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

(N/A) The combustion of carbon to form $CO_{2}$ is represented as:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \, kJ \, mol^{-1}$
The molar mass of $CO_{2}$ is $12 + (2 \times 16) = 44 \, g \, mol^{-1}$.
Heat released for the formation of $44 \, g$ of $CO_{2} = 393.5 \, kJ$.
Therefore,heat released for the formation of $35.2 \, g$ of $CO_{2}$ is:
$= \frac{393.5 \, kJ}{44 \, g} \times 35.2 \, g$
$= 8.943 \times 35.2 \, kJ$
$= 314.8 \, kJ$.

(D) $\Delta_r H$ for a reaction is defined as the difference between the sum of $\Delta_f H$ values of products and the sum of $\Delta_f H$ values of reactants.
$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$
For the given reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$
$\Delta_r H = [\Delta_f H(N_2O) + 3 \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 \Delta_f H(CO)]$
Substituting the given values:
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)] \, kJ \, mol^{-1}$
$\Delta_r H = [81 - 1179] - [9.7 - 330] \, kJ \, mol^{-1}$
$\Delta_r H = -1098 - (-320.3) \, kJ \, mol^{-1}$
$\Delta_r H = -777.7 \, kJ \, mol^{-1}$

(C) The standard enthalpy of formation of a compound is the change in enthalpy that occurs during the formation of $1 \, mol$ of a substance in its standard state from its constituent elements in their standard states.
Rewriting the given equation for the formation of $1 \, mol$ of $NH_{3(g)}$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
Therefore,the standard enthalpy of formation of $NH_{3(g)}$ is:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$

(N/A) The reaction for the formation of $CH_{3}OH_{(l)}$ is:
$C_{(graphite)} + 2 H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow CH_{3}OH_{(l)}$
Using Hess's Law,we manipulate the given equations:
$(i) CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$(ii) C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$(iii) H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$
Target reaction = $(ii) + 2 \times (iii) - (i)$
$\Delta_{f} H^{\ominus} = (-393) + 2(-286) - (-726)$
$\Delta_{f} H^{\ominus} = -393 - 572 + 726$
$\Delta_{f} H^{\ominus} = -239 \ kJ \ mol^{-1}$

The chemical equations for the given enthalpy values are:
$(i) \quad CCl_{4(l)} \longrightarrow CCl_{4(g)} \quad \Delta_{vap}H^{\theta} = 30.5 \, kJ \, mol^{-1}$
$(ii) \quad C_{(s)} \longrightarrow C_{(g)} \quad \Delta_{a}H^{\theta} = 715.0 \, kJ \, mol^{-1}$
$(iii) \quad Cl_{2(g)} \longrightarrow 2Cl_{(g)} \quad \Delta_{a}H^{\theta} = 242 \, kJ \, mol^{-1}$
$(iv) \quad C_{(s)} + 2Cl_{2(g)} \longrightarrow CCl_{4(l)} \quad \Delta_{f}H^{\theta} = -135.5 \, kJ \, mol^{-1}$
To find the enthalpy change for $CCl_{4(g)} \longrightarrow C_{(g)} + 4Cl_{(g)}$,we rearrange the equations:
$\Delta H = \Delta_{a}H^{\theta}(C) + 2\Delta_{a}H^{\theta}(Cl_{2}) - \Delta_{vap}H^{\theta} - \Delta_{f}H^{\theta}$
$\Delta H = 715.0 + 2(242) - 30.5 - (-135.5)$
$\Delta H = 715.0 + 484 - 30.5 + 135.5 = 1304 \, kJ \, mol^{-1}$
Bond enthalpy of $C-Cl$ bond in $CCl_{4(g)} = \frac{1304}{4} = 326 \, kJ \, mol^{-1}$.

(N/A) The positive value of $\Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$ indicates that heat is absorbed during the formation of $NO_{(g)}$ from its elements.
This means that $NO_{(g)}$ has higher energy than the reactants ($N_{2(g)}$ and $O_{2(g)}$),making it thermodynamically unstable.
Conversely,the negative value of $\Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$ for the reaction $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}$ indicates that heat is evolved,leading to a more stable product $NO_{2(g)}$.
Therefore,$NO_{(g)}$ is thermodynamically unstable and tends to convert into $NO_{2(g)}$.

