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

(B) The heat of formation of $Ca(OH)_{2(s)}$ is the enthalpy change for the reaction: $Ca_{(s)} + O_{2(g)} + H_{2(g)} \to Ca(OH)_{2(s)}$.
We are given:
$(i) CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_1 = -15.26\,K\,cal$
$(ii) H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_2 = -68.37\,K\,cal$
$(iii) Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_3 = -151.80\,K\,cal$
Adding equations $(i)$,$(ii)$,and $(iii)$:
$CaO_{(s)} + H_2O_{(l)} + H_{2(g)} + \frac{1}{2}O_{2(g)} + Ca_{(s)} + \frac{1}{2}O_{2(g)} \to Ca(OH)_{2(s)} + H_2O_{(l)} + CaO_{(s)}$
Canceling common terms on both sides:
$Ca_{(s)} + O_{2(g)} + H_{2(g)} \to Ca(OH)_{2(s)}$
The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-15.26) + (-68.37) + (-151.80) = -235.43\,K\,cal$.

(B) The formation reaction for $CO$ is: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$.
Given equations:
$(1) \ C + O_2 \rightarrow CO_2$ ; $\Delta H = -x \ kJ$
$(2) \ 2CO + O_2 \rightarrow 2CO_2$ ; $\Delta H = -y \ kJ$
Divide equation $(2)$ by $2$:
$CO + \frac{1}{2}O_2 \rightarrow CO_2$ ; $\Delta H = -\frac{y}{2} \ kJ$ $(3)$
Subtract equation $(3)$ from equation $(1)$:
$(C + O_2) - (CO + \frac{1}{2}O_2) \rightarrow CO_2 - CO_2$
$C + \frac{1}{2}O_2 \rightarrow CO$
Enthalpy change $\Delta H_f = (-x) - (-\frac{y}{2}) = \frac{y}{2} - x = \frac{y - 2x}{2} \ kJ$.

(D) According to Hess's Law,the total enthalpy change for a chemical reaction is independent of the path taken,meaning it does not depend on the number of steps in the reaction.

(C) Let the bond dissociation energies be $E_{XY} = x$,$E_{X_2} = x$,and $E_{Y_2} = 0.5x$.
The reaction for the formation of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g)$.
The enthalpy of formation is given by: $\Delta_fH = \sum E_{\text{reactants}} - \sum E_{\text{products}}$.
Substituting the values: $-200 = [\frac{1}{2} E_{X_2} + \frac{1}{2} E_{Y_2}] - E_{XY}$.
$-200 = [\frac{1}{2}(x) + \frac{1}{2}(0.5x)] - x$.
$-200 = 0.5x + 0.25x - x$.
$-200 = -0.25x$.
$x = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.
Thus,the bond dissociation energy of $X_2$ is $800 \ kJ \ mol^{-1}$.

(B) The heat of neutralization of any strong acid by a strong base is constant and is equal to $-13.7 \ kcal$ per equivalent of $H^+$ ions neutralized by $OH^-$ ions.
Since the heat of neutralization is defined per equivalent of water formed,it remains $-13.7 \ kcal/equivalent$ regardless of whether the acid is monobasic or dibasic.

(D) The given equations are:
$(i) \ Zn + \frac{1}{2}O_2 \rightarrow ZnO; \ \Delta H_1 = -84000 \ cal$
$(ii) \ Hg + \frac{1}{2}O_2 \rightarrow HgO; \ \Delta H_2 = -21700 \ cal$
We need to find $\Delta H$ for the reaction:
$(iii) \ Zn + HgO \rightarrow ZnO + Hg$
Applying Hess's Law,we can obtain equation $(iii)$ by subtracting equation $(ii)$ from equation $(i)$:
$(iii) = (i) - (ii)$
$\Delta H = \Delta H_1 - \Delta H_2$
$\Delta H = -84000 - (-21700)$
$\Delta H = -84000 + 21700 = -62300 \ cal$

