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

(A) The reaction is: $H-H + O=O \rightarrow H-O-O-H$.
The enthalpy change is calculated as: $\Delta_{r}H = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$.
$\Delta_{r}H = [BE_{H-H} + BE_{O=O}] - [BE_{O-O} + 2 \times BE_{O-H}]$.
$\Delta_{r}H = [436 + 499] - [142 + 2 \times 460]$.
$\Delta_{r}H = 935 - [142 + 920]$.
$\Delta_{r}H = 935 - 1062 = -127 \ kJ$.

(C) The bond energy (or bond dissociation enthalpy) is defined as the amount of energy required to break one mole of similar bonds in the gaseous state to separate the bonded atoms.

(C) The equation given in option $(A)$ is the Arrhenius equation.
The equation given in option $(B)$ is the Clausius-Clapeyron equation.
The equation given in option $(C)$ is Kirchoff's equation,which relates the change in enthalpy of a reaction to the change in temperature using the heat capacity difference $\Delta C_{p}$.
The equation given in option $(D)$ is the van't Hoff equation.

(C) According to the given thermochemical equation,$74.8 \ kJ$ of heat is evolved when $1 \ mol$ of methane $(CH_4)$ is formed.
The molar mass of $CH_4$ is $12 + (4 \times 1) = 16 \ g/mol$.
Thus,$74.8 \ kJ$ of heat corresponds to the formation of $16 \ g$ of $CH_4$.
For $149.6 \ kJ$ of heat,the amount of $CH_4$ formed is:
$x = \frac{149.6 \ kJ \times 16 \ g}{74.8 \ kJ} = 32 \ g$.

(B) The standard enthalpy of formation is defined as the change in enthalpy when $1 \text{ mole}$ of a compound is formed from its constituent elements in their most stable physical states at $298 \text{ K}$ and $1 \text{ bar}$ pressure.
Sodium $(Na)$ exists as a solid $(s)$ at room temperature,and chlorine $(Cl_2)$ exists as a gas $(g)$.
Therefore,the correct reaction is: $Na_{(s)} + \frac{1}{2} Cl_{2(g)} \longrightarrow NaCl_{(s)}$.

(C) The standard enthalpy of reaction is calculated using bond enthalpies as: $\Delta H^o_{\text{reaction}} = \sum H^o_{\text{reactant bonds}} - \sum H^o_{\text{product bonds}}$.
Option $C$ states the reverse of this relationship,making it the $INVALID$ equation.

(C) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g \ mol^{-1}$.
Number of moles of benzene $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{13 \ g}{78 \ g \ mol^{-1}} = \frac{1}{6} \ mol$.
The enthalpy of vaporisation $(\Delta_{vap}H)$ is given by the heat supplied per mole: $\Delta_{vap}H = \frac{q}{n}$.
$\Delta_{vap}H = \frac{5.1 \ kJ}{1/6 \ mol} = 5.1 \times 6 \ kJ \ mol^{-1} = 30.6 \ kJ \ mol^{-1}$.

(B) The molar mass of ethanol $(C_2H_5OH)$ is calculated as: $(2 \times 12.01) + (6 \times 1.008) + (1 \times 16.00) = 46.07 \ g \ mol^{-1}$,which is approximately $46 \ g \ mol^{-1}$.
Number of moles of ethanol $(n) = \frac{\text{mass}}{\text{molar mass}} = \frac{11.5 \ g}{46 \ g \ mol^{-1}} = 0.25 \ mol$.
The enthalpy of vaporisation $(\Delta_{vap}H)$ is the heat required to vaporise $1 \ mol$ of a substance.
$\Delta_{vap}H = \frac{\text{Heat supplied}}{\text{Number of moles}} = \frac{11.8 \ kJ}{0.25 \ mol} = 47.2 \ kJ \ mol^{-1}$.
Therefore,the correct option is $B$.

(A) The standard enthalpy change of the reaction is calculated using the formula:
$\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$
For the reaction: $C_2H_{4(g)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 2H_2O_{(\ell)}$
$\Delta_{r}H^{\circ} = [2 \times \Delta_{f}H^{\circ}(CO_2) + 2 \times \Delta_{f}H^{\circ}(H_2O)] - [1 \times \Delta_{f}H^{\circ}(C_2H_4) + 3 \times \Delta_{f}H^{\circ}(O_2)]$
Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state):
$\Delta_{r}H^{\circ} = [2(-393.5) + 2(-285.8)] - [52 + 0]$
$\Delta_{r}H^{\circ} = [-787 - 571.6] - 52$
$\Delta_{r}H^{\circ} = -1358.6 - 52 = -1410.6 \ kJ \ mol^{-1}$
The closest value is $-1411.1 \ kJ \ mol^{-1}$.