(N/A) Exothermic reactions: Reactions which evolve heat are known as exothermic reactions. Evolution of heat can be shown using a '$+$' sign or by a $\Delta H$ with a '$-$' sign along with the product.
Example: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} + 393.5 \ kJ$
Endothermic reactions: Reactions which absorb heat are called endothermic reactions. Absorbed heat can be shown as a '$-$' sign or in terms of $\Delta H$ with a '$+$' sign.
Example: $C_{(s)} + H_{2}O_{(g)} \longrightarrow CO_{(g)} + H_{2(g)} - 131.4 \ kJ$

(N/A) Bond enthalpy is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in the gaseous state.
The unit of bond enthalpy is $kJ \ mol^{-1}$.
For diatomic molecules,it is the bond dissociation enthalpy. For example,for $H_2$:
$H_{2(g)} \rightarrow H_{(g)} + H_{(g)} ; \Delta_{a} H^{\ominus} = 435.8 \ kJ \ mol^{-1}$
In polyatomic molecules like $H_2O$,the energy required to break the two $O-H$ bonds is not the same due to the change in the chemical environment after the first bond breaks:
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} ; \Delta_{a} H_{1}^{\ominus} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)} ; \Delta_{a} H_{2}^{\ominus} = 427 \ kJ \ mol^{-1}$
The average bond enthalpy for $O-H$ in water is the mean of these values: $(502 + 427) / 2 = 464.5 \ kJ \ mol^{-1}$.
This difference in energy values occurs because the electronic environment of the $O-H$ bond changes once one hydrogen atom is removed.

In a chemical reaction,reactants are converted into products,represented as: $\text{Reactants} \rightarrow \text{Products}$.
The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction is denoted by the symbol $\Delta_{r} H$.
$\Delta_{r} H = (\sum \text{Enthalpies of products}) - (\sum \text{Enthalpies of reactants}) = \sum a_{i} H_{\text{products}} - \sum b_{i} H_{\text{reactants}}$
Here,the symbol $\sum$ represents summation,and $a_{i}$ and $b_{i}$ are the stoichiometric coefficients of the products and reactants,respectively,in the balanced chemical equation.
For example,for the reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)}$
$\Delta_{r} H = [H_{m}(CO_{2}, g) + 2H_{m}(H_{2}O, l)] - [H_{m}(CH_{4}, g) + 2H_{m}(O_{2}, g)]$
where $H_{m}$ represents the molar enthalpy.

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called standard molar enthalpy of formation. Its symbol is $\Delta_{f} H^{\ominus}$.
Elements are considered in their most stable states of aggregation. For example,$H_{2}$ and $O_{2}$ are in their gaseous state at $298 \ K$ temperature and $1 \ bar$ pressure.
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} ; \Delta_{f} H^{\ominus} = -285.8 \ kJ \ mol^{-1}$
$C_{\text{(graphite,s)}} + 2H_{2(g)} \rightarrow CH_{4(g)} ; \Delta_{f} H^{\ominus} = -74.81 \ kJ \ mol^{-1}$
In these examples,one mole of a compound is formed from its constituent elements.
In contrast,consider the enthalpy change for the following exothermic reaction:
$CaO_{(s)} + CO_{2(g)} \rightarrow CaCO_{3(s)} ; \Delta_{r} H^{\ominus} = -178.3 \ kJ \ mol^{-1}$
This is not an enthalpy of formation of calcium carbonate,since calcium carbonate is formed from other compounds,not from its constituent elements.
Similarly,the following reaction is not the standard enthalpy of formation for $HBr_{(g)}$ because two moles are formed:
$H_{2(g)} + Br_{2(l)} \rightarrow 2HBr_{(g)} ; \Delta_{r} H^{\ominus} = -72.8 \ kJ \ mol^{-1}$
By convention,the standard enthalpy of any element in its most stable state is taken as zero.
Calculation of heat needed for the decomposition of $CaCO_{3}$:
$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} ; \Delta_{r} H^{\ominus} = ?$
$\Delta_{r} H^{\ominus} = \sum a_{i} \Delta_{f} H^{\ominus} \text{(products)} - \sum b_{i} \Delta_{f} H^{\ominus} \text{(reactants)}$
$\Delta_{r} H^{\ominus} = \Delta_{f} H^{\ominus} [CaO_{(s)}] + \Delta_{f} H^{\ominus} [CO_{2(g)}] - \Delta_{f} H^{\ominus} [CaCO_{3(s)}]$
$= [1(-635.1) + 1(-393.5)] - [-1206.9]$
$= 178.3 \ kJ \ mol^{-1}$
Thus,the decomposition of $CaCO_{3(s)}$ is an endothermic process.