(B) For the thermochemical equation $Fe_2O_{3(s)} + 3CO_{(g)} = 2Fe_{(s)} + 3CO_{2(g)}$,the standard heat of reaction $\Delta H_{rxn}^o = -6.6 \, kcal$.
Using the formula: $\Delta H_{rxn}^o = [\Sigma \Delta H_f^o (products) - \Sigma \Delta H_f^o (reactants)]$
Substituting the values:
$-6.6 = [2 \times \Delta H_f^o(Fe_{(s)}) + 3 \times \Delta H_f^o(CO_{2(g)})] - [\Delta H_f^o(Fe_2O_{3(s)}) + 3 \times \Delta H_f^o(CO_{(g)})]$
Since $\Delta H_f^o$ of an element in its standard state is $0$,$\Delta H_f^o(Fe_{(s)}) = 0$.
$-6.6 = [2(0) + 3(-94)] - [\Delta H_f^o(Fe_2O_{3(s)}) + 3(-26.4)]$
$-6.6 = [-282] - [\Delta H_f^o(Fe_2O_{3(s)}) - 79.2]$
$-6.6 = -282 - \Delta H_f^o(Fe_2O_{3(s)}) + 79.2$
$-6.6 = -202.8 - \Delta H_f^o(Fe_2O_{3(s)})$
$\Delta H_f^o(Fe_2O_{3(s)}) = -202.8 + 6.6 = -196.2 \, kcal/mol$.

(A) The heat of neutralization of a strong acid with a strong base is always $-57.3 \, kJ/mol$.
For a weak acid like $HCN$,the heat of neutralization is given by:
$\Delta H_{neut} = \Delta H_{ionization} + \Delta H_{neutralization(strong \, acid-strong \, base)}$
$-12.1 = \Delta H_{ionization} + (-57.3)$
$\Delta H_{ionization} = 57.3 - 12.1 = 45.2 \, kJ/mol$.

(D) The combustion reaction of $CS_2$ is: $CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)$.
The enthalpy of combustion is given by: $\Delta H_c = [\Delta H_f(CO_2) + 2 \times \Delta H_f(SO_2)] - [\Delta H_f(CS_2) + 3 \times \Delta H_f(O_2)]$.
Given: $\Delta H_c = -110.2 \, kJ \, mol^{-1}$,$\Delta H_f(CO_2) = -394.5 \, kJ \, mol^{-1}$,$\Delta H_f(SO_2) = -297.4 \, kJ \, mol^{-1}$,and $\Delta H_f(O_2) = 0$.
Substituting the values: $-110.2 = [-394.5 + 2 \times (-297.4)] - [\Delta H_f(CS_2) + 0]$.
$-110.2 = [-394.5 - 594.8] - \Delta H_f(CS_2)$.
$-110.2 = -989.3 - \Delta H_f(CS_2)$.
$\Delta H_f(CS_2) = -989.3 + 110.2 = -879.1 \, kJ \, mol^{-1}$.

(A) The heat of formation of $H_2SO_4$ is the enthalpy change for the reaction: $S(s) + H_2(g) + 2O_2(g) \rightarrow H_2SO_4(l)$.
By adding the given thermochemical equations:
$(1) \ S + O_2 \rightarrow SO_2, \Delta H_1 = -298.2 \ kJ$
$(2) \ SO_2 + 1/2 \ O_2 \rightarrow SO_3, \Delta H_2 = -98.7 \ kJ$
$(3) \ SO_3 + H_2O \rightarrow H_2SO_4, \Delta H_3 = -130.2 \ kJ$
$(4) \ H_2 + 1/2 \ O_2 \rightarrow H_2O, \Delta H_4 = -287.3 \ kJ$
Summing these equations:
$(S + O_2) + (SO_2 + 1/2 \ O_2) + (SO_3 + H_2O) + (H_2 + 1/2 \ O_2) \rightarrow SO_2 + SO_3 + H_2SO_4 + H_2O$
Canceling common terms on both sides gives:
$S + H_2 + 2O_2 \rightarrow H_2SO_4$
Therefore,$\Delta H_f(H_2SO_4) = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$
$\Delta H_f = -298.2 + (-98.7) + (-130.2) + (-287.3) = -814.4 \ kJ$.

