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

(D) The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements in their standard states.
The equation representing the standard enthalpy of formation of $H_2O_{(l)}$ is the reaction where $1 \ mol$ of $H_2O_{(l)}$ is formed from its constituent elements,$H_{2_{(g)}}$ and $O_{2_{(g)}}$.
This is given by reaction $(ii)$: $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$.
Therefore,the enthalpy of formation of $H_2O_{(l)}$ is $-X_2 \ kJ \ mol^{-1}$.

(B) The formation reaction for $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$.
The enthalpy of formation is given by: $\Delta H_f = [\sum \text{Bond Energy of Reactants}] - [\sum \text{Bond Energy of Products}]$.
Substituting the given values: $-90 = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - [BE(H-Cl)]$.
$-90 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 240] - BE(H-Cl)$.
$-90 = [215 + 120] - BE(H-Cl)$.
$-90 = 335 - BE(H-Cl)$.
$BE(H-Cl) = 335 + 90 = 425 \ kJ \ mol^{-1}$.

(D) Energy required to break one mole of $Cl_2$ molecules is $242 \, kJ \, mol^{-1} = 242 \times 10^3 \, J \, mol^{-1}.$
Energy required to break a single $Cl-Cl$ bond is $E = \frac{242 \times 10^3}{6.02 \times 10^{23}} \, J.$
Using the relation $E = \frac{hc}{\lambda}$,the wavelength is $\lambda = \frac{hc}{E}.$
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s \times 3 \times 10^8 \, m \cdot s^{-1} \times 6.02 \times 10^{23} \, mol^{-1}}{242 \times 10^3 \, J \cdot mol^{-1}}.$
$\lambda \approx 4.94 \times 10^{-7} \, m = 494 \times 10^{-9} \, m = 494 \, nm.$

(C) The standard enthalpy of formation of $CH_4$ is given by the reaction:
$C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
To calculate the average $C-H$ bond energy,we need to consider the atomization of the reactants:
$1.$ Enthalpy of sublimation of carbon: $C_{(s)} \rightarrow C_{(g)}$
$2.$ Dissociation energy of hydrogen: $2H_{2(g)} \rightarrow 4H_{(g)}$
By using Hess's Law,we combine these with the enthalpy of formation to find the total bond dissociation energy of $CH_4$,which is then divided by $4$ to get the average $C-H$ bond energy.
Therefore,the dissociation energy of $H_2$ and the enthalpy of sublimation of carbon are required.

(A) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow ICl_{(g)}$
The enthalpy of formation is calculated using the bond dissociation energies and sublimation energy:
$\Delta H_f^o = [\frac{1}{2} \Delta H_{sub}(I_2) + \frac{1}{2} BE(I-I) + \frac{1}{2} BE(Cl-Cl)] - BE(I-Cl)$
Substituting the given values:
$\Delta H_f^o = [\frac{1}{2}(62.76) + \frac{1}{2}(151.0) + \frac{1}{2}(242.3)] - 211.3$
$\Delta H_f^o = [31.38 + 75.5 + 121.15] - 211.3$
$\Delta H_f^o = 228.03 - 211.3 = 16.73 \ kJ \ mol^{-1}$
Rounding to one decimal place,we get $+16.8 \ kJ \ mol^{-1}$.

(A) Given,for reaction:
$(I) \quad H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \quad \Delta H_r = 57.32 \, kJ$
$(II) \quad H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \quad \Delta H_r = -286.20 \, kJ$
For reaction $(I)$:
$\Delta H_r = \Delta H_f^o(H^{+}, aq) + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell)$
Since $\Delta H_f^o(H^{+}, aq) = 0$,we have:
$57.32 = 0 + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell) \quad \dots (III)$
For reaction $(II)$:
$\Delta H_f^o(H_2O, \ell) = -286.20 \, kJ$
Substituting this value in equation $(III)$:
$57.32 = \Delta H_f^o(OH^{-}, aq) - (-286.20)$
$\Delta H_f^o(OH^{-}, aq) = 57.32 - 286.20$
$\Delta H_f^o(OH^{-}, aq) = -228.88 \, kJ$