(C) The standard enthalpy of formation $(\Delta_{f}H^0)$ 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: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the enthalpy change is $\Delta_{r}H^0 = -92.0 \ kJ$ for the production of $2 \ mol$ of $NH_3$.
To find the enthalpy of formation per mole of $NH_3$,we divide the reaction enthalpy by the stoichiometric coefficient of $NH_3$:
$\Delta_{f}H^0(NH_3) = \frac{\Delta_{r}H^0}{2} = \frac{-92.0 \ kJ}{2} = -46.0 \ kJ \ mol^{-1}$.

(C) The standard enthalpy change of reaction $\Delta_rH^{\circ}$ is calculated using the formula:
$\Delta_rH^{\circ} = \sum \Delta_fH^{\circ}(\text{products}) - \sum \Delta_fH^{\circ}(\text{reactants})$
For the reaction $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(\ell)}$:
$\Delta_rH^{\circ} = [2 \times \Delta_fH^{\circ}(CO_2) + 1 \times \Delta_fH^{\circ}(H_2O)] - [1 \times \Delta_fH^{\circ}(C_2H_2) + \frac{5}{2} \times \Delta_fH^{\circ}(O_2)]$
Since $\Delta_fH^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element):
$\Delta_rH^{\circ} = [2(-393) + (-286)] - [227 + 0]$
$\Delta_rH^{\circ} = [-786 - 286] - 227$
$\Delta_rH^{\circ} = -1072 - 227 = -1299 \ kJ \ mol^{-1}$

(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})$
For the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$,the expression is:
$\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_2) + 2 \times \Delta_{f} H^{\circ}(H_2O)] - [\Delta_{f} H^{\circ}(CH_4) + 2 \times \Delta_{f} H^{\circ}(O_2)]$
Since $O_2$ is an element in its standard state,$\Delta_{f} H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$.
Substituting the given values:
$\Delta_{r} H^{\circ} = [-394 + 2 \times (-286)] - [-75 + 0]$
$\Delta_{r} H^{\circ} = [-394 - 572] - [-75]$
$\Delta_{r} H^{\circ} = -966 + 75 = -891 \ kJ \ mol^{-1}$

(A) The enthalpy of solution $(\Delta_{sol} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Delta_{hyd} H)$.
$\Delta_{sol} H = \Delta_{L} H + \Delta_{hyd} H$
Given:
$\Delta_{L} H = 700 \ kJ \ mol^{-1}$
$\Delta_{hyd} H = -680 \ kJ \ mol^{-1}$
Substituting the values:
$\Delta_{sol} H = 700 \ kJ \ mol^{-1} + (-680 \ kJ \ mol^{-1})$
$\Delta_{sol} H = 20 \ kJ \ mol^{-1}$
Therefore,the correct option is $A$.

(C) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$.
For the reaction $C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$,the expression is:
$\Delta_{r}H^{\circ} = [2 \times \Delta_{f}H^{\circ}(CO_2) + 3 \times \Delta_{f}H^{\circ}(H_2O)] - [\Delta_{f}H^{\circ}(C_2H_5OH) + 3 \times \Delta_{f}H^{\circ}(O_2)]$.
Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we have:
$\Delta_{r}H^{\circ} = [2(-390) + 3(-285)] - [-280 + 3(0)]$.
$\Delta_{r}H^{\circ} = [-780 - 855] - [-280]$.
$\Delta_{r}H^{\circ} = -1635 + 280 = -1355 \ kJ \ mol^{-1}$.

(A) An exothermic reaction is one that releases heat to the surroundings,characterized by a negative enthalpy change $(\Delta H < 0)$.
$A$. The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is a neutralization reaction. Neutralization reactions are always exothermic because they involve the formation of water from $H^+$ and $OH^-$ ions,releasing energy.
$B$. The melting of ice is an endothermic process as it requires the absorption of heat to break intermolecular hydrogen bonds.
$C$. The dissolution of $NaCl$ in water is an endothermic process (or slightly endothermic) as it requires energy to overcome the lattice enthalpy of the crystal.
$D$. The formation of nitrogen dioxide from nitrogen and oxygen is an endothermic reaction,requiring energy input.
Therefore,the correct option is $A$.