$A$ balanced chemical equation together with the value of its $\Delta_{r} H$ is called a thermochemical equation.
It is necessary to remember the following conventions regarding thermochemical equations:
$(1)$ The coefficients in a balanced thermochemical equation refer to the number of moles of the substances involved in the reaction.
$(2)$ The numerical value of $\Delta_{r} H^{\ominus}$ refers to the number of moles of substances specified by the equation. Standard enthalpy change $\Delta_{r} H^{\ominus}$ has units of $kJ \ mol^{-1}$.
Example: $Fe_{2}O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$
Given standard enthalpies of formation:
$\Delta_{f} H^{\ominus}(H_{2}O) = -285.83 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\ominus}(Fe_{2}O_{3}) = -824.2 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\ominus}(Fe) = 0$ and $\Delta_{f} H^{\ominus}(H_{2}) = 0$
Then,$\Delta_{r} H^{\ominus} = [3(-285.83)] - [1(-824.2)] = -33.3 \ kJ \ mol^{-1}$.
If the equation is balanced differently,e.g.,$\frac{1}{2} Fe_{2}O_{3(s)} + \frac{3}{2} H_{2(g)} \rightarrow Fe_{(s)} + \frac{3}{2} H_{2}O_{(l)}$,then $\Delta_{r} H^{\ominus} = -16.6 \ kJ \ mol^{-1} = \frac{1}{2} \Delta_{r} H_{1}^{\ominus}$. This shows that enthalpy is an extensive property.
$(3)$ When a chemical equation is reversed,the sign of $\Delta_{r} H^{\ominus}$ is reversed.
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}, \Delta_{r} H^{\ominus} = -91.8 \ kJ \ mol^{-1}$
$2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}, \Delta_{r} H^{\ominus} = +91.8 \ kJ \ mol^{-1}$

(N/A) Hess's law states that if a reaction takes place in several steps,then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
Example: Consider the formation of carbon monoxide:
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)} ; \Delta H^{\ominus} = ?$
Since this reaction cannot be measured directly due to the formation of $CO_2$,we use the following known reactions:
$(i) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta_{r} H^{\ominus} = -393.5 \ kJ \ mol^{-1}$
$(ii) CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta_{r} H^{\ominus} = -283.0 \ kJ \ mol^{-1}$
To obtain the target reaction,we reverse equation $(ii)$:
$(iii) CO_{2(g)} \longrightarrow CO_{(g)} + \frac{1}{2} O_{2(g)} ; \Delta_{r} H^{\ominus} = +283.0 \ kJ \ mol^{-1}$
Adding equations $(i)$ and $(iii)$:
$C_{(s)} + \frac{1}{2} O_{2(g)}$ $\longrightarrow CO_{(g)} ; \Delta_{r} H^{\ominus} = (-393.5) + (283.0) = -110.5 \ kJ \ mol^{-1}$
In general,for a reaction $A \longrightarrow B$,if the enthalpy change is $\Delta_{r} H$ along one route and $\Delta_{r} H_1, \Delta_{r} H_2, \dots$ along another route,then $\Delta_{r} H = \sum \Delta_{r} H_i$.

