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

(C) The 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.
The target reaction is: $2C_{(s)} + 3H_{2(g)} + \frac{1}{2}O_{2(g)} \to C_2H_5OH_{(l)}, \Delta H_f = ?$
Given equations:
$(i) \ 2C_{(s)} + 2O_{2(g)} \to 2CO_{2(g)}, \Delta H = 2 \times (-94) = -188 \ kcal$
$(ii) \ 3H_{2(g)} + \frac{3}{2}O_{2(g)} \to 3H_2O_{(l)}, \Delta H = 3 \times (-68) = -204 \ kcal$
$(iii) \ 2CO_{2(g)} + 3H_2O_{(l)} \to C_2H_5OH_{(l)} + 3O_{2(g)}, \Delta H = +327 \ kcal$
Adding equations $(i)$,$(ii)$,and $(iii)$:
$\Delta H_f = (-188) + (-204) + 327 = -392 + 327 = -65 \ kcal/mol$.

(D) An exothermic reaction is a process that releases energy,characterized by a negative change in internal energy $(\Delta U < 0)$ or enthalpy $(\Delta H < 0)$.
In option $A$,$180 \, kJ$ is added to the reactants,indicating an endothermic process.
In option $B$,the negative sign on the product side indicates energy is absorbed,making it endothermic.
In option $C$,$\Delta U = +181 \, kJ$ indicates a positive value,meaning energy is absorbed (endothermic).
In option $D$,$\Delta U = -314 \, kJ$ indicates a negative value,meaning energy is released,which characterizes an exothermic reaction.

(C) The transition reaction is: $C_{\text{(diamond)}} \rightarrow C_{\text{(graphite)}}$.
Given $\Delta H = -453.5 \ \text{cal}$.
Since $\Delta H$ is negative,the reaction is exothermic,meaning the product (graphite) has lower energy than the reactant (diamond).
In thermodynamics,a substance with lower internal energy is more stable.
Therefore,graphite is more stable than diamond.

(D) The combustion reaction of benzoic acid is: $C_6H_5COOH_{(s)} + \frac{15}{2} O_{2_{(g)}} \rightarrow 7CO_{2_{(g)}} + 3H_2O_{(l)}$
The enthalpy of combustion $\Delta H_c^o$ is given by: $\Delta H_c^o = [7 \times \Delta H_f^o(CO_2) + 3 \times \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_6H_5COOH)]$
$\Delta H_c^o = [7 \times (-393) + 3 \times (-286)] - [-408]$
$\Delta H_c^o = [-2751 - 858] + 408 = -3609 + 408 = -3201 \, kJ \, mol^{-1}$
For the reaction,$\Delta n_g = n_{products(g)} - n_{reactants(g)} = 7 - 7.5 = -0.5 \, mol$
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta U$ is the heat of combustion at constant volume:
$-3201 = \Delta U + (-0.5) \times (8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}) \times 300 \, K$
$-3201 = \Delta U - 1.2471$
$\Delta U = -3201 + 1.2471 = -3199.7529 \, kJ \, mol^{-1}$

(A) The reaction is $2H_2O_2(\ell) \to 2H_2O(\ell) + O_2(g)$.
The enthalpy change of the reaction is given by: $\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$.
Given: $\Delta H_f(H_2O_2) = -188 \, kJ/mol$,$\Delta H_f(H_2O) = -286 \, kJ/mol$,and $\Delta H_f(O_2) = 0 \, kJ/mol$ (standard state).
$\Delta H = [2 \times \Delta H_f(H_2O) + 1 \times \Delta H_f(O_2)] - [2 \times \Delta H_f(H_2O_2)]$.
$\Delta H = [2 \times (-286) + 0] - [2 \times (-188)]$.
$\Delta H = -572 - (-376) = -572 + 376 = -196 \, kJ/mol$.