(B) The enthalpy change of the reaction is calculated using the formula: $\Delta H = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$.
For the reaction $H_2C=CH_{2(g)} + H_{2(g)} \rightarrow H_3C-CH_{3(g)}$:
Bonds broken: $1 \times (C=C)$ and $1 \times (H-H)$.
Bonds formed: $1 \times (C-C)$ and $2 \times (C-H)$.
Note: The reactants have $4 \times (C-H)$ bonds and the product has $6 \times (C-H)$ bonds. The net change is the formation of $2 \times (C-H)$ bonds.
$\Delta H = [1(B.E.)_{C=C} + 4(B.E.)_{C-H} + 1(B.E.)_{H-H}] - [1(B.E.)_{C-C} + 6(B.E.)_{C-H}]$
$\Delta H = (615 + 4 \times 414 + 435) - (347 + 6 \times 414)$
$\Delta H = (615 + 1656 + 435) - (347 + 2484)$
$\Delta H = 2706 - 2831 = -125 \, kJ \, mol^{-1}$.

(D) The combustion reactions are:
$(i) \, C_{(s)} + O_{2(g)} \to CO_{2(g)} \, ; \, \Delta H_1 = -393.5 \, kJ/mol$
$(ii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \, ; \, \Delta H_2 = -283 \, kJ/mol$
We need the enthalpy of formation of $CO_{(g)}$,which is the reaction:
$C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$
This can be obtained by subtracting equation $(ii)$ from equation $(i)$:
$\Delta H_f = \Delta H_1 - \Delta H_2$
$\Delta H_f = -393.5 - (-283) = -393.5 + 283 = -110.5 \, kJ/mol$

(B) The enthalpy of reaction is given by the formula: $\Delta H = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$.
Given: $\Delta H = -182 \ kJ \ mol^{-1}$,$(B.E.)_{H-H} = 430 \ kJ \ mol^{-1}$,and $(B.E.)_{Cl-Cl} = 242 \ kJ \ mol^{-1}$.
The reaction is $H_2 + Cl_2 \rightarrow 2HCl$.
Substituting the values: $-182 = [1 \times (B.E.)_{H-H} + 1 \times (B.E.)_{Cl-Cl}] - [2 \times (B.E.)_{H-Cl}]$.
$-182 = (430 + 242) - 2 \times (B.E.)_{H-Cl}$.
$-182 = 672 - 2 \times (B.E.)_{H-Cl}$.
$2 \times (B.E.)_{H-Cl} = 672 + 182 = 854$.
$(B.E.)_{H-Cl} = \frac{854}{2} = 427 \ kJ \ mol^{-1}$.

(C) Given reactions:
$(1) \, C_{\text{diamond}} + O_2 \to CO_2; \, \Delta H_1 = -395.4 \, kJ$
$(2) \, C_{\text{graphite}} + O_2 \to CO_2; \, \Delta H_2 = -393.5 \, kJ$
To find the enthalpy of transformation: $C_{\text{diamond}} \to C_{\text{graphite}}$,we subtract equation $(2)$ from equation $(1)$:
$(C_{\text{diamond}} + O_2) - (C_{\text{graphite}} + O_2) = (-395.4) - (-393.5)$
$C_{\text{diamond}} - C_{\text{graphite}} = -395.4 + 393.5$
$C_{\text{diamond}} \to C_{\text{graphite}}; \, \Delta H = -1.9 \, kJ$

(C) The formation reaction of $CO$ is given by:
$C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}, \Delta H = -110.5 \ kJ/mol$
For this reaction,$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$
$\Delta n_g = 1 - \frac{1}{2} = 0.5$
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta U$ is the change in internal energy at constant volume:
$-110.5 = \Delta U + (0.5) \times (8.314 \times 10^{-3} \ kJ/K \cdot mol) \times (298 \ K)$
$-110.5 = \Delta U + 1.2388$
$\Delta U = -110.5 - 1.2388 = -111.7388 \ kJ/mol$
Rounding to two decimal places,we get $-111.73 \ kJ/mol$.