(A) The synthesis of ammonia gas is represented by the equation:
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
To obtain this equation,we add the two given reactions:
Reaction $i$: $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
Reaction $ii$: $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$
Adding reaction $i$ and reaction $ii$:
$(2 H_{2(g)} + N_{2(g)}) + (N_{2}H_{4(g)} + H_{2(g)}) \longrightarrow N_{2}H_{4(g)} + 2 NH_{3(g)}$
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
The total enthalpy change is $\Delta_{r}H^{0} = \Delta_{r}H_{1}^{0} + \Delta_{r}H_{2}^{0}$
$\Delta_{r}H^{0} = 95.4 \ kJ + (-187.6 \ kJ) = -92.2 \ kJ$
This is the enthalpy change for the production of $2 \ moles$ of $NH_{3}$.
Therefore,the correct option is $A$.

(A) The standard enthalpy change of reaction $\Delta_{r} H^{\circ}$ is calculated using the formula: $\Delta_{r} H^{\circ} = \sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})$.
For the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$,the expression is: $\Delta_{r} H^{\circ} = [\Delta_{f} H^{\circ}(CO_{2}) + 2 \times \Delta_{f} H^{\circ}(H_{2}O)] - [\Delta_{f} H^{\circ}(CH_{4}) + 2 \times \Delta_{f} H^{\circ}(O_{2})]$.
Since $\Delta_{f} H^{\circ}(O_{2}) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we substitute the given values:
$\Delta_{r} H^{\circ} = [-390 + 2 \times (-286)] - [-75 + 2 \times 0]$
$\Delta_{r} H^{\circ} = [-390 - 572] - [-75]$
$\Delta_{r} H^{\circ} = -962 + 75 = -887 \ kJ \ mol^{-1}$.

(A) For the reaction:
$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$
$\Delta_{r}H^{\circ} = \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$
$\Delta_{r}H^{\circ} = [4 \times \Delta H_{(C-H)} + 1 \times \Delta H_{(C=C)} + 1 \times \Delta H_{(H-H)}] - [6 \times \Delta H_{(C-H)} + 1 \times \Delta H_{(C-C)}]$
$\Delta_{r}H^{\circ} = [4 \times 414 + 615 + 435] - [6 \times 414 + 347]$
$\Delta_{r}H^{\circ} = [1656 + 615 + 435] - [2484 + 347]$
$\Delta_{r}H^{\circ} = 2706 - 2831 = -125 \ kJ$

(B) The bond dissociation energy is the energy required to break one mole of bonds in a gaseous substance.
For the reaction $H_{2(g)} \longrightarrow 2H_{(g)}$,the enthalpy change $\Delta_{r} H^{\circ}$ is equal to the bond dissociation energy.
Since the bond formation energy of $H-H$ is $-433 \ kJ \ mol^{-1}$,the bond dissociation energy for $1 \ mol$ of $H_{2(g)}$ is $+433 \ kJ \ mol^{-1}$.
Therefore,the bond dissociation energy for $0.5 \ mol$ of $H_{2(g)}$ is $0.5 \ mol \times 433 \ kJ \ mol^{-1} = 216.5 \ kJ$.

(A) The reaction $C_{(g)} + 4H_{(g)} \longrightarrow CH_{4(g)}$ represents the formation of $4$ moles of $C-H$ bonds from gaseous atoms.
The enthalpy change $\Delta H^{\circ} = -1665 \ kJ$ corresponds to the energy released during the formation of these $4$ bonds.
Bond energy is defined as the energy required to break one mole of a specific bond.
$BE_{C-H} = \frac{|\Delta H^{\circ}|}{4} = \frac{1665 \ kJ}{4} = 416.25 \ kJ \ mol^{-1}$.

(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Adding equations $(ii)$ and $(iii)$:
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
Summing these gives:
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} + CO_{(g)} - CO_{(g)}$ $\longrightarrow CO_{(g)} + CO_{2_{(g)}} - CO_{(g)}$
$C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$
Therefore,$Q = (-x) + (-y) = -(x+y) \ kJ$.