(C) The required reaction is:
$6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$
Given thermochemical equations:
$(1) \ C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)} \quad \Delta H = -3268 \ kJ$
$(2) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$
To obtain the required reaction, we use Hess's Law:
$\Delta H_f(C_6H_6) = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - \Delta H_{comb}(C_6H_6)$
$\Delta H_f(C_6H_6) = [6 \times (-393.5) + 3 \times (-285.8)] - (-3268)$
$\Delta H_f(C_6H_6) = [-2361 - 857.4] + 3268$
$\Delta H_f(C_6H_6) = -3218.4 + 3268 = 49.6 \ kJ$

(C) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \, g/mol$.
Heat released by $1 \, mol$ of glucose is $2900 \, KJ$.
Therefore,the calorific value (heat released per gram) is calculated as:
$\text{Calorific Value} = \frac{\text{Heat released}}{\text{Molar mass}} = \frac{2900 \, KJ}{180 \, g} \approx 16.11 \, KJ/g$.

(B) The combustion reactions are:
$(1)$ $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$; $\Delta H = x$
$(2)$ $CH_3OH(l) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(l)$; $\Delta H = y$
Subtracting equation $(2)$ from $(1)$:
$(CH_4 + 2O_2) - (CH_3OH + \frac{3}{2}O_2) = x - y$
$CH_4 + \frac{1}{2}O_2 \to CH_3OH$; $\Delta H = x - y$
Since the reaction is exothermic,$\Delta H < 0$,which implies $x - y < 0$,or $x < y$.

(D) The enthalpy change of the reaction is calculated using the formula: $\Delta H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$.
Reactants: $1 \times (C=C) + 4 \times (C-H) + 1 \times (H-H) = 615 + 4(414) + 435 = 615 + 1656 + 435 = 2706 \, kJ \, mol^{-1}$.
Products: $1 \times (C-C) + 6 \times (C-H) = 347 + 6(414) = 347 + 2484 = 2831 \, kJ \, mol^{-1}$.
$\Delta H = 2706 - 2831 = -125 \, kJ \, mol^{-1}$.

(B) Hess's law states that the total enthalpy change for a reaction is the same,whether it occurs in one step or in several steps. This is a direct consequence of the $1^{st}$ law of thermodynamics,which states that energy is conserved. Since enthalpy is a state function,its value depends only on the initial and final states,not on the path taken.

(B) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow ICl_{(g)}$
Step $1$: Sublimation of $I_{2(s)}$: $\frac{1}{2} I_{2(s)} \rightarrow \frac{1}{2} I_{2(g)}$,$\Delta H_1 = \frac{1}{2} \times 62.76 = 31.38 \, kJ \, mol^{-1}$
Step $2$: Dissociation of $I_{2(g)}$: $\frac{1}{2} I_{2(g)} \rightarrow I_{(g)}$,$\Delta H_2 = \frac{1}{2} \times 151.0 = 75.5 \, kJ \, mol^{-1}$
Step $3$: Dissociation of $Cl_{2(g)}$: $\frac{1}{2} Cl_{2(g)} \rightarrow Cl_{(g)}$,$\Delta H_3 = \frac{1}{2} \times 242.3 = 121.15 \, kJ \, mol^{-1}$
Step $4$: Formation of $ICl_{(g)}$ from atoms: $I_{(g)} + Cl_{(g)} \rightarrow ICl_{(g)}$,$\Delta H_4 = -211.3 \, kJ \, mol^{-1}$
Total enthalpy of formation $\Delta H_f = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 = 31.38 + 75.5 + 121.15 - 211.3 = 16.73 \, kJ \, mol^{-1}$.

(D) The heat of reaction $(\Delta H)$ is calculated using the formula: $\Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$.
Given: $\Delta H_f^\circ(CH_4) = -17.9 \ kcal/mol$, $\Delta H_f^\circ(C_2H_4) = 12.5 \ kcal/mol$, and $\Delta H_f^\circ(C_3H_8) = -24.8 \ kcal/mol$.
For the reaction $CH_4 + C_2H_4 \rightarrow C_3H_8$:
$\Delta H = \Delta H_f^\circ(C_3H_8) - [\Delta H_f^\circ(CH_4) + \Delta H_f^\circ(C_2H_4)]$.
$\Delta H = -24.8 - [-17.9 + 12.5]$.
$\Delta H = -24.8 - [-5.4]$.
$\Delta H = -24.8 + 5.4 = -19.4 \ kcal$.