(C) The combustion of graphite $(C_{(s)})$ is represented by the reaction:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
The heat of combustion of graphite is equal to the standard enthalpy of formation of $CO_2$ because the reaction represents the formation of $1 \, mol$ of $CO_2$ from its constituent elements in their standard states.
Given,$\Delta H_f^{\circ} (CO_2) = -394 \, kJ \, mol^{-1}$.
Therefore,the heat of combustion of graphite is $-394 \, kJ \, mol^{-1}$.

(A) According to $Hess's \ Law$ of constant heat summation,the total enthalpy change for a chemical reaction is the same,whether the reaction takes place in one step or in several steps.
Therefore,the enthalpy change depends only on the initial and final states of the system and is independent of the path taken or the nature of the intermediate reaction steps.
However,it does depend on the physical states of the reactants and products,the temperature,and the specific reactants used to form the products.

(D) The reaction is: $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$; $\Delta H = 42 \, kJ$.
The enthalpy of reaction is given by the formula: $\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$.
$\Delta H = [\Delta H_f(CaO) + \Delta H_f(CO_2)] - \Delta H_f(CaCO_3)$.
Substituting the given values: $42 = [-152 + (-94)] - \Delta H_f(CaCO_3)$.
$42 = -246 - \Delta H_f(CaCO_3)$.
$\Delta H_f(CaCO_3) = -246 - 42 = -288 \, kJ$.

(B) The reaction is: $CO_{2(g)} + H_{2(g)} \rightarrow CO_{(g)} + H_2O_{(g)}$
The standard enthalpy change of the reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$
Substituting the given values:
$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$
Since the standard enthalpy of formation of an element in its standard state,such as $H_{2(g)}$,is $0 \ kJ \ mol^{-1}$,we have:
$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$
$\Delta H^o = -352.3 + 393.5 = +41.2 \ kJ \ mol^{-1}$

(A) The formation reaction of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$.
The enthalpy of formation $\Delta_f H$ is calculated as: $\Delta_f H = [\sum \text{Bond Enthalpies of Reactants}] - [\sum \text{Bond Enthalpies of Products}]$.
$\Delta_f H = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - [BE(H-Cl)]$.
$\Delta_f H = [\frac{1}{2} \times 434 + \frac{1}{2} \times 242] - 431$.
$\Delta_f H = [217 + 121] - 431 = 338 - 431 = -93 \, kJ/mol$.

(B) The formation reaction of ammonia is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightarrow NH_{3(g)}$; $\Delta H_f = -46 \ kJ \ mol^{-1}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the enthalpy change is $\Delta H = 2 \times (-46) = -92 \ kJ \ mol^{-1}$.
The given reaction $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$ is the reverse of the above reaction.
Therefore,the enthalpy change for the reverse reaction is $\Delta H = -(-92) = +92 \ kJ \ mol^{-1}$.

(B) The enthalpy of formation of $H_2O_{(l)}$ is given as $\Delta H_f(H_2O, l) = -286.20 \ KJ/mol$.
For the reaction: $H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)}$,the enthalpy change is $\Delta H_r = 57.32 \ KJ/mol$.
The enthalpy of reaction is calculated as:
$\Delta H_r = [\Delta H_f(H^+, aq) + \Delta H_f(OH^-, aq)] - \Delta H_f(H_2O, l)$
By convention,the enthalpy of formation of $H^+_{(aq)}$ is $0 \ KJ/mol$.
Substituting the values:
$57.32 = [0 + \Delta H_f(OH^-, aq)] - (-286.20)$
$57.32 = \Delta H_f(OH^-, aq) + 286.20$
$\Delta H_f(OH^-, aq) = 57.32 - 286.20 = -228.88 \ KJ/mol$.