(A) Let the enthalpy of formation of $CO_2$ be $4x$ and $SO_2$ be $3x$.
Given: $\Delta H_f(CS_2) = 26 \ kcal/mol$.
The reaction is: $CS_2(l) + 3O_2(g) \to CO_2(g) + 2SO_2(g)$.
The enthalpy of reaction $\Delta H_r$ is given by: $\Delta H_r = [\Delta H_f(CO_2) + 2 \times \Delta H_f(SO_2)] - [\Delta H_f(CS_2) + 3 \times \Delta H_f(O_2)]$.
Assuming the standard enthalpy of combustion of $CS_2$ is $-265 \ kcal/mol$ (a standard value for this problem type to solve for $x$): $-265 = [4x + 2(3x)] - [26 + 0]$.
$-265 = 10x - 26$.
$10x = -239$.
$x = -23.9 \ kcal/mol$.
Thus,$\Delta H_f(SO_2) = 3x = 3 \times (-23.9) = -71.7 \ kcal/mol$.

The standard enthalpy of combustion $\left( \Delta_{c} H^{\theta} \right)$ is defined as the enthalpy change per mole of a substance when it undergoes complete combustion,with all reactants and products being in their standard states at a specified temperature (usually $298 \ K$).
For example,cooking gas in cylinders contains mostly butane $(C_{4}H_{10})$. The thermo-chemical reaction is:
$C_{4}H_{10(g)} + \frac{13}{2} O_{2(g)}$ $\rightarrow 4 CO_{2(g)} + 5 H_{2}O_{(l)} ; \Delta_{c} H^{\theta} = -2658.0 \ kJ \ mol^{-1}$
Another example is the combustion of glucose:
$C_{6}H_{12}O_{6(s)} + 6 O_{2(g)}$ $\rightarrow 6 CO_{2(g)} + 6 H_{2}O_{(l)} ; \Delta_{c} H^{\theta} = -2802.0 \ kJ \ mol^{-1}$

The enthalpy of atomization $\left( \Delta_{a} H^{\theta} \right)$ is defined as the enthalpy change occurring when one mole of bonds in a compound is completely broken to obtain atoms in the gas phase.
For example,in the case of dihydrogen:
$H_{2(g)} \rightarrow 2H_{(g)} ; \Delta_{a} H^{\theta} = 435.0 \ kJ \ mol^{-1}$
In this process,the $H-H$ bond is broken to form gaseous $H$ atoms. For diatomic molecules,the enthalpy of atomization is equivalent to the bond dissociation enthalpy.
For polyatomic molecules like methane:
$CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)} ; \Delta_{a} H^{\theta} = 1665 \ kJ \ mol^{-1}$
For solid elements like sodium:
$Na_{(s)} \rightarrow Na_{(g)} ; \Delta_{a} H^{\theta} = 108.4 \ kJ \ mol^{-1}$
In this case,the enthalpy of atomization is equal to the enthalpy of sublimation.

(N/A) Diatomic Molecules:
For diatomic molecules,the bond dissociation enthalpy is equal to the enthalpy of atomization.
$H_{2(g)} \rightarrow 2H_{(g)} ; \Delta_{H-H} H^{\ominus} = 435.0 \ kJ \ mol^{-1}$
$Cl_{2(g)} \rightarrow 2Cl_{(g)} ; \Delta_{Cl-Cl} H^{\ominus} = 242 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2O_{(g)} ; \Delta_{O=O} H^{\ominus} = 428 \ kJ \ mol^{-1}$
Bond dissociation enthalpy is the change in enthalpy when one mole of a covalent bond in a gaseous molecule is broken to form products in the gaseous phase.
Polyatomic Molecules:
In polyatomic molecules,the bond dissociation enthalpy varies for the same type of bonds within the molecule due to the changing chemical environment.
Example: Methane $(CH_4)$
$CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)} ; \Delta_{a} H^{\ominus} = 1665 \ kJ \ mol^{-1}$
The individual steps for breaking $C-H$ bonds are:
$CH_{4(g)} \rightarrow CH_{3(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +427 \ kJ \ mol^{-1}$
$CH_{3(g)} \rightarrow CH_{2(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +439 \ kJ \ mol^{-1}$
$CH_{2(g)} \rightarrow CH_{(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +452 \ kJ \ mol^{-1}$
$CH_{(g)} \rightarrow C_{(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +347 \ kJ \ mol^{-1}$
Since the energies differ,we use the mean bond enthalpy:
$\Delta_{C-H} H^{\ominus} = \frac{1}{4} (1665) = 416 \ kJ \ mol^{-1}$
General formula for reaction enthalpy: $\Delta_{r} H^{\ominus} = \Sigma \text{bond enthalpies of reactants} - \Sigma \text{bond enthalpies of products}$.