(B) The reaction for the formation of $2 \ mol$ of $NH_3$ is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
The enthalpy change for this reaction is $\Delta H = 2 \times \Delta_f H^{\circ}(NH_3) = 2 \times (-46.0) = -92 \ kJ \ mol^{-1}$.
Using bond enthalpies,$\Delta H = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$.
Given bond energies (as enthalpy of formation from atoms is negative of bond dissociation energy): $BE(N \equiv N) = 712 \ kJ \ mol^{-1}$ and $BE(H-H) = 436 \ kJ \ mol^{-1}$.
Let $x$ be the bond enthalpy of $N-H$ bond. There are $6$ $N-H$ bonds in $2 \ mol$ of $NH_3$.
$-92 = [BE(N \equiv N) + 3 \times BE(H-H)] - 6x$.
$-92 = [712 + 3 \times 436] - 6x$.
$-92 = [712 + 1308] - 6x$.
$-92 = 2020 - 6x$.
$6x = 2020 + 92 = 2112$.
$x = 2112 / 6 = +352 \ kJ \ mol^{-1}$.

(B) The heat of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.
Given reactions:
$1) C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H_1 = -393.5 \ kJ \ mol^{-1}$
$2) CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H_2 = -283.5 \ kJ \ mol^{-1}$
To obtain the target reaction,subtract equation $(2)$ from equation $(1)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
Therefore,$\Delta H_f = \Delta H_1 - \Delta H_2 = -393.5 - (-283.5) = -110 \ kJ \ mol^{-1}$.

(D) Subtract the second equation from the first:
$(C (diamond) + O_2$ $\rightarrow CO_{2(g)}) - (C (graphite) + O_2$ $\rightarrow CO_{2(g)})$
$\Delta H = -97.6 - (-94.3) = -3.3 \ kcal$
So,$C (diamond) \rightarrow C (graphite)$; $\Delta H = -3.3 \ kcal$ for $1 \ mole$ $(12 \ g)$ of carbon.
Heat change for $12 \ g$ of diamond to graphite $= -3.3 \ kcal$.
Heat change for $1 \ g$ of diamond to graphite $= \frac{-3.3 \ kcal}{12 \ g} = -0.275 \ kcal$.
The magnitude of the heat change is $0.275 \ kcal$.

(D) The target equation for the formation of acetylene is: $2C_{(s)} + H_{2(g)} \to C_2H_{2(g)}$
Let the given equations be:
$(i) 2C_{(s)} + 2O_{2(g)} \to 2CO_{2(g)}$; $\Delta H_1 = -787 \ kJ$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}$; $\Delta H_2 = -286 \ kJ$
$(iii) C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \to 2CO_{2(g)} + H_2O_{(l)}$; $\Delta H_3 = -1301 \ kJ$
To obtain the target equation,perform: $(i) + (ii) - (iii)$
$\Delta H_f = \Delta H_1 + \Delta H_2 - \Delta H_3$
$\Delta H_f = (-787) + (-286) - (-1301)$
$\Delta H_f = -1073 + 1301$
$\Delta H_f = +228 \ kJ$

(C) The reaction for the trimerisation of ethyne to benzene is:
$3C_2H_{2(g)} \to C_6H_{6(g)}$
The standard heat of reaction $(\Delta H_{r}^o)$ is calculated using the formula:
$\Delta H_{r}^o = \sum \Delta H_{f(\text{products})}^o - \sum \Delta H_{f(\text{reactants})}^o$
$\Delta H_{r}^o = \Delta H_{f(C_6H_6)(g)}^o - 3 \times \Delta H_{f(C_2H_2)(g)}^o$
Substituting the given values:
$\Delta H_{r}^o = 85 - 3(230)$
$\Delta H_{r}^o = 85 - 690$
$\Delta H_{r}^o = -605 \ kJ \ mol^{-1}$

(B) The heat of formation is the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
The target equation is: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} \to PCl_{5(s)}$.
From the first given equation:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
Dividing by $2$ gives: $\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)} \to PCl_{3(l)} \;; \Delta H = -317.5 \ kJ$.
From the second given equation:
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$.
Adding these two equations:
$(\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)}) + (PCl_{3(l)} + Cl_{2(g)}) \to PCl_{3(l)} + PCl_{5(s)}$
$\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} \to PCl_{5(s)}$
$\Delta H_{f} = -317.5 \ kJ + (-137 \ kJ) = -454.5 \ kJ \ mol^{-1}$.

(D) The enthalpy of formation is defined as the heat change when $1 \text{ mole}$ of a substance is formed from its constituent elements in their most stable standard states.
For carbon,the most stable allotropic form is graphite,not diamond.
For sulfur,the most stable allotropic form is rhombic sulfur,not diamond (diamond is an allotrope of carbon,not sulfur).
Therefore,neither reaction represents the standard enthalpy of formation.

