1}$
To find the enthalpy of formation of water vapor,we add the two equations:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$
Applying Hess's Law:
$\Delta H_f^{\circ} (H_2O_{(g)}) = \Delta H_1 + \Delta H_2$
$\Delta H_f^{\circ} (H_2O_{(g)}) = -285.8 \ kJ + 44 \ kJ = -241.8 \ kJ \ mol^{-1}$

(B) The chemical equation for the formation of ferric oxide $(Fe_2O_3)$ is: $2Fe(s) + \frac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)$.
Given that $4 \ g$ of iron releases $29.28 \ kJ$ of heat.
The molar mass of iron $(Fe)$ is $56 \ g \ mol^{-1}$.
For $2 \ moles$ of iron,the mass is $2 \times 56 = 112 \ g$.
Since heat is evolved,the enthalpy change $(\Delta H)$ is negative.
For $4 \ g$ of $Fe$,$\Delta H = -29.28 \ kJ$.
For $112 \ g$ of $Fe$ (which corresponds to $1 \ mole$ of $Fe_2O_3$),the enthalpy of formation is:
$\Delta H_f = \frac{-29.28 \ kJ}{4 \ g} \times 112 \ g = -819.8 \ kJ \ mol^{-1}$.

(B) The molar mass of $Fe$ is $56 \ g/mol$ and $S$ is $32 \ g/mol$. The molar mass of $FeS$ is $56 + 32 = 88 \ g/mol$.
Given that $2.1 \ g$ of $Fe$ releases $3.77 \ kJ$ of heat.
To find the heat of formation per mole of $FeS$,we calculate the heat released for $1 \ mole$ $(56 \ g)$ of $Fe$.
Heat of formation $\Delta H_f = -(\frac{3.77 \ kJ}{2.1 \ g} \times 56 \ g/mol) = -100.5 \ kJ/mol$.
Since the heat is evolved,the enthalpy change is negative.

(D) The heat of formation is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $H_2 + Cl_2 \to 2HCl$,the enthalpy change is $\Delta H = -194 \ kJ$ for the production of $2 \ mol$ of $HCl$.
Therefore,the heat of formation for $1 \ mol$ of $HCl$ is $\Delta H_f = \frac{-194 \ kJ}{2} = -97 \ kJ \ mol^{-1}$.
Thus,the correct option is $D$.

(A) The standard enthalpy of neutralization for the reaction between a strong acid and a strong base is always approximately $-57.32 \, kJ \, mol^{-1}$.
$HNO_3$ is a strong acid and $LiOH$ is a strong base.
$HCOOH$,$CH_3COOH$,and $NH_4OH$ are weak electrolytes,so their heat of neutralization will be less than $57.32 \, kJ$ due to the energy consumed in the dissociation of the weak species.
Therefore,the correct pair is $HNO_3 + LiOH$.

(B) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g \ mol^{-1}$.
Given that the heat evolved for $1 \ mol$ $(78 \ g)$ of benzene is $781.0 \ kcal$.
Therefore,the heat evolved for $39 \ g$ of benzene is calculated as:
$\text{Heat} = \frac{781.0 \ kcal}{78 \ g} \times 39 \ g = 390.5 \ kcal$.

(B) Given equations:
$(i) \ H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H_1 = -44 \ kcal$
$(ii) \ 2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H_2 = -152 \ kcal$
To find $\Delta H$ for $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}$,we add equations $(i)$ and $(ii)$ and divide by $2$:
$\frac{1}{2} \times [(i) + (ii)] \implies \frac{1}{2} \times [2Na_{(s)} + Cl_{2(g)} \to 2NaCl_{(s)}], \Delta H = \frac{1}{2} \times (-44 - 152) \ kcal$
$\Delta H = \frac{1}{2} \times (-196) \ kcal = -98 \ kcal$.

(C) The heat of solution $(\Delta H_s)$ represents the enthalpy change when a substance dissolves in a solvent.
If $\Delta H_s$ is positive,the process is endothermic,meaning heat is absorbed from the surroundings.
If $\Delta H_s$ is negative,the process is exothermic,meaning heat is released.
To absorb the maximum quantity of heat,the compound must have the most positive value of $\Delta H_s$.
Comparing the given values: $HNO_3$ $(-33)$,$KCl$ $(+17.64)$,$NH_4NO_3$ $(+25.5)$,and $HCl$ $(-74.1)$.
The compound $NH_4NO_3$ has the highest positive value of $\Delta H_s$ $(+25.5 \, kJ/mole)$,therefore it absorbs the maximum amount of heat.

(D) The reaction $C(s) + 2S(s) \to CS_2(l)$ represents the formation of $1 \ mol$ of $CS_2$ from its constituent elements in their standard states.
Therefore,the enthalpy change $\Delta H$ associated with this reaction is defined as the heat of formation of $CS_2$.

(A) The reaction is $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$.
For this reaction,the change in the number of moles of gaseous species is $\Delta n_g = n_p - n_r = 1 - 2 = -1$.
Given $\Delta H = -18500 \ cal$,$T = 25 + 273 = 298 \ K$,and $R = 2 \ cal \ K^{-1} \ mol^{-1}$.
The relation between $\Delta H$ and $\Delta U$ (or $\Delta E$) is $\Delta H = \Delta U + \Delta n_g RT$.
Substituting the values: $-18500 = \Delta U + (-1) \times 2 \times 298$.
$-18500 = \Delta U - 596$.
$\Delta U = -18500 + 596 = -17904 \ cal$.
Note: The question asks for the heat of reaction at constant volume,which is $\Delta U$. Based on the provided options and standard calculation,the magnitude is $17904 \ cal$.

(B) The molar mass of benzene $({C_6H_6})$ is $(6 \times 12) + (6 \times 1) = 78 \, g/mol$.
Number of moles of benzene = $\frac{\text{mass}}{\text{molar mass}} = \frac{0.39 \, g}{78 \, g/mol} = 0.005 \, mol$.
Heat evolved = $\text{moles} \times \text{enthalpy of combustion} = 0.005 \, mol \times 3250 \, kJ/mol = 16.25 \, kJ$.