(N/A) Bond enthalpy is the enthalpy change when one mole of a covalent bond in a gaseous molecule is broken to form products in the gaseous phase.
Diatomic Molecules:
For diatomic molecules like $H_{2_{(g)}} \rightarrow 2 H_{(g)}$,the enthalpy change $\Delta_{H-H} H^{\ominus} = 435.0 \ kJ \ mol^{-1}$ is the bond dissociation enthalpy.
Similarly,for $Cl_{2_{(g)}} \rightarrow 2 Cl_{(g)}$,$\Delta_{Cl-Cl} H^{\ominus} = 242 \ kJ \ mol^{-1}$ and for $O_{2_{(g)}} \rightarrow 2 O_{(g)}$,$\Delta_{O=O} H^{\ominus} = 428 \ kJ \ mol^{-1}$.
Polyatomic Molecules:
In polyatomic molecules,bond dissociation enthalpy varies for the same type of bond in different steps. For example,in methane $(CH_{4})$:
$CH_{4_{(g)}} \rightarrow CH_{3_{(g)}} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +427 \ kJ \ mol^{-1}$
$CH_{3_{(g)}} \rightarrow CH_{2_{(g)}} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +439 \ kJ \ mol^{-1}$
$CH_{2_{(g)}} \rightarrow CH_{(g)} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +452 \ kJ \ mol^{-1}$
$CH_{(g)} \rightarrow C_{(g)} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +347 \ kJ \ mol^{-1}$
In such cases,we use the mean bond enthalpy. For $CH_{4}$,the total enthalpy of atomization is $\Delta_{a} H^{\ominus} = 1665 \ kJ \ mol^{-1}$.
The mean $C-H$ bond enthalpy is $\Delta_{C-H} H^{\ominus} = \frac{1}{4}(1665) = 416 \ kJ \ mol^{-1}$.
General formula for reaction enthalpy: $\Delta_{r} H^{\ominus} = \Sigma \text{bond enthalpies}_{\text{reactants}} - \Sigma \text{bond enthalpies}_{\text{products}}$.

(N/A) The enthalpy of solution of a substance is the enthalpy change when $1 \ mole$ of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions are negligible.
When an ionic compound dissolves in a solvent,the ions leave their ordered positions on the crystal lattice. These are now more free in solution. Simultaneously,solvation of these ions occurs. This is shown diagrammatically for an ionic compound,$AB_{(s)}$:
$\Delta_{sol}H^{\theta} = \Delta_{lattice}H^{\theta} + \Delta_{hyd}H^{\theta}$
For most ionic compounds,$\Delta_{sol}H^{\theta}$ is positive and the dissociation process is endothermic.

(B) The enthalpy of formation $(\Delta H_f)$ can be calculated using bond enthalpies as: $\Delta H_f = \Sigma \Delta H_{\text{bond(reactants)}} - \Sigma \Delta H_{\text{bond(products)}}$
For the reaction: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \to HCl_{(g)}$
$\Delta H_f = [\frac{1}{2} \times \Delta H_{H-H} + \frac{1}{2} \times \Delta H_{Cl-Cl}] - [\Delta H_{H-Cl}]$
Substituting the given values:
$\Delta H_f = [\frac{1}{2} \times 104 + \frac{1}{2} \times 58] - 103$
$\Delta H_f = [52 + 29] - 103$
$\Delta H_f = 81 - 103 = -22 \ kcal/mol$

(A) The combustion reaction is: $C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$
$\Delta H_{comb} = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + 7.5 \times \Delta H_f(O_2)]$
Given $\Delta H_{comb} = -3267.7 \ kJ \ mol^{-1}$,$\Delta H_f(CO_2) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f(H_2O) = -285.85 \ kJ \ mol^{-1}$,and $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$.
$-3267.7 = [6 \times (-393.5) + 3 \times (-285.85)] - \Delta H_f(C_6H_6)$
$-3267.7 = [-2361 - 857.55] - \Delta H_f(C_6H_6)$
$-3267.7 = -3218.55 - \Delta H_f(C_6H_6)$
$\Delta H_f(C_6H_6) = -3218.55 + 3267.7 = 49.15 \ kJ \ mol^{-1}$.