(A) Step $1$: Determine the enthalpy of vaporization $(\Delta H_{vap})$ for $1 \ mol$ of water.
Subtract the second equation from the first:
$H_2O_{(l)} \longrightarrow H_2O_{(g)} \quad \Delta H_{vap} = (-57 \ kCal) - (-68.3 \ kCal) = 11.3 \ kCal/mol$.
Step $2$: Calculate the number of moles in $9 \ g$ of water.
Molar mass of $H_2O = 18 \ g/mol$.
$\text{Moles} = \frac{9 \ g}{18 \ g/mol} = 0.5 \ mol$.
Step $3$: Calculate the total heat required.
$\text{Heat} = \text{Moles} \times \Delta H_{vap} = 0.5 \ mol \times 11.3 \ kCal/mol = 5.65 \ kCal$.
Therefore,the correct option is $A$.

(C) The given reaction is: $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2 O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$.
This represents the enthalpy of formation for $2 \ mol$ of water vapor.
To find the heat of decomposition for $1 \ mol$ of water,we reverse the reaction and divide by $2$:
$2 H_2 O_{(g)} \longrightarrow 2 H_{2(g)} + O_{2(g)}$,$\Delta H^{\circ} = +573.2 \ kJ$.
For $1 \ mol$ of water:
$H_2 O_{(g)} \longrightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$,$\Delta H^{\circ} = \frac{573.2}{2} \ kJ = 286.6 \ kJ$.
Thus,the heat of decomposition is $286.6 \ kJ \ mol^{-1}$.

(B) The correct answer is $(B)$ Total number of steps in which the reaction occurs.
According to Hess's Law of Constant Heat Summation,the total enthalpy change $(\Delta H)$ for a chemical reaction is the same,whether the reaction takes place in one step or in several steps.
This is because enthalpy is a state function,meaning it depends only on the initial and final states of the system,not on the path taken.
Therefore,the heat of reaction does not depend on the number of steps involved.

(C) The reaction for the formation of methane $(CH_4)$ from its constituent elements is:
$C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
Given the standard enthalpy of formation $(\Delta H_f^{\circ})$ of methane is $-75 \ kJ \ mol^{-1}$.
This means the formation of $1 \ mol$ of $CH_4$ releases $75 \ kJ$ of energy.
First,calculate the number of moles of carbon $(C)$ used:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$
Since $1 \ mol$ of $C$ produces $1 \ mol$ of $CH_4$,the enthalpy change $(\Delta H)$ is:
$\Delta H = n \times \Delta H_f^{\circ} = 1 \ mol \times (-75 \ kJ \ mol^{-1}) = -75 \ kJ$
Therefore,the enthalpy change is $-75 \ kJ$.

(A) For the phase transition process,$H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$.
Under similar conditions of $0^{\circ}C$ and $1 \ atm$ pressure,the enthalpy of fusion $(\Delta_{fus}H)$ is $+6.01 \ kJ \ mol^{-1}$.
The enthalpy of freezing $(\Delta_{free}H)$ is the reverse process,$H_2O_{(l)} \rightleftharpoons H_2O_{(s)}$,which has a value of $-6.01 \ kJ \ mol^{-1}$.
Thus,the enthalpy of freezing is exactly opposite to the enthalpy of fusion.

(C) The relationship between enthalpy of solution,lattice enthalpy,and hydration enthalpy is given by:
$\Delta_{soln} H = \Delta_{lattice} H + \Delta_{hyd} H$
Rearranging the formula to solve for hydration enthalpy:
$\Delta_{hyd} H = \Delta_{soln} H - \Delta_{lattice} H$
Substituting the given values:
$\Delta_{hyd} H = 4 \ kJ \ mol^{-1} - 790 \ kJ \ mol^{-1}$
$\Delta_{hyd} H = -786 \ kJ \ mol^{-1}$

(D) The chemical equation for the formation of ethane is:
$2 C_{(s)} + 3 H_{2(g)} \longrightarrow C_2 H_{6(g)} ; \Delta_{f} H^{\circ} = ?$
The molar mass of ethane $(C_2 H_6)$ is $(2 \times 12) + (6 \times 1) = 30 \ g \ mol^{-1}$.
The number of moles in $3 \ g$ of $C_2 H_6$ is $n = \frac{3 \ g}{30 \ g \ mol^{-1}} = 0.1 \ mol$.
Given that $8.84 \ kJ$ of heat is liberated for $0.1 \ mol$ of $C_2 H_6$,the heat liberated for $1 \ mol$ is:
$\Delta_{f} H^{\circ} = \frac{-8.84 \ kJ}{0.1 \ mol} = -88.4 \ kJ \ mol^{-1}$.
Since heat is liberated,the enthalpy change is negative.