(A) The enthalpy of formation of $H_2SO_4$ corresponds to the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$.
To obtain this,we add the given equations:
$(1) S + O_2 \rightarrow SO_2 ; \Delta H_1 = -298.2 \ kJ$
$(2) SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H_2 = -98.7 \ kJ$
$(3) SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H_3 = -130.2 \ kJ$
$(4) H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H_4 = -287.3 \ kJ$
Summing these equations:
$(S + O_2) + (SO_2 + \frac{1}{2} O_2) + (SO_3 + H_2O) + (H_2 + \frac{1}{2} O_2) \rightarrow SO_2 + SO_3 + H_2SO_4 + H_2O$
Canceling common terms on both sides:
$S + H_2 + (1 + \frac{1}{2} + \frac{1}{2}) O_2 \rightarrow H_2SO_4$
$S + H_2 + 2O_2 \rightarrow H_2SO_4$
The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$
$\Delta H = (-298.2) + (-98.7) + (-130.2) + (-287.3) = -814.4 \ kJ$.

(D) The neutralization reaction between a strong acid and a strong base involves the combination of $H^+$ ions and $OH^-$ ions to form water: $H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$.
Since strong acids and strong bases are completely dissociated in aqueous solution,the net reaction is always the same.
The enthalpy change for this reaction is constant at approximately $-57.1 \ kJ \ mol^{-1}$ at $298 \ K$.

(D) The bond dissociation energy (or bond energy) is defined as the energy required to break one mole of bonds in a gaseous molecule to produce gaseous atoms.
For $HCl_{(g)}$,the reaction representing the breaking of the $H-Cl$ bond into its constituent gaseous atoms is:
$HCl_{(g)} \rightarrow H_{(g)} + Cl_{(g)}$
Therefore,the correct option is $D$.

(B) The stability of a compound is inversely proportional to its standard enthalpy of formation $(\Delta_fH^\circ)$.
More negative (lower) the value of $\Delta_fH^\circ$,more stable is the compound.
The given values are:
$A (NH_3) = -46.19 \, kJ/mol$
$B (CO_2) = -393.4 \, kJ/mol$
$C (HI) = +24.94 \, kJ/mol$
$D (SO_2) = -296.9 \, kJ/mol$
Comparing the values: $+24.94 > -46.19 > -296.9 > -393.4$.
Therefore,the increasing order of stability is $C < A < D < B$.

(B) The combustion reactions are given as:
$(i)$ $C(s) + O_2(g) \rightarrow CO_2(g); \Delta H_1 = -393.5 \, kJ \, mol^{-1}$
(ii) $CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g); \Delta H_2 = -283 \, kJ \, mol^{-1}$
We need to find the enthalpy of formation of $CO(g)$,which corresponds to the reaction:
(iii) $C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g); \Delta H_f = ?$
Subtracting equation (ii) from equation $(i)$:
$(C(s) + O_2(g)) - (CO(g) + \frac{1}{2} O_2(g)) = CO_2(g) - CO_2(g)$
$C(s) + \frac{1}{2} O_2(g) - CO(g) = 0$
$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$
Therefore,$\Delta H_f = \Delta H_1 - \Delta H_2$
$\Delta H_f = -393.5 - (-283) = -393.5 + 283 = -110.5 \, kJ \, mol^{-1}$

(B) The enthalpy of combustion $(\Delta H_c)$ is defined as the enthalpy change when $1 \ mole$ of a substance is completely burnt in excess of oxygen.
Since combustion reactions involve the breaking of bonds and the release of energy in the form of heat and light,they are inherently spontaneous and release energy to the surroundings.
Therefore,the enthalpy of combustion is always negative,meaning the process is always exothermic.