(B) To find the bond enthalpy of $C-H$,we need the enthalpy change for the reaction: $CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)}$.
Adding the given equations:
$1) CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)};$ $\Delta H_1 = -890 \, kJ$
$2) CO_{2(g)} \rightarrow C_{(graphite)} + O_{2(g)};$ $\Delta H_2 = 393 \, kJ$
$3) 2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)};$ $\Delta H_3 = 571 \, kJ$
$4) 2H_{2(g)} \rightarrow 4H_{(g)};$ $\Delta H_4 = 871 \, kJ$
$5) C_{(graphite)} \rightarrow C_{(g)};$ $\Delta H_5 = 716 \, kJ$
Summing these reactions gives:
$CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)}$
$\Delta H_{total} = (-890) + 393 + 571 + 871 + 716 = 1661 \, kJ$
Since there are $4$ $C-H$ bonds in $CH_4$,the bond enthalpy is:
$C-H \text{ bond enthalpy} = \frac{1661 \, kJ}{4} = 415.25 \, kJ/mol$.

(A) The enthalpy of reaction is calculated using the formula: $\Delta H_r = \Sigma B.E._{\text{reactants}} - \Sigma B.E._{\text{products}}$
Reactants: $3 \times (C-H) + 1 \times (C-O) + 1 \times (O-H) + 1 \times (H-Cl)$
$= 3(414) + 335 + 463 + 431 = 1242 + 335 + 463 + 431 = 2471 \ kJ/mol$
Products: $3 \times (C-H) + 1 \times (C-Cl) + 2 \times (O-H)$
$= 3(414) + 326 + 2(463) = 1242 + 326 + 926 = 2494 \ kJ/mol$
$\Delta H_r = 2471 - 2494 = -23 \ kJ/mol$

(A) Let the bond enthalpies of $H_2$,$X_2$,and $HX$ be $2x$,$x$,and $2x$ respectively.
The chemical equation for the formation of $HX$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} X_2(g) \rightarrow HX(g)$.
The enthalpy of reaction is given by: $\Delta H_f = \Sigma B.E.(\text{reactants}) - \Sigma B.E.(\text{products})$.
Substituting the values: $-50 = [\frac{1}{2} \times B.E.(H_2) + \frac{1}{2} \times B.E.(X_2)] - [B.E.(HX)]$.
$-50 = [\frac{1}{2} \times 2x + \frac{1}{2} \times x] - [2x]$.
$-50 = x + 0.5x - 2x = -0.5x$.
$x = 100 \ kJ \ mol^{-1}$.
Therefore,the bond enthalpy of $H_2$ is $2x = 2 \times 100 = 200 \ kJ \ mol^{-1}$.

(B) The neutralization reaction is $HNO_3 + KOH \rightarrow KNO_3 + H_2O$.
Since $HNO_3$ is a strong acid and $KOH$ is a strong base,the heat of neutralization for $1 \, mol$ of $H^+$ reacting with $1 \, mol$ of $OH^-$ is $57.0 \, kJ$.
Here,$0.5 \, mol$ of $HNO_3$ and $0.2 \, mol$ of $KOH$ are mixed.
The limiting reagent is $KOH$ $(0.2 \, mol)$,which determines the amount of water formed.
Therefore,$0.2 \, mol$ of $H_2O$ is produced.
Heat released $= 57.0 \, kJ \, mol^{-1} \times 0.2 \, mol = 11.4 \, kJ$.

(D) The standard enthalpy of combustion,$\Delta H_c^ \circ$,is defined as the enthalpy change when $1 \text{ mole}$ of a substance is completely burnt in oxygen under standard conditions ($298 \text{ K}$ and $1 \text{ bar}$ pressure).
Option $A$ represents the formation of water vapor,not the standard state of water.
Option $B$ represents the enthalpy of formation of ammonia.
Option $C$ represents the combustion of diamond,but the standard state of carbon is graphite.
Option $D$ represents the combustion of $1 \text{ mole}$ of $H_{2(g)}$ to form $1 \text{ mole}$ of $H_2O_{(l)}$,which is the standard state of water at $298 \text{ K}$.