(B) The reaction $C(s) + O_2(g) \rightarrow CO_2(g)$ is a combustion reaction.
Combustion reactions are exothermic in nature,meaning heat is released during the process.
For exothermic reactions,the enthalpy change $(\Delta H)$ is always negative $(\Delta H < 0)$.

(B) The combustion reaction of benzene is: $C_6H_6(l) + \frac{15}{2}O_2(g) \to 6CO_2(g) + 3H_2O(l)$.
The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.
Heat produced by the combustion of $1 \ mol$ $(78 \ g)$ of $C_6H_6 = 3264.6 \ kJ$.
Heat produced by the combustion of $39 \ g$ of $C_6H_6 = \frac{3264.6 \ kJ}{78 \ g} \times 39 \ g = \frac{3264.6}{2} = 1632.3 \ kJ$.

(D) The combustion reactions are as follows:
$1$) $P_{\text{yellow}} + O_2 \rightarrow P_4O_{10} \quad \Delta H_1 = -9.91 \, kJ$
$2$) $P_{\text{red}} + O_2 \rightarrow P_4O_{10} \quad \Delta H_2 = -8.78 \, kJ$
We want the transition: $P_{\text{yellow}} \rightarrow P_{\text{red}}$
This can be obtained by subtracting equation $(2)$ from equation $(1)$:
$(P_{\text{yellow}} + O_2) - (P_{\text{red}} + O_2) = -9.91 - (-8.78)$
$P_{\text{yellow}} - P_{\text{red}} = -9.91 + 8.78$
$P_{\text{yellow}} \rightarrow P_{\text{red}} \quad \Delta H_{\text{transition}} = -1.13 \, kJ$

(A) 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 reaction is given by: $\Delta H_f = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - [BE(H-Cl)]$
Substituting the given values:
$-90 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 240] - BE(H-Cl)$
$-90 = [215 + 120] - BE(H-Cl)$
$-90 = 335 - BE(H-Cl)$
$BE(H-Cl) = 335 + 90 = 425 \ kJ \ mol^{-1}$

(A) The heat of neutralization for a strong acid and a strong base is $56 \text{ kJ/mol}$ for $1 \text{ mole}$ of $H^{+}$ ions.
$H_2SO_4$ is a dibasic acid,meaning $1 \text{ mole}$ of $H_2SO_4$ provides $2 \text{ moles}$ of $H^{+}$ ions.
Therefore,the heat of neutralization for $1 \text{ mole}$ of $H_2SO_4$ is $56 \text{ kJ/mol} \times 2 = 112 \text{ kJ}$.

(B) Hess's Law of Constant Heat Summation states that the total enthalpy change for a chemical reaction is the same,whether the reaction takes place in one step or in several steps.
This implies that the enthalpy change depends only on the initial state of the reactants and the final state of the products,and is independent of the path taken.
Therefore,the correct option is $B$.

(C) The standard enthalpy of formation $(\Delta H_f^o)$ is defined as the enthalpy change when $1 \text{ mole}$ of a compound is formed from its constituent elements in their standard states.
In option $C$,$Xe_{(g)} + 2F_{2(g)} \rightarrow XeF_{4(g)}$,$1 \text{ mole}$ of $XeF_4$ is formed from its elements $Xe$ and $F_2$ in their standard states.
Therefore,for this reaction,$\Delta H_{reaction}^o = \Delta H_f^o(XeF_4)$.
In other options,either more than $1 \text{ mole}$ of product is formed or the reactants are not in their standard elemental states.