(A) For the reaction: $2 H_2 + O_2 \rightarrow 2 H_2 O$ (which contains $4 \times O-H$ bonds).
$\Delta_{r} H = [2 \times BE(H-H) + BE(O=O)] - [4 \times BE(O-H)]$
$-571 = [2 \times 435 + 498] - 4 \times BE(O-H)$
$-571 = [870 + 498] - 4 \times BE(O-H)$
$-571 = 1368 - 4 \times BE(O-H)$
$4 \times BE(O-H) = 1368 + 571$
$4 \times BE(O-H) = 1939$
$BE(O-H) = \frac{1939}{4} \approx 484.75 \ kJ/mol$
Thus,the average bond energy is approximately $484 \ kJ/mol$.

(D) The combustion reactions are:
$C$(graphite) $+ O_{2(g)} \rightarrow CO_{2(g)}$; $\Delta H_1 = -393.8 \ kJ \ mol^{-1}$ $(eq. I)$
$C$(diamond) $+ O_{2(g)} \rightarrow CO_{2(g)}$; $\Delta H_2 = -395.3 \ kJ \ mol^{-1}$ $(eq. II)$
To find the enthalpy of conversion of $C$(graphite) to $C$(diamond),we subtract $eq. II$ from $eq. I$:
$C$(graphite) $\rightarrow C$(diamond)
$\Delta H = \Delta H_1 - \Delta H_2$
$\Delta H = -393.8 - (-395.3) \ kJ \ mol^{-1}$
$\Delta H = 1.5 \ kJ \ mol^{-1}$

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species.
For the reaction: $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$.
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - 3 = -1$.
Substituting this value into the equation: $\Delta H = \Delta U + (-1)RT = \Delta U - RT$.

(B) The enthalpy of reaction is given by the formula:
$\Delta H_{r}^{\circ} = \sum (B.E.)_{\text{reactants}} - \sum (B.E.)_{\text{products}}$
Substituting the given values:
$-185 = [B.E._{H-H} + B.E._{Cl-Cl}] - [2 \times B.E._{H-Cl}]$
$-185 = [435 + 244] - 2x$
$-185 = 679 - 2x$
$2x = 679 + 185$
$2x = 864$
$x = 432 \ kJ \ mol^{-1}$
Therefore,the bond enthalpy of the $H-Cl$ bond is $432 \ kJ \ mol^{-1}$.

(C) The hydrogenation reaction of ethylene is: $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$.
The enthalpy change of the reaction is calculated as: $\Delta H = \sum \text{Bond Enthalpies of Reactants} - \sum \text{Bond Enthalpies of Products}$.
Reactants: $1 \times (C=C) + 4 \times (C-H) + 1 \times (H-H) = 600 + 4(410) + 400 = 600 + 1640 + 400 = 2640 \ kJ \ mol^{-1}$.
Products: $1 \times (C-C) + 6 \times (C-H) = 360 + 6(410) = 360 + 2460 = 2820 \ kJ \ mol^{-1}$.
$\Delta H = 2640 - 2820 = -180 \ kJ \ mol^{-1}$.

(A) We need to find the enthalpy of formation for the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \ (I)$
Given equations:
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H_1 = -393.3 \ kJ \ mol^{-1} \ (II)$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H_2 = -282.2 \ kJ \ mol^{-1} \ (III)$
Subtracting equation $(III)$ from equation $(II)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
Therefore,$\Delta H_f = \Delta H_1 - \Delta H_2$
$\Delta H_f = -393.3 - (-282.2) = -111.1 \ kJ \ mol^{-1}$

(B) The heat of formation $(\Delta_{f}H)$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
In the given equation: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$,the enthalpy change for the formation of $2 \ mol$ of $HCl$ is $-194 \ kJ$.
Therefore,for $1 \ mol$ of $HCl_{(g)}$,the heat of formation is $\Delta_{f}H = \frac{-194 \ kJ}{2} = -97 \ kJ$.