(D) The heat of atomisation is the enthalpy change when one mole of a substance is completely converted into its constituent gaseous atoms.
For $P_4O_{6(s)}$,the process involves two steps:
$1$. Sublimation of solid $P_4O_6$ to gaseous $P_4O_6$: $P_4O_{6(s)} \rightarrow P_4O_{6(g)}$,with enthalpy change $\Delta H_1 = x \ kJ/mol$.
$2$. Breaking all $P-O$ bonds in gaseous $P_4O_6$ to form gaseous atoms: $P_4O_{6(g)} \rightarrow 4P_{(g)} + 6O_{(g)}$.
The structure of $P_4O_6$ contains $12$ $P-O$ bonds. Since the energy required to break one $P-O$ bond is $y \ kJ/mol$,the energy required to break $12$ $P-O$ bonds is $12y \ kJ/mol$.
Therefore,the total heat of atomisation is the sum of the heat of sublimation and the total bond dissociation energy:
$\Delta H_{atomisation} = x + 12y \ kJ/mol$.

(A) According to Hess's Law,the enthalpy change for the reaction $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$ can be calculated using the dissolution enthalpies.
Let $\Delta H_{hyd}$ be the enthalpy of hydration.
We have the following processes:
$1$) $BaCl_2(s) \to BaCl_2(aq)$,$\Delta H_1 = -20.6 \ kJ \ mol^{-1}$
$2$) $BaCl_2 \cdot 2H_2O(s) \to BaCl_2(aq)$,$\Delta H_2 = 8.8 \ kJ \ mol^{-1}$
The target reaction is: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$
This can be written as: $(1) - (2)$
$\Delta H_{hyd} = \Delta H_1 - \Delta H_2$
$\Delta H_{hyd} = -20.6 - 8.8 = -29.4 \ kJ \ mol^{-1}$

(D) The target reaction for the formation of benzene is: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)} \; \Delta H_f = ?$
Given combustion reactions:
$1$) $C_6H_{6(l)} + \frac{15}{2} O_2 \rightarrow 6CO_2 + 3H_2O \; \Delta H = Q_1$
$2$) $C_{(s)} + O_2 \rightarrow CO_2 \; \Delta H = Q_2$
$3$) $H_{2(g)} + \frac{1}{2} O_2 \rightarrow H_2O \; \Delta H = Q_3$
To obtain the target reaction,we perform: $6 \times (2) + 3 \times (3) - (1)$:
$\Delta H_f = 6Q_2 + 3Q_3 - Q_1$

(B) The target equation for the heat of formation of $C_xH_y$ is:
$x C + \frac{y}{2} H_2 \to C_xH_y \quad \Delta H_f = ?$
Given combustion reactions:
$(i) \ C + O_2 \to CO_2 \quad \Delta H = b$
$(ii) \ H_2 + \frac{1}{2} O_2 \to H_2O \quad \Delta H = c$
$(iii) \ C_xH_y + (x + \frac{y}{4}) O_2 \to x CO_2 + \frac{y}{2} H_2O \quad \Delta H = a$
To obtain the target equation,perform: $x \times (i) + \frac{y}{2} \times (ii) - (iii)$:
$\Delta H_f = x(b) + \frac{y}{2}(c) - a$
$\Delta H_f = \left( xb + \frac{yc}{2} - a \right) \ cal$

(D) The formation reaction is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightarrow NH_{3(g)}$
Using the relation: $\Delta_f H = \sum BE_{reactants} - \sum BE_{products}$
Here,$\sum BE_{products}$ is the enthalpy of atomisation $(\Delta_a H)$ of $NH_3$.
$-45 = [\frac{1}{2} \times 940 + \frac{3}{2} \times 436] - \Delta_a H$
$-45 = [470 + 654] - \Delta_a H$
$-45 = 1124 - \Delta_a H$
$\Delta_a H = 1124 + 45 = 1169 \ kJ \ mol^{-1}$

(B) By definition,the standard molar enthalpy of formation $(\Delta_f H^\circ)$ of an element in its most stable physical state at $1 \ bar$ and $298 \ K$ is zero.
For hydrogen,the most stable state is the diatomic gas $H_{2(g)}$.
However,the question asks for the substance whose standard enthalpy of formation is zero among the given options.
By convention,the standard enthalpy of formation of $H^{ }_{(aq)}$ is defined as $0 \ kJ \ mol^{-1}$ at all temperatures.
Therefore,the correct option is $B$.