(D) The given reaction is: $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH_1 = -111 \ kJ$.
We want to find the enthalpy change for the reaction: $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(s)$.
The sublimation process is $N_2O_5(s) \rightarrow N_2O_5(g)$, with $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$.
For $2 \ mol$ of $N_2O_5$, the enthalpy change for condensation (reverse of sublimation) is $\Delta H_{cond} = -2 \times 54 \ kJ = -108 \ kJ$.
Adding the two reactions:
$(4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)) + (2N_2O_5(g) \rightarrow 2N_2O_5(s))$
$\Delta_rH_2 = \Delta_rH_1 + \Delta H_{cond} = -111 \ kJ + (-108 \ kJ) = -219 \ kJ$.

(C) Given reactions:
$(i) H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$
Dividing by $2$: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \rightarrow HCl_{(g)}$; $\Delta H = -22 \, Kcal$
$(ii) 2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$
Dividing by $2$: $Na_{(s)} + HCl_{(g)} \rightarrow NaCl_{(s)} + \frac{1}{2}H_{2(g)}$; $\Delta H = -76 \, Kcal$
Adding the two resulting equations:
$(Na_{(s)} + HCl_{(g)}) + (\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)})$ $\rightarrow (NaCl_{(s)} + \frac{1}{2}H_{2(g)}) + HCl_{(g)}$
$Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$
$\Delta H = (-22) + (-76) = -98 \, Kcal$

(A) The formation reaction of $C_2H_5OH$ is: $2C(s) + 3H_2(g) + 1/2 O_2(g) \rightarrow C_2H_5OH(l)$.
Given equations:
$(1) \ H_2 + 1/2 O_2 \rightarrow H_2O, \Delta H_1 = -68.4 \ \text{kcal}$
$(2) \ C + O_2 \rightarrow CO_2, \Delta H_2 = -94.0 \ \text{kcal}$
$(3) \ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O, \Delta H_3 = -327.0 \ \text{kcal}$
To get the formation reaction,perform: $2 \times (2) + 3 \times (1) - (3)$:
$\Delta H_f = 2(-94.0) + 3(-68.4) - (-327.0)$
$\Delta H_f = -188.0 - 205.2 + 327.0$
$\Delta H_f = -393.2 + 327.0 = -66.2 \ \text{kcal}$.

(B) The standard enthalpy of formation $(\Delta_fH^{\circ})$ is defined as the enthalpy change when $1 \text{ mole}$ of a compound is formed from its constituent elements in their most stable standard states at $298 \text{ K}$ and $1 \text{ bar}$ pressure.
For methane $(CH_4)$,the constituent elements are carbon and hydrogen.
The most stable form of carbon at standard conditions is graphite.
The most stable form of hydrogen is diatomic gas $(H_2)$.
Therefore,the correct equation is: $C(\text{graphite}) + 2H_{2(g)} \rightarrow CH_{4(g)}$.

(D) The stability of a compound is inversely related to its enthalpy of formation $(\Delta H_f)$.
Compounds with a large negative value of $\Delta H_f$ are highly stable (exothermic formation).
Compounds with a positive value of $\Delta H_f$ are unstable or less stable because energy must be absorbed to form them from their elements.
Among the given options,$+64.8 \ kcal$ is the most positive value,indicating the highest energy state and thus the least stability.

(A) The reaction $2C(\text{graphite}) + 3H_{2(g)} \rightarrow C_2H_{6(g)}$ represents the formation of $1 \text{ mole}$ of ethane $(C_2H_6)$ from its constituent elements in their standard states (carbon as graphite and hydrogen as $H_2$ gas).
By definition,the enthalpy change accompanying the formation of $1 \text{ mole}$ of a compound from its elements in their standard states is known as the standard enthalpy of formation $(\Delta_fH^\circ)$.

(B) Bond dissociation energy is the energy required to break a chemical bond in a molecule in the gaseous state.
Since energy must be supplied to the system to break a bond,the process is endothermic.
Therefore,the enthalpy change for bond dissociation is always positive $(> 0)$.