(D) The given reaction is: $H_{2(g)} + Cl_{2(g)} \longrightarrow 2 HCl_{(g)}, \Delta H = -194 \ kJ$
The heat of formation $(\Delta H_f)$ is defined as the enthalpy change when $1 \ mole$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $\Delta H_{reaction} = 2 \Delta H_{f(HCl)} - [\Delta H_{f(H_2)} + \Delta H_{f(Cl_2)}]$
Since $H_{2(g)}$ and $Cl_{2(g)}$ are elements in their standard states,their $\Delta H_f = 0$.
$-194 \ kJ = 2 \Delta H_{f(HCl)} - 0 - 0$
$\Delta H_{f(HCl)} = -194 / 2 \ kJ \ mol^{-1} = -97 \ kJ \ mol^{-1}$

(C) The combustion reaction for acetylene is: $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \longrightarrow 2CO_{2(g)} + H_2O_{(\ell)}$; $\Delta_{c}H^{\circ} = -1300 \ kJ \ mol^{-1}$.
The molar mass of acetylene $(C_2H_2)$ is $(2 \times 12) + (2 \times 1) = 26 \ g \ mol^{-1}$.
Since $26 \ g$ of acetylene releases $1300 \ kJ$ of energy upon complete combustion,the enthalpy change for $39 \ g$ is calculated as:
$\Delta H = \left( \frac{-1300 \ kJ \ mol^{-1}}{26 \ g \ mol^{-1}} \right) \times 39 \ g = -50 \times 39 = -1950 \ kJ$.

(A) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g / mol$.
Number of moles of $CH_4$ in $24 \ g$ is calculated as $n = \frac{24 \ g}{16 \ g / mol} = 1.5 \ mol$.
The enthalpy of formation for $1 \ mol$ of $CH_4$ is given as $-75 \ kJ / mol$.
Therefore,the enthalpy change for the formation of $1.5 \ mol$ of $CH_4$ is $\Delta H = 1.5 \ mol \times (-75 \ kJ / mol) = -112.5 \ kJ$.

(C) The formation reaction for $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
The enthalpy of formation is calculated using bond enthalpies as: $\Delta H_f = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$
$\Delta H_f = [\frac{1}{2} BE_{(N \equiv N)} + \frac{3}{2} BE_{(H-H)}] - [3 BE_{(N-H)}]$
Substituting the given values:
$\Delta H_f = [\frac{1}{2} \times 941 + \frac{3}{2} \times 436] - [3 \times 389]$
$\Delta H_f = [470.5 + 654] - 1167$
$\Delta H_f = 1124.5 - 1167 = -42.5 \ kJ/mol$

(D) The chemical equation for the formation of $NO_{2(g)}$ is: $\frac{1}{2} N_{2(g)} + O_{2(g)} \longrightarrow NO_{2(g)}$.
In an endothermic process,heat is absorbed from the surroundings during the reaction.
By definition,for an endothermic reaction,the change in enthalpy is positive,i.e.,$\Delta H > 0$.

(A) The standard enthalpy of formation,$\Delta_{f} H^{\circ}$,is defined as the enthalpy change when $1 \text{ mole}$ of a compound is formed from its constituent elements in their standard states.
For $\Delta_{f} H^{\circ} = \Delta H^{\circ}$,the reaction must form exactly $1 \text{ mole}$ of the product from its elements in their standard states.
In option $A$,$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_{2}O_{(l)}$ represents the formation of $1 \text{ mole}$ of liquid water from its elements $H_2$ and $O_2$ in their standard states,thus $\Delta_{f} H^{\circ} = \Delta H^{\circ}$.

(D) Sublimation is the process of direct conversion of a solid into a gas.
According to Hess's Law,the enthalpy change for a process is the same whether it occurs in one step or multiple steps.
Solid to gas can be represented as:
$1. \text{Solid} \rightarrow \text{Liquid} \quad (\Delta_{fus} H)$
$2. \text{Liquid} \rightarrow \text{Gas} \quad (\Delta_{vap} H)$
Therefore,the total enthalpy change for sublimation is the sum of the enthalpy of fusion and the enthalpy of vaporization:
$\Delta_{sub} H = \Delta_{fus} H + \Delta_{vap} H$.

(C) Average bond enthalpy is the mean of the bond dissociation enthalpies of all bonds of the same type.
For water $(H_2O)$,the average bond enthalpy is given by $\Delta H_{avg} = \frac{\Delta H_1 + \Delta H_2}{2}$.
Given $\Delta H_{avg} = 464.5 \text{ kJ mol}^{-1}$ and $\Delta H_1 = 502 \text{ kJ mol}^{-1}$.
Substituting the values: $464.5 = \frac{502 + \Delta H_2}{2}$.
$929 = 502 + \Delta H_2$.
$\Delta H_2 = 929 - 502 = 427 \text{ kJ mol}^{-1}$.