(D) 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 law is a direct consequence of the fact that enthalpy is a state function.
It is widely used to calculate the enthalpy of reactions that are difficult to measure directly,such as the enthalpy of formation,enthalpy of combustion,enthalpy of neutralization,and other complex reaction enthalpies.
Therefore,all the given options can be determined using Hess's Law.

(A) The reaction is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
The enthalpy change of the reaction is given by: $\Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$
Given: $\Delta H_f^\circ (CH_{4(g)}) = -75 \ kJ \ mol^{-1}$,$\Delta H_f^\circ (CO_{2(g)}) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f^\circ (H_2O_{(l)}) = -286 \ kJ \ mol^{-1}$,and $\Delta H_f^\circ (O_{2(g)}) = 0 \ kJ \ mol^{-1}$ (standard state).
Substituting the values:
$\Delta H^\circ = [(-393.5) + 2(-286)] - [(-75) + 2(0)]$
$\Delta H^\circ = [-393.5 - 572] - [-75]$
$\Delta H^\circ = -965.5 + 75$
$\Delta H^\circ = -890.5 \ kJ$

(D) The enthalpy of formation reaction for sucrose is: $12C(s) + 11H_2(g) + \frac{11}{2}O_2(g) \to C_{12}H_{22}O_{11}(s)$.
From the given equations:
$(i) C_{12}H_{22}O_{11} + 12O_2 \to 12CO_2 + 11H_2O, \Delta H_1 = -5200.7 \, kJ \, mol^{-1}$
$(ii) C + O_2 \to CO_2, \Delta H_2 = -394.5 \, kJ \, mol^{-1}$
$(iii) H_2 + \frac{1}{2}O_2 \to H_2O, \Delta H_3 = -285.8 \, kJ \, mol^{-1}$
To get the formation reaction,we perform: $12 \times (ii) + 11 \times (iii) - (i)$:
$\Delta H_f = 12(-394.5) + 11(-285.8) - (-5200.7)$
$\Delta H_f = -4734 - 3143.8 + 5200.7$
$\Delta H_f = -7877.8 + 5200.7 = -2677.1 \, kJ \, mol^{-1}$.

(B) The neutralization reaction between a strong acid and a strong base is an exothermic process,meaning heat is released.
Therefore,the enthalpy change $(\Delta H)$ is always negative.
For example,$H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)$; $\Delta H = -57.1 \ kJ \ mol^{-1}$.

(B) The combustion reactions are:
$1$) $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta H_1 = -212.4 \, kcal$
$2$) $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$; $\Delta H_2 = -94.0 \, kcal$
$3$) $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}$; $\Delta H_3 = -68.4 \, kcal$
We need the heat of formation for $CH_4$: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
This can be obtained by: $\Delta H_f = \Delta H_2 + 2(\Delta H_3) - \Delta H_1$
$\Delta H_f = -94.0 + 2(-68.4) - (-212.4)$
$\Delta H_f = -94.0 - 136.8 + 212.4$
$\Delta H_f = -230.8 + 212.4 = -18.4 \, kcal$

(D) The formation reaction of methane is: $C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$.
To calculate the average $C-H$ bond energy,we need to consider the atomization of the reactants into gaseous atoms: $C_{(s)} \rightarrow C_{(g)}$ (enthalpy of sublimation) and $2H_{2(g)} \rightarrow 4H_{(g)}$ (dissociation energy of $H_2$).
Then,the enthalpy of formation is related to the bond energies by the equation: $\Delta_fH^o = [\Delta_{sub}H(C) + 2 \times BE(H-H)] - [4 \times BE(C-H)]$.
Thus,we need the enthalpy of sublimation of carbon and the dissociation energy of $H_2$.