(B) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Step $1$: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H_1 = -110.5 \, kJ$
Step $2$: $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H_2 = -283.2 \, kJ$
Adding the two equations:
$C(\text{graphite}) + \frac{1}{2}O_2 + CO + \frac{1}{2}O_2 \rightarrow CO + CO_2$
$C(\text{graphite}) + O_2 \rightarrow CO_2$
The total enthalpy change $\Delta H = \Delta H_1 + \Delta H_2$
$\Delta H = -110.5 \, kJ + (-283.2 \, kJ) = -393.7 \, kJ$.

(B) By definition,the standard molar enthalpy of formation $(\Delta_fH^\circ)$ of an element in its most stable state at $298 \ K$ and $1 \ bar$ pressure is zero.
Among the given options,$Cl_{2(g)}$ is an element in its most stable diatomic gaseous state at $298 \ K$,therefore its $\Delta_fH^\circ = 0$.
$Br_{2(l)}$ is the stable state of bromine at $298 \ K$,but the option provided was $Br_{2(l)}$ (liquid),which is correct for zero enthalpy,however,$Cl_{2(g)}$ is also a standard element in its stable state.
$H_2O_{(g)}$ and $CH_{4(g)}$ are compounds,so their standard enthalpy of formation is not zero.

(A) The enthalpy of neutralization is defined as the heat evolved when $1 \ gram \ equivalent$ of an acid is neutralized by $1 \ gram \ equivalent$ of a base in dilute solution.
For strong acids and strong bases,the enthalpy of neutralization is constant at approximately $-57.1 \ kJ \ mol^{-1}$ because it represents the heat of formation of water from $H^+$ and $OH^-$ ions $(H^+ + OH^- \rightarrow H_2O)$.
If either the acid or the base is weak,some energy is consumed in the dissociation of the weak electrolyte,making the enthalpy of neutralization less than $-57.1 \ kJ \ mol^{-1}$.
Therefore,the value is most significant and constant when both the acid and the base are strong.

(A) The heat of hydration of a salt is calculated by the difference between the heat of solution of the anhydrous salt and the heat of solution of the hydrated salt.
$\Delta H_{\text{hydration}} = \Delta H_{\text{sol}}(\text{anhydrous}) - \Delta H_{\text{sol}}(\text{hydrated})$
$\Delta H_{\text{hydration}} = -15.89 \, kcal \, mol^{-1} - 2.80 \, kcal \, mol^{-1}$
$\Delta H_{\text{hydration}} = -18.69 \, kcal \, mol^{-1}$

(B) Hess's law of constant heat summation states that the total enthalpy change for a reaction is the same,whether it occurs in one step or in several steps. This law is used to calculate the enthalpy of reactions that are difficult or impossible to measure experimentally.

(C) The formation reaction for $CS_{2(\ell)}$ is: $C_{(s)} + 2S_{(s)} \to CS_{2(\ell)}$
The standard heat of formation $\Delta H_f^\circ$ is calculated using the heats of combustion of reactants and products:
$\Delta H_f^\circ = \sum \Delta H_c^\circ(\text{reactants}) - \sum \Delta H_c^\circ(\text{products})$
$\Delta H_f^\circ = [\Delta H_c^\circ(C) + 2 \times \Delta H_c^\circ(S)] - [\Delta H_c^\circ(CS_{2(\ell)})]$
Substituting the given values:
$\Delta H_f^\circ = [-393.3 + 2 \times (-293.72)] - [-1108.76]$
$\Delta H_f^\circ = [-393.3 - 587.44] + 1108.76$
$\Delta H_f^\circ = -980.74 + 1108.76 = +128.02 \, kJ \, mol^{-1}$

(B) The enthalpy of hydration is the enthalpy change when anhydrous salt converts to its hydrated form.
$CuSO_4(s) + aq \to CuSO_4(aq)$,$\Delta H_1 = -15.9 \, kcal$ $(1)$
$CuSO_4 \cdot 5H_2O(s) + aq \to CuSO_4(aq)$,$\Delta H_2 = 2.8 \, kcal$ $(2)$
To find the enthalpy of hydration for the reaction $CuSO_4(s) + 5H_2O(l) \to CuSO_4 \cdot 5H_2O(s)$,we subtract equation $(2)$ from equation $(1)$:
$\Delta H_{hydration} = \Delta H_1 - \Delta H_2$
$\Delta H_{hydration} = -15.9 - 2.8 = -18.7 \, kcal$
The magnitude of the enthalpy of hydration is $18.7 \, kcal$.