(B) The combustion reaction for $C_3H_6$ (propene or cyclopropane) is:
$C_3H_6 (g) + 4.5 O_2 (g) \to 3 CO_2 (g) + 3 H_2O (l)$
First,calculate the enthalpy of combustion for propene:
$\Delta H^o_{comb} [propene] = [3 \times \Delta H_f^o (CO_2) + 3 \times \Delta H_f^o (H_2O)] - [\Delta H_f^o (propene)]$
$\Delta H^o_{comb} [propene] = [3 \times (-394) + 3 \times (-286)] - [20]$
$\Delta H^o_{comb} [propene] = [-1182 - 858] - 20 = -2040 - 20 = -2060 \, kJ/mol$
Given the isomerisation reaction: $cyclopropane (g) \to propene (g)$ with $\Delta H^o_{isomerisation} = -33 \, kJ/mol$.
Using Hess's Law:
$\Delta H^o_{comb} [cyclopropane] = \Delta H^o_{comb} [propene] + \Delta H^o_{isomerisation}$
$\Delta H^o_{comb} [cyclopropane] = -2060 + (-33) = -2093 \, kJ/mol$.

(B) The reaction is $N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)}$ with $\Delta H = -142 \ kJ/mol$.
First,convert $N_2H_{4(l)}$ to $N_2H_{4(g)}$: $N_2H_{4(l)} \to N_2H_{4(g)}$,$\Delta H = 18 \ kJ/mol$.
Adding this to the reaction: $N_2H_{4(g)} + H_{2(g)} \to 2NH_{3(g)}$,$\Delta H = -142 - 18 = -160 \ kJ/mol$.
The enthalpy of reaction is $\Delta H = \sum \text{Bond Energy}_{\text{reactants}} - \sum \text{Bond Energy}_{\text{products}}$.
For $N_2H_{4(g)} + H_{2(g)} \to 2NH_{3(g)}$:
Reactants: $1 \times (N-N) + 4 \times (N-H) + 1 \times (H-H)$.
Products: $6 \times (N-H)$.
$\Delta H = [BE_{N-N} + 4(393) + 436] - [6(393)] = -160$.
$BE_{N-N} + 1572 + 436 - 2358 = -160$.
$BE_{N-N} - 350 = -160$.
$BE_{N-N} = 350 - 160 = 190 \ kJ/mol$.

(C) The enthalpy change of a reaction is calculated as: $\Delta H = \sum \text{Bond Energies of Reactants} - \sum \text{Bond Energies of Products}$.
Reactants: $1 \times e_{H-H} + 1 \times e_{C=C} + 4 \times e_{C-H} = 103 + 145 + 4(99) = 103 + 145 + 396 = 644 \ kcal \ mol^{-1}$.
Products: $1 \times e_{C-C} + 6 \times e_{C-H} = 80 + 6(99) = 80 + 594 = 674 \ kcal \ mol^{-1}$.
$\Delta H = 644 - 674 = -30 \ kcal \ mol^{-1}$.

(D) The reaction $C_{2}H_{6(g)} \to 2C_{(g)} + 6H_{(g)}$ represents the atomization of ethane.
The enthalpy change $(X)$ for this reaction is equal to the sum of all bond dissociation energies in the molecule.
$X = 1 \times BE(C-C) + 6 \times BE(C-H)$.
Since the bond energy of the $C-C$ bond is not provided,we have one equation with two unknowns ($BE(C-C)$ and $BE(C-H)$).
Therefore,the bond energy of the $C-H$ bond cannot be determined from the given information.

(A) The hydrogenation reaction is: $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$
$\Delta H = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$
$\Delta H = [BE(C=C) + 4 \times BE(C-H) + BE(H-H)] - [BE(C-C) + 6 \times BE(C-H)]$
$\Delta H = [600 + 4(410) + 400] - [350 + 6(410)]$
$\Delta H = [600 + 1640 + 400] - [350 + 2460]$
$\Delta H = 2640 - 2810 = -170 \ kJ \cdot mol^{-1}$.

(C) The neutralization reaction is: $H^{+}_{(aq)} + OH^{-}_{(aq)} \longrightarrow H_{2}O_{(l)}$
Given $\Delta H_{neut}^{\circ} = -57.32 \ kJ/mol$.
The enthalpy of reaction is given by: $\Delta H_{neut}^{\circ} = \Delta H_{f}^{\circ}(H_{2}O_{(l)}) - [\Delta H_{f}^{\circ}(H^{+}_{(aq)}) + \Delta H_{f}^{\circ}(OH^{-}_{(aq)})]$
By convention,$\Delta H_{f}^{\circ}(H^{+}_{(aq)}) = 0 \ kJ/mol$.
Substituting the values: $-57.32 = -285.84 - [0 + \Delta H_{f}^{\circ}(OH^{-}_{(aq)})]$
$\Delta H_{f}^{\circ}(OH^{-}_{(aq)}) = -285.84 + 57.32 = -228.52 \ kJ/mol$.