(A) The energy required to break one mole of $Cl-Cl$ bonds is $242 \, kJ \, mol^{-1}$.
To break a single $Cl-Cl$ bond,the energy required $(E)$ is: $E = \frac{242 \times 10^3 \, J}{6.022 \times 10^{23}} \approx 4.0186 \times 10^{-19} \, J$.
Using the relation $E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is given by $\lambda = \frac{hc}{E}$.
Substituting the values: $\lambda = \frac{(6.626 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m/s)}{4.0186 \times 10^{-19} \, J}$.
$\lambda \approx 4.947 \times 10^{-7} \, m = 494.7 \times 10^{-9} \, m = 494.7 \, nm$.
Rounding to the nearest integer,we get $494 \, nm$.

(B) The heat of neutralization is defined as the heat released when $1 \ gram \ equivalent$ of an acid is neutralized by $1 \ gram \ equivalent$ of a base.
Since the volumes and molarities are equal,the number of moles of $HCl$ is $n$ and $H_2SO_4$ is $n$.
$HCl$ provides $n \ moles$ of $H^+$ ions,while $H_2SO_4$ provides $2n \ moles$ of $H^+$ ions.
Therefore,the heat released by $H_2SO_4$ $(y)$ will be twice the heat released by $HCl$ $(x)$.
Thus,$y = 2x$ or $x = \frac{y}{2}$.

(C) The enthalpy of vaporization of water is positive,i.e.,$H_2O_{(l)} \to H_2O_{(g)}$,$\Delta H_{vap} > 0$.
We can write the two reactions as:
$1) H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)} \quad \Delta H_2$
$2) H_2O_{(l)} \to H_2O_{(g)} \quad \Delta H_{vap}$
Adding these two reactions gives:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)} \quad \Delta H_1 = \Delta H_2 + \Delta H_{vap}$
Since $\Delta H_{vap} > 0$,it follows that $\Delta H_1 > \Delta H_2$.

(A) The structure of hydrazine is $H_2N-NH_2$,which contains $4$ $N-H$ bonds and $1$ $N-N$ bond.
The total enthalpy change for the dissociation is given by: $\Delta H = 4 \times BE(N-H) + 1 \times BE(N-N)$.
Given $\Delta H = 1724 \ kJ \ mol^{-1}$ and $BE(N-H) = 391 \ kJ \ mol^{-1}$.
Substituting the values: $1724 = 4 \times 391 + BE(N-N)$.
$1724 = 1564 + BE(N-N)$.
$BE(N-N) = 1724 - 1564 = 160 \ kJ \ mol^{-1}$.

(D) Energy absorbed by molecule $(A_2) = 4.4 \times 10^{-19} \ J$
Bond energy per molecule $(A_2) = 4.0 \times 10^{-19} \ J$
Excess energy available as kinetic energy $(K.E.) = \text{Energy absorbed} - \text{Bond energy}$
$K.E. = 4.4 \times 10^{-19} - 4.0 \times 10^{-19} = 0.4 \times 10^{-19} \ J$
Since the molecule is $(A_2)$,it contains $2$ atoms.
$K.E.$ per atom $= \frac{0.4 \times 10^{-19} \ J}{2} = 0.2 \times 10^{-19} \ J = 2 \times 10^{-20} \ J$
Thus,the kinetic energy per atom is $2 \times 10^{-20} \ J$.

(C) The combustion of carbon to $CO_2$ is represented as:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ/mol$
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Heat released for the formation of $44 \ g$ of $CO_2$ is $393.5 \ kJ$.
Therefore,the heat released for the formation of $35.2 \ g$ of $CO_2$ is:
$\text{Heat} = \frac{393.5 \ kJ}{44 \ g} \times 35.2 \ g = 314.8 \ kJ$.
Since the process is exothermic,the heat released is $314.8 \ kJ$ (or $\Delta H = -314.8 \ kJ$).