(B) Hess's Law of Constant Heat Summation 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 enthalpy change depends only on the initial and final states of the reactants and products,not on the path taken.

(D) The given reactions are:
$1$) $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}$; $\Delta H_1 = -68.32 \, kcal$
$2$) $H_2O_{(l)} \rightarrow H_2O_{(g)}$; $\Delta H_2 = 10.52 \, kcal$
Adding these two equations gives the formation reaction for water vapor:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(g)}$
Applying Hess's Law:
$\Delta H_f(H_2O_{(g)}) = \Delta H_1 + \Delta H_2$
$\Delta H_f(H_2O_{(g)}) = -68.32 + 10.52 = -57.8 \, kcal$

(D) The enthalpy of neutralization is defined as the heat evolved when $1 \text{ gram equivalent}$ of an acid is neutralized by $1 \text{ gram equivalent}$ of a base.
For strong acids and strong bases,the enthalpy of neutralization is constant at approximately $-57.1 \text{ kJ/mol}$ because it involves the reaction $H^+ + OH^- \rightarrow H_2O$.
In the case of weak acids or weak bases,some energy is consumed in the dissociation of the weak electrolyte,resulting in a lower value of enthalpy of neutralization.
Since $NaOH$ is a strong base and $HCl$ is a strong acid,the reaction between them releases the maximum amount of energy.

(C) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2}I_{2(s)} + \frac{1}{2}Cl_{2(g)} \to ICl_{(g)}$,$\Delta H_f = ?$
From the given equations:
$(i) \frac{1}{2}Cl_{2(g)} \to Cl_{(g)}$,$\Delta H = \frac{242.3}{2} = 121.15 \text{ kJ mol}^{-1}$
$(ii) \frac{1}{2}I_{2(g)} \to I_{(g)}$,$\Delta H = \frac{151.0}{2} = 75.5 \text{ kJ mol}^{-1}$
$(iii) I_{(g)} + Cl_{(g)} \to ICl_{(g)}$,$\Delta H = -211.3 \text{ kJ mol}^{-1}$
$(iv) \frac{1}{2}I_{2(s)} \to \frac{1}{2}I_{2(g)}$,$\Delta H = \frac{62.76}{2} = 31.38 \text{ kJ mol}^{-1}$
Adding these equations:
$\Delta H_f = 121.15 + 75.5 - 211.3 + 31.38 = 16.73 \approx 16.8 \text{ kJ mol}^{-1}$.

(C) According to Hess's Law,the total enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Step $1$: $C + \frac{1}{2}O_2 \to CO$ ; $\Delta H_1 = -12 \ kJ$
Step $2$: $CO + \frac{1}{2}O_2 \to CO_2$ ; $\Delta H_2 = -10 \ kJ$
Adding the two equations:
$(C + \frac{1}{2}O_2) + (CO + \frac{1}{2}O_2) \to CO + CO_2$
$C + O_2 \to CO_2$
Therefore,$\Delta H = \Delta H_1 + \Delta H_2 = (-12) + (-10) = -22 \ kJ$.
Thus,$x = -22$.

(A) The standard enthalpy of formation,$\Delta H_f^o$,represents the heat change when one mole of a substance is formed from its constituent elements in their standard states.
For an explosive substance like $NCl_3$,the formation reaction is endothermic,meaning it absorbs energy from the surroundings to form the unstable compound.
Therefore,the enthalpy of formation $\Delta H_f^o$ for $NCl_3$ is positive $(> 0)$.