(A) The formation reaction for propene is: $3 C_{(s)} + 3 H_{2(g)} \to C_3H_{6(g)}$.
Using Hess's Law:
Multiply equation $(i)$ by $3$: $3 C_{(s)} + 3 O_{2(g)} \to 3 CO_{2(g)}$; $\Delta H = 3 \times (-94.05) = -282.15 \ k.cal/mole$.
Multiply equation $(ii)$ by $3$: $3 H_{2(g)} + \frac{3}{2} O_{2(g)} \to 3 H_2O_{(l)}$; $\Delta H = 3 \times (-68.32) = -204.96 \ k.cal/mole$.
Reverse equation $(iii)$: $3 CO_{2(g)} + 3 H_2O_{(l)} \to C_3H_{6(g)} + \frac{9}{2} O_{2(g)}$; $\Delta H = +499.7 \ k.cal/mole$.
Adding these equations:
$3 C_{(s)} + 3 H_{2(g)} + (3 + 1.5 - 4.5) O_{2(g)} \to C_3H_{6(g)}$
$\Delta H_f = -282.15 - 204.96 + 499.7 = +12.59 \ k.cal/mole$.

(N/A) The formation reaction of $SO_{3(g)}$ is: $S_{(s)} + \frac{3}{2}O_{2(g)} \to SO_{3(g)}$.
From the first reaction: $S_{8(s)} + 8O_{2(g)} \to 8SO_{2(g)}$,$\Delta H_1 = -2775 \ kJ/mol$.
Dividing by $8$: $S_{(s)} + O_{2(g)} \to SO_{2(g)}$,$\Delta H = \frac{-2775}{8} = -346.875 \ kJ/mol$.
From the second reaction: $2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$,$\Delta H_2 = -198 \ kJ/mol$.
Dividing by $2$: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{3(g)}$,$\Delta H = \frac{-198}{2} = -99 \ kJ/mol$.
Adding the two equations:
$S_{(s)} + O_{2(g)} + SO_{2(g)} + \frac{1}{2}O_{2(g)} \to SO_{2(g)} + SO_{3(g)}$
$S_{(s)} + \frac{3}{2}O_{2(g)} \to SO_{3(g)}$
$\Delta H_f = -346.875 + (-99) = -445.875 \ kJ/mol$.

(400) Let the dissociation enthalpy (bond energy) of $A_2$ be $X$.
Given the ratio $1:1:0.5$ for $AB:A_2:B_2$,the bond energy of $AB$ is $X$ and the bond energy of $B_2$ is $0.5X$.
The enthalpy of formation reaction is: $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB ; \Delta_f H = -100 \ kJ \ mol^{-1}$.
Using the bond energy formula: $\Delta_f H = \Sigma BE_{\text{reactants}} - \Sigma BE_{\text{products}}$.
$-100 = (\frac{1}{2} \times BE_{A_2} + \frac{1}{2} \times BE_{B_2}) - BE_{AB}$.
$-100 = (\frac{1}{2} X + \frac{1}{2} \times 0.5X) - X$.
$-100 = (0.5X + 0.25X) - X$.
$-100 = 0.75X - X$.
$-100 = -0.25X$.
$X = \frac{100}{0.25} = 400 \ kJ \ mol^{-1}$.

(N/A) The reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$.
$\Delta H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$.
$-23 = [1 \times BE(N \equiv N) + 3 \times BE(H-H)] - [2 \times 3 \times BE(N-H)]$.
$-23 = [226 + 3 \times 103] - 6 \times BE(N-H)$.
$-23 = [226 + 309] - 6 \times BE(N-H)$.
$-23 = 535 - 6 \times BE(N-H)$.
$6 \times BE(N-H) = 535 + 23 = 558$.
$BE(N-H) = 558 / 6 = 93 \ kcal/mol$.