(B) We need to find the enthalpy of formation for $H_2O_{2(l)}$,which corresponds to the reaction: $H_{2(g)} + O_{2(g)} \to H_2O_{2(l)}$.
Let the given reactions be:
$(1) N_2H_{4(l)} + 2H_2O_{2(l)} \to N_{2(g)} + 4H_2O_{(l)}$; $\Delta_r H_1^o = -818 \ kJ/mol$
$(2) N_2H_{4(l)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(l)}$; $\Delta_r H_2^o = -622 \ kJ/mol$
$(3) H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta_r H_3^o = -285 \ kJ/mol$
To obtain the target reaction,we perform: $(2) + 2 \times (3) - (1)$ divided by $2$:
$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [\Delta_r H_2^o + 2 \times \Delta_r H_3^o - \Delta_r H_1^o]$
$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-622 + 2(-285) - (-818)]$
$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-622 - 570 + 818]$
$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-374] = -187 \ kJ/mol$.

(B) The given reaction is $4H_{(g)} \to 2H_{2(g)}$ with $\Delta H = -869.6\, kJ$.
Bond dissociation energy is defined as the energy required to break one mole of bonds.
The reverse reaction represents the breaking of two moles of $H-H$ bonds: $2H_{2(g)} \to 4H_{(g)}$,where $\Delta H = +869.6\, kJ$.
Since $2$ moles of $H-H$ bonds are broken,$2 \times BE_{H-H} = 869.6\, kJ$.
Therefore,$BE_{H-H} = \frac{869.6}{2} = 434.8\, kJ$.

(A) The standard enthalpy of formation $(\Delta_fH^\circ)$ of an element in its most stable allotropic form at standard state ($298 \ K$ and $1 \ bar$) is defined as zero.
$F_{2(g)}$ is the most stable elemental form of fluorine under standard conditions.
Therefore,the standard enthalpy of formation of $F_{2(g)}$ is $0 \ kJ/mol$.

(A) The reaction $C_{(graphite)} + O_{2(g)} \to CO_{2(g)}$ satisfies all the given conditions:
$1.$ Heat of combustion: It represents the combustion of $1 \ mol$ of $C_{(graphite)}$.
$2.$ Heat of formation: It represents the formation of $1 \ mol$ of $CO_{2(g)}$ from its constituent elements in their standard states ($C_{(graphite)}$ and $O_{2(g)}$).
$3.$ Exothermic reaction: Combustion reactions are inherently exothermic $(\Delta H < 0)$.
$4.$ Not a neutralization reaction: This is a combustion reaction,not an acid-base neutralization.

(B) The process $I$ represents the direct dissolution of the solid $AB(s)$ into its aqueous ions,which is the enthalpy of solution,$\Delta_{sol}H^o$.
The process $II$ represents the breaking of the ionic lattice into gaseous ions,which is the lattice enthalpy,$\Delta_{lattice}H^o$.
The process $III$ represents the hydration of gaseous ions to form aqueous ions,which is the enthalpy of hydration,$\Delta_{hyd}H^o$.
According to Hess's Law,$\Delta_{sol}H^o = \Delta_{lattice}H^o + \Delta_{hyd}H^o$.
Comparing this with the diagram:
$I = \Delta_{sol}H^o$
$II = \Delta_{lattice}H^o$
$III = \Delta_{hyd}H^o$
Therefore,the correct sequence is $\Delta_{sol}H^o, \Delta_{lattice}H^o, \Delta_{hyd}H^o$.

(C) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
The enthalpy of combustion $\Delta H_c$ is calculated as: $\Delta H_c = [\Delta H_f(CO_2) + 2 \times \Delta H_f(H_2O)] - [\Delta H_f(CH_4) + 2 \times \Delta H_f(O_2)]$.
Given $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$,we have:
$\Delta H_c = -400 + 2(-240) - (-75) = -400 - 480 + 75 = -805 \ kJ \ mol^{-1}$.
For $0.4 \ mol$ of $CH_4$,the heat evolved is: $0.4 \ mol \times 805 \ kJ \ mol^{-1} = 322 \ kJ$.