(A) Subtracting the second equation from the first:
$(S_{\text{(rhombic)}} + O_{2(g)}) - (S_{\text{(monoclinic)}} + O_{2(g)}) = -297.5 - (-300) \, kJ$
$S_{\text{(rhombic)}} - S_{\text{(monoclinic)}} = +2.5 \, kJ$
$S_{\text{(monoclinic)}} \rightarrow S_{\text{(rhombic)}}; \Delta H = -2.5 \, kJ$
Since the transition from monoclinic to rhombic releases energy (negative $\Delta H$),rhombic sulfur is at a lower energy state and is more stable than monoclinic sulfur.

(A) The heat of neutralization of a strong acid with a strong base is always constant at approximately $-13.6 \ kcal/mol$ or $-57.1 \ kJ/mol$.
This is because the reaction essentially involves the formation of water from $H^+$ and $OH^-$ ions: $H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$.
Among the given options,$HCl$ is a strong acid and $NaOH$ is a strong base.
Therefore,the heat of neutralization for $HCl$ and $NaOH$ is $-13.6 \ kcal/mol$.

(C) The chemical equation for the formation of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \to NH_{3(g)}$; $\Delta H_f = -46 \, kJ \, mol^{-1}$.
Using the bond enthalpy formula: $\Delta H_f = [\frac{1}{2} \times (B.E.)_{N \equiv N} + \frac{3}{2} \times (B.E.)_{H-H}] - [3 \times (B.E.)_{N-H}]$.
Substituting the given values: $-46 = [\frac{1}{2} \times 712 + \frac{3}{2} \times 436] - 3 \times (B.E.)_{N-H}$.
$-46 = [356 + 654] - 3 \times (B.E.)_{N-H}$.
$-46 = 1010 - 3 \times (B.E.)_{N-H}$.
$3 \times (B.E.)_{N-H} = 1010 + 46 = 1056$.
$(B.E.)_{N-H} = \frac{1056}{3} = 352 \, kJ \, mol^{-1}$.

(A) The given thermochemical equations are:
$(i)$ $C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H_1 = -x \, kJ$
(ii) $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)} \quad \Delta H_2 = -y \, kJ$
(iii) $CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H_3 = -z \, kJ$
We need to find the heat of formation of $CH_4$ for the reaction:
$C_{(s)} + 2H_{2(g)} \to CH_{4(g)} \quad \Delta H_f = ?$
Applying Hess's Law:
$\Delta H_f = \Delta H_1 + 2 \times \Delta H_2 - \Delta H_3$
$\Delta H_f = (-x) + 2(-y) - (-z)$
$\Delta H_f = (-x - 2y + z) \, kJ$

(A) The combustion reaction for $2 \, mol$ of $C_2H_6$ is: $2C_2H_6(g) + 7O_2(g) \to 4CO_2(g) + 6H_2O(l)$.
Given $\Delta H_{combustion} = -3129 \, kJ$.
The formula for enthalpy of reaction is: $\Delta H = \sum \Delta H_f(products) - \sum \Delta H_f(reactants)$.
Substituting the values: $-3129 = [4 \times (-395) + 6 \times (-286)] - [2 \times \Delta H_f(C_2H_6) + 7 \times 0]$.
$-3129 = [-1580 - 1716] - 2 \times \Delta H_f(C_2H_6)$.
$-3129 = -3296 - 2 \times \Delta H_f(C_2H_6)$.
$2 \times \Delta H_f(C_2H_6) = -3296 + 3129 = -167 \, kJ$.
$\Delta H_f(C_2H_6) = -167 / 2 = -83.5 \, kJ/mol$.

(B) The standard enthalpy of formation $(\Delta_f H^\circ)$ is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their standard states.
Therefore,the enthalpy of $1 \ mol$ of a compound is equal to its heat of formation.

(C) The reaction between a strong acid $(HCl)$ and a strong base releases $13.7 \ kcal$ of heat,which is the standard enthalpy of neutralization.
$NH_4OH$ is a weak base,while $HCl$ is a strong acid.
Part of the heat released during the neutralization process is consumed in the dissociation of the weak base $(NH_4OH)$.
Therefore,the total heat released is less than $13.7 \ kcal$.