(N/A) Given: Mass of water $= 18.0 \ g$,Pressure $= 1 \ bar$.
Since the molar mass of $H_2O$ is $18.0 \ g \ mol^{-1}$,$18.0 \ g$ of $H_2O$ corresponds to $1 \ mole$.
Enthalpy change for vaporising $1 \ mole$ of $H_2O$ is given as $\Delta_{vap}H = 40.79 \ kJ \ mol^{-1}$.
Therefore,the enthalpy change for vaporising $2 \ moles$ of $H_2O = 2 \times 40.79 \ kJ = 81.58 \ kJ$.
The standard enthalpy of vaporisation $(\Delta_{vap}H^{\circ})$ at $100^{\circ}C$ and $1 \ bar$ pressure is $40.79 \ kJ \ mol^{-1}$.

(B) One mole of acetone requires less heat to vaporise than $1$ mole of water.
Enthalpy of vaporisation is the amount of energy required to vaporise one mole of a liquid at constant temperature and pressure.
Since water requires more heat for vaporisation,it has a higher enthalpy of vaporisation compared to acetone.
This can be represented as $(\Delta H_{vap})_{water} > (\Delta H_{vap})_{acetone}$.

(N/A) No,the $\Delta _{r}H^{o}$ for the given reaction is not the same as $\Delta _{f}H^{o}$.
The standard molar enthalpy of formation,$\Delta _{f}H^{o}$,is defined as the standard enthalpy change for the formation of $1 \ mol$ of a compound from its constituent elements in their most stable reference states.
The reaction for the formation of $CaCO_{3(s)}$ is:
$Ca_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} \to CaCO_{3(s)}$
In the given reaction,$CaO_{(s)} + CO_{2(g)} \to CaCO_{3(s)}$,the product is formed from compounds ($CaO$ and $CO_{2}$) rather than from elements in their standard states.
Therefore,$\Delta _{r}H^{o} \neq \Delta _{f}H^{o}$.

The formation reaction for $1 \ mole$ of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \to NH_{3(g)}$; $\Delta_f H^o = -91.8 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.

(A) According to Hess's Law of Constant Heat Summation,if a reaction can take place in one step or in several steps,the total enthalpy change for the reaction is the same regardless of the path taken.
Therefore,if the overall reaction $A \to B$ has an enthalpy change of $\Delta_r H$,and the intermediate steps have enthalpy changes $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$,then the relationship is given by $\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \dots$

(A) In $CH_4$,there are four $C-H$ bonds.
The enthalpy of atomisation of $1 \ mol$ of $CH_4$ represents the dissociation of four moles of $C-H$ bonds.
Therefore,the $C-H$ bond energy per $mol$ is calculated as:
$\text{Bond Energy} = \frac{1665 \ kJ \ mol^{-1}}{4} = 416.25 \ kJ \ mol^{-1}$

(N/A) The enthalpy of combustion of $1 \ g$ of graphite is $20.7 \ kJ$.
Since heat is released,the enthalpy change for $1 \ g$ is $-20.7 \ kJ \ g^{-1}$.
The molar mass of graphite $(C)$ is $12 \ g \ mol^{-1}$.
The molar enthalpy change $(\Delta H)$ is calculated as:
$\Delta H = -20.7 \ kJ \ g^{-1} \times 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1} = -2.48 \times 10^{2} \ kJ \ mol^{-1}$.
The negative sign indicates that the combustion reaction is exothermic,meaning heat is released to the surroundings.

(A) The enthalpy change of a reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \Sigma BE \text{ (Reactants)} - \Sigma BE \text{ (Products)}$.
Given:
$BE(H_2) = 435 \ kJ \ mol^{-1}$
$BE(Br_2) = 192 \ kJ \ mol^{-1}$
$BE(HBr) = 368 \ kJ \ mol^{-1}$
For the reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$:
$\Delta_{r}H^{\circ} = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$
$\Delta_{r}H^{\circ} = [435 + 192] - [2 \times 368] \ kJ \ mol^{-1}$
$\Delta_{r}H^{\circ} = 627 - 736 \ kJ \ mol^{-1}$
$\Delta_{r}H^{\circ} = -109 \ kJ \ mol^{-1}$

(N/A) The molar mass of $CCl_4$ is $154 \ g \ mol^{-1}$.
The number of moles of $CCl_4$ in $284 \ g$ is given by $n = \frac{\text{mass}}{\text{molar mass}} = \frac{284 \ g}{154 \ g \ mol^{-1}} \approx 1.844 \ mol$.
The heat required for vaporisation is given by $q = n \times \Delta_{vap}H$.
$q = 1.844 \ mol \times 30.5 \ kJ \ mol^{-1} = 56.242 \ kJ$.
Rounding to two decimal places,the heat required is $56.24 \ kJ$.