(A) The molar mass of $O_2$ is $32 \ g/mol$.
Given mass of $O_2$ is $16 \ g$.
Number of moles of $O_2 = \frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$.
Energy required to dissociate $0.5 \ mol$ of $O_2$ is $x \ kJ$.
Bond enthalpy is defined as the energy required to dissociate $1 \ mol$ of bonds.
Energy required for $1 \ mol$ of $O_2 = \frac{x \ kJ}{0.5 \ mol} = 2x \ kJ/mol$.
Therefore,the bond enthalpy of $O=O$ bond is $2x \ kJ$.

(C) The balanced chemical equation is $C(s) + 2S(s) \rightarrow CS_2(l) \quad \Delta H = 92 \ kJ \ mol^{-1}$.
This reaction is endothermic,meaning $92 \ kJ$ of heat is absorbed for every $1 \ mol$ $(12 \ g)$ of carbon reacted.
To find the heat for $6 \ g$ of carbon:
Heat $= (92 \ kJ \ mol^{-1} / 12 \ g \ mol^{-1}) \times 6 \ g = 46 \ kJ$.

(C) The target reaction is the hydration of ethene: $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.
We can obtain this by subtracting equation $(i)$ from equation $(ii)$:
$(ii) - (i): (C_2H_{4_{(g)}} + 3O_{2_{(g)}}) - (C_2H_5OH_{(l)} + 3O_{2_{(g)}})$ $\longrightarrow (2CO_{2_{(g)}} + 2H_2O_{(l)}) - (2CO_{2_{(g)}} + 3H_2O_{(l)})$
This simplifies to: $C_2H_{4_{(g)}} - C_2H_5OH_{(l)} \longrightarrow -H_2O_{(l)}$,which rearranges to $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.
The enthalpy change is $\Delta H^{\circ} = \Delta H^{\circ}_{(ii)} - \Delta H^{\circ}_{(i)} = -1410 \ kJ - (-1368 \ kJ) = -42 \ kJ$.

(B) The balanced chemical equation is: $2ClF_{(g)} + O_{2(g)} \longrightarrow Cl_2O_{(g)} + OF_{2(g)}$.
Given that $6.0 \ g$ of $O_2$ corresponds to $n = \frac{6.0 \ g}{32.0 \ g/mol} = 0.1875 \ mol$ of $O_2$.
The heat absorbed for $0.1875 \ mol$ of $O_2$ is $\Delta H = +38.55 \ kJ$.
The standard enthalpy of reaction $(\Delta H^0)$ is defined for the reaction as written,which involves $1 \ mol$ of $O_2$.
Therefore,$\Delta H^0 = \frac{38.55 \ kJ}{0.1875 \ mol} = 205.6 \ kJ/mol$.

(D) The thermochemical equation for the formation of water is:
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_{2}O_{(\ell)} \quad \Delta H_{f}^{0} = -286 \ kJ \ mol^{-1}$
This means $1 \ mol$ of $H_{2}O$ $(18 \ g)$ releases $286 \ kJ$ of energy.
Given mass of water $= 1800 \ mg = 1.8 \ g$.
Number of moles of $H_{2}O = \frac{1.8 \ g}{18 \ g \ mol^{-1}} = 0.1 \ mol$.
Energy liberated $= 0.1 \ mol \times 286 \ kJ \ mol^{-1} = 28.6 \ kJ$.

(C) The enthalpy change of a reaction is given by the sum of bond enthalpies of reactants minus the sum of bond enthalpies of products:
$\Delta H^{\circ} = \sum \Delta H^{\circ}_{(\text{reactants})} - \sum \Delta H^{\circ}_{(\text{products})}$
For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$:
$\Delta H^{\circ} = [\Delta H^{\circ}_{(N \equiv N)} + 3 \Delta H^{\circ}_{(H-H)}] - [6 \Delta H^{\circ}_{(N-H)}]$
Substituting the given values:
$-83 = \Delta H^{\circ}_{(N \equiv N)} + 3(435) - 6(389)$
$-83 = \Delta H^{\circ}_{(N \equiv N)} + 1305 - 2334$
$-83 = \Delta H^{\circ}_{(N \equiv N)} - 1029$
$\Delta H^{\circ}_{(N \equiv N)} = 1029 - 83 = 946 \ kJ \text{ mol}^{-1}$