(B) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.
Adding reaction $(ii)$ and reaction $(iii)$:
$(C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}) + (CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)})$
This simplifies to:
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}$
Therefore,the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual steps:
$x = y + z$

(A) The given reaction is $4H_{(g)} \rightarrow 2H_{2_{(g)}}$ with $\Delta H = -869.6 \ kJ$.
This reaction represents the formation of $2$ moles of $H_2$ molecules from $4$ moles of $H$ atoms.
The dissociation energy of a bond is defined as the energy required to break $1$ mole of bonds in the gaseous state.
First,reverse the reaction to represent the dissociation of $2$ moles of $H_2$ molecules:
$2H_{2_{(g)}} \rightarrow 4H_{(g)}$; $\Delta H = +869.6 \ kJ$.
Now,for $1$ mole of $H_2$ dissociation:
$H_{2_{(g)}} \rightarrow 2H_{(g)}$; $\Delta H = \frac{869.6}{2} = 434.8 \ kJ$.
Thus,the dissociation energy of the $H-H$ bond is $434.8 \ kJ$.

(B) To obtain the target reaction $B + D \rightarrow E + 2C$,we manipulate the given equations:
$1. \ 2 \times (I): A \rightarrow 2B, \Delta H = 2 \times 150 = 300 \ kJ/mol$
$2. \ (II): 3B \rightarrow 2C + D, \Delta H = -125 \ kJ/mol$
$3. \ -(III): 2D \rightarrow E + A, \Delta H = -350 \ kJ/mol$
Adding these equations:
$(A + 3B + 2D) \rightarrow (2B + 2C + D + E + A)$
Simplifying by canceling common terms ($A$ and $2B$ and $D$ from both sides):
$B + D \rightarrow E + 2C$
$\Delta H = 300 - 125 - 350 = -175 \ kJ/mol$

(D) Given reactions:
$(I) \ Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H_1 = -26.8 \ kJ$
$(II) \ FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H_2 = -16.5 \ kJ$
We need to find $\Delta H$ for:
$(III) \ Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$
To obtain reaction $(III)$,we perform the operation: $(I) - 2 \times (II)$
$\Delta H = \Delta H_1 - 2(\Delta H_2)$
$\Delta H = -26.8 - 2(-16.5)$
$\Delta H = -26.8 + 33.0 = +6.2 \ kJ$

(B) The enthalpy of reaction $(\Delta H)$ is calculated using the formula:
$\Delta H = \sum \text{Bond Energy (Reactants)} - \sum \text{Bond Energy (Products)}$
For the reaction: $CH_2=CH_2 + H-H \to CH_3-CH_3$
Reactants have: $1 \times (C=C)$,$4 \times (C-H)$,and $1 \times (H-H)$ bonds.
Products have: $1 \times (C-C)$ and $6 \times (C-H)$ bonds.
$\Delta H = [BE(C=C) + 4 \times BE(C-H) + BE(H-H)] - [BE(C-C) + 6 \times BE(C-H)]$
$\Delta H = BE(C=C) + BE(H-H) - BE(C-C) - 2 \times BE(C-H)$
$\Delta H = 606.10 + 431.37 - 336.49 - 2(410.50)$
$\Delta H = 1037.47 - 336.49 - 821.00$
$\Delta H = -120.02 \approx -120 \text{ kJ mol}^{-1}$.

(A) The chemical equation for the formation of $1 \ mol$ of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$
The enthalpy of reaction $(\Delta H_f)$ can be calculated using bond dissociation enthalpies as:
$\Delta H_f = \Sigma (B.E.)_{\text{reactants}} - \Sigma (B.E.)_{\text{products}}$
$\Delta H_f = [\frac{1}{2} \times B.E.(H-H) + \frac{1}{2} \times B.E.(Cl-Cl)] - [B.E.(H-Cl)]$
$\Delta H_f = [\frac{1}{2} \times 434 + \frac{1}{2} \times 242] - [431]$
$\Delta H_f = [217 + 121] - 431$
$\Delta H_f = 338 - 431 = -93 \ kJ \ mol^{-1}$