(C) The dimerisation reaction of $NO_2$ is represented as: $2NO_2 \to N_2O_4$.
We are given:
$(i) \ N_2 + 2O_2 \to 2NO_2, \Delta H_1 = 67.9 \ kJ$
$(ii) \ N_2 + 2O_2 \to N_2O_4, \Delta H_2 = 9.3 \ kJ$
To obtain the target equation,we perform: $(ii) - (i)$:
$(N_2 + 2O_2) - (N_2 + 2O_2) \to N_2O_4 - 2NO_2$
$0 \to N_2O_4 - 2NO_2$
$2NO_2 \to N_2O_4$
Therefore,the enthalpy change $\Delta H = \Delta H_2 - \Delta H_1$
$\Delta H = 9.3 \ kJ - 67.9 \ kJ = -58.6 \ kJ$.

(C) The formation reaction of methane is: $C(s) + 2H_2(g) \rightarrow CH_4(g)$.
The standard enthalpy of formation is given by the formula: $\Delta H_f^\circ(CH_4) = \Delta H_c^\circ(C) + 2 \times \Delta H_c^\circ(H_2) - \Delta H_c^\circ(CH_4)$.
Substituting the given values:
$\Delta H_f^\circ(CH_4) = -393.5 + 2(-284.8) - (-890.4)$.
$\Delta H_f^\circ(CH_4) = -393.5 - 569.6 + 890.4$.
$\Delta H_f^\circ(CH_4) = -72.7 \ kJ/mole$.

(A) The formation reaction of $CS_2$ is: $C(s) + 2S(s) \rightarrow CS_2(l)$
The heat of formation is given by: $\Delta_{f} H^{\circ} = \sum \Delta_{c} H^{\circ}(\text{reactants}) - \sum \Delta_{c} H^{\circ}(\text{products})$
$\Delta_{f} H^{\circ} = [\Delta_{c} H^{\circ}(C) + 2 \times \Delta_{c} H^{\circ}(S)] - \Delta_{c} H^{\circ}(CS_2)$
$\Delta_{f} H^{\circ} = [-393.3 + 2(-293.7)] - (-1108.76)$
$\Delta_{f} H^{\circ} = [-393.3 - 587.4] + 1108.76$
$\Delta_{f} H^{\circ} = -980.7 + 1108.76 = +128.06 \ kJ$

(B) The heat of neutralisation is highest when a strong acid reacts with a strong base.
In the given options,$Ba(OH)_2$ is a strong base and $H_2SO_4$ is a strong acid.
When $1 \ g$-equivalent of a strong acid reacts with $1 \ g$-equivalent of a strong base,the enthalpy of neutralisation is approximately $-57.1 \ kJ \ mol^{-1}$.
Other options involve weak acids or weak bases,which consume some energy for dissociation,resulting in a lower net heat of neutralisation.

(B) The given reaction is the formation of ammonia: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ with $\Delta_r H^{\circ} = -91.8 \, kJ \, mol^{-1}$.
When a chemical reaction is reversed,the sign of the enthalpy change $(\Delta_r H^{\circ})$ is also reversed.
The decomposition reaction is the reverse of the formation reaction: $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$.
Therefore,the enthalpy of reaction for the decomposition is $\Delta_r H^{\circ} = -(-91.8 \, kJ \, mol^{-1}) = +91.8 \, kJ \, mol^{-1}$.

(D) The heat of neutralisation is defined as the enthalpy change when $1 \text{ equivalent}$ of an acid is neutralised by $1 \text{ equivalent}$ of a base.
For strong acids and strong bases,the reaction is simply $H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)$,which releases approximately $-57.1 \text{ kJ/mol}$.
When either the acid or the base is weak,some energy is consumed in the dissociation of the weak electrolyte.
When both the acid and the base are weak,the energy required for the dissociation of both the weak acid and the weak base is significant,resulting in the minimum net heat of neutralisation.

(A) Given equations:
$(1) \ H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$(2) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$(3) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
We need to find $\Delta H$ for the reaction:
$(4) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
By adding equations $(1)$,$(2)$,and $(3)$:
$(H_{2}O_{(g)} + C_{(s)}) + (CO_{(g)} + \frac{1}{2} O_{2(g)}) + (H_{2(g)} + \frac{1}{2} O_{2(g)}) \to (CO_{(g)} + H_{2(g)}) + CO_{2(g)} + H_{2}O_{(g)}$
Canceling common terms on both sides:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
Therefore,$X = 131 + (-282) + (-242) = 131 - 524 = -393 \ kJ$.

(A) 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.
In option $A$,$CaCO_3$ is formed from $CaO$ and $CO_2$,which are compounds,not elements. Therefore,this reaction represents the enthalpy of reaction,not the enthalpy of formation.
In options $B$,$C$,and $D$,the products are formed from their constituent elements in their standard states ($Ca_{(s)}$,$C_{(graphite)}$,$O_{2_{(g)}}$,$H_{2_{(g)}}$,$S_{(rhombic)}$),which correctly represents the enthalpy of formation.

(C) The standard heat of reaction $\Delta H_r^\circ$ is calculated using the formula: $\Delta H_r^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$.
Given:
$\Delta H_f^\circ(CO_{2(g)}) = -393.5 \ kJ/mol$ (since it is the heat of combustion of $C_{(s)}$).
$\Delta H_f^\circ(H_2O_{(l)}) = -285 \ kJ/mol$ (since it is the heat of combustion of $H_{2(g)}$).
$\Delta H_f^\circ(C_2H_{6(g)}) = -88.2 \ kJ/mol$.
$\Delta H_f^\circ(O_{2(g)}) = 0 \ kJ/mol$ (standard state).
Substituting the values:
$\Delta H_r^\circ = [2 \times (-393.5) + 3 \times (-285)] - [-88.2 + 0]$
$\Delta H_r^\circ = [-787 - 855] + 88.2$
$\Delta H_r^\circ = -1642 + 88.2 = -1553.8 \ kJ/mol$.

(C) The standard heat of formation $(\Delta H_f^\circ)$ 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.
In option $C$,$\frac{1}{2} H_{2_{(g)}} + \frac{1}{2} Br_{2(\ell)} \to HBr_{(g)}$,$1 \text{ mole}$ of $HBr$ is formed from its constituent elements ($H_2$ and $Br_2$) in their standard states ($H_2$ gas and $Br_2$ liquid). This satisfies the definition of standard heat of formation.
Option $A$ involves a compound decomposing into an element.
Option $B$ involves ions,not elements in their standard states.
Option $D$ involves $P_4O_{11}$,which is not a standard stable oxide of phosphorus,and the stoichiometry does not form $1 \text{ mole}$ of the product correctly.

(B) The molar mass of oxygen gas $(O_2)$ is $32 \ g \ mol^{-1}$.
Given mass of $O_2 = 16 \ g$.
Number of moles of $O_2 = \frac{16 \ g}{32 \ g \ mol^{-1}} = 0.5 \ mol$.
Energy required to dissociate $0.5 \ mol$ of $O_2$ is $x \ kJ$.
Heat of atomisation is defined as the energy required to produce $1 \ mol$ of gaseous atoms from $1 \ mol$ of the substance.
The dissociation reaction is: $O_2(g) \rightarrow 2O(g)$.
Since $0.5 \ mol$ of $O_2$ requires $x \ kJ$,then $1 \ mol$ of $O_2$ requires $2x \ kJ$.
Thus,the heat of atomisation for $1 \ mol$ of $O_2$ is $2x \ kJ \ mol^{-1}$.

(C) The standard enthalpy of formation $\Delta H_f^o$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
The given reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ with $\Delta H_r = -183.6 \ kJ/mol$.
This reaction represents the formation of $2 \ mol$ of $NH_{3(g)}$.
To find the enthalpy of formation for $1 \ mol$ of $NH_{3(g)}$,we divide the enthalpy change of the reaction by $2$:
$\Delta H_f^o[NH_{3(g)}] = \frac{\Delta H_r}{2} = \frac{-183.6 \ kJ/mol}{2} = -91.8 \ kJ/mol$.

(D) The enthalpy of formation of $PCl_5(s)$ is the enthalpy change for the reaction: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} \to PCl_{5(s)}$.
Step $1$: Divide the first equation by $2$:
$\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)} \to PCl_3(\ell) ; \Delta H = \frac{-635}{2} \ kJ = -317.5 \ kJ$
Step $2$: Add the second equation:
$PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$
Step $3$: Add the two resulting equations:
$\frac{1}{4} P_{4(s)} + (\frac{3}{2} + 1) Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -317.5 + (-137) = -454.5 \ kJ$.