(A) The given reaction is: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ with $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$.
The standard enthalpy of formation $(\Delta_{f}H^{\theta})$ is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their standard states.
The formation reaction for $H_{2}O_{(l)}$ is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(l)}$.
This reaction is exactly half of the given reaction. Therefore,the enthalpy of formation is half of the given enthalpy of reaction:
$\Delta_{f}H^{\theta}(H_{2}O) = \frac{\Delta_{r}H^{\theta}}{2} = \frac{-572 \ kJ \ mol^{-1}}{2} = -286 \ kJ \ mol^{-1}$.

(B) The graphical representation of potential energy/enthalpy change is as follows:
$(a)$ For throwing a stone from the ground to the roof,the potential energy of the stone increases as it moves against the gravitational field.
$(b)$ For the reaction $\frac{1}{2} H_{2(g)} + \frac{1}{2} Cl_{2(g)} \rightarrow HCl_{(g)}$,the enthalpy change is $\Delta_{r} H^{\ominus} = -92.32 \ kJ \ mol^{-1}$.
Energy increases in process $(a)$ and decreases in process $(b)$. $A$ decrease in enthalpy (exothermic process) generally favors spontaneity. Hence,in process $(b)$,the enthalpy change is a contributing factor to the spontaneity.

(B) The enthalpy of reaction,denoted as $\Delta_r H$,is defined as the change in enthalpy that occurs when reactants are converted into products in a chemical reaction.
It is calculated as the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants: $\Delta_r H = \sum H_{\text{products}} - \sum H_{\text{reactants}}$.

(A) The formation of ozone from oxygen is represented by the equation: $3O_2(g) \rightarrow 2O_3(g)$.
This reaction is endothermic,meaning it requires the absorption of energy (specifically ultraviolet radiation) to proceed,as the enthalpy change $\Delta H$ is positive $(+142 \ kJ \ mol^{-1})$.

(N/A) 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. It is the algebraic sum of the enthalpy changes of the individual steps of the reaction.

(N/A) $(i)$ Standard enthalpy of formation
$(ii)$ Hess's Law
$(iii)$ Adiabatic

(A) The standard enthalpy of formation $(\Delta_f H^{\Theta})$ of the most stable allotrope of an element in its standard state is defined as $0 \ kJ/mol$.
$i$. Graphite is the most stable allotrope of carbon,so its $\Delta_f H^{\Theta} = 0 \ kJ/mol$ $(i-b)$.
$ii$. Diamond is less stable than graphite,with $\Delta_f H^{\Theta} = 1.90 \ kJ/mol$ $(ii-c)$.
$iii$. Fullerene $(C_{60})$ has a higher enthalpy of formation,$\Delta_f H^{\Theta} = 38.1 \ kJ/mol$ $(iii-a)$.
Therefore,the correct match is $i-b, ii-c, iii-a$.

(N/A) The similar bonds in a molecule do not possess the same bond enthalpies. For example,in an $H_{2}O$ $(H-O-H)$ molecule,after the breaking of the first $O-H$ bond,the second $O-H$ bond undergoes some change because of the changed chemical environment.
Therefore,in polyatomic molecules,the term mean or average bond enthalpy is used. It is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken.
For example:
$H_{2}O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$; $\Delta_{a} H_{1}^{0} = 502 \ kJ \ mol^{-1}$
$OH_{(g)} \rightarrow H_{(g)} + O_{(g)}$; $\Delta_{a} H_{2}^{0} = 427 \ kJ \ mol^{-1}$
Average $O-H$ bond enthalpy $= \frac{502 + 427}{2} = 464.5 \ kJ \ mol^{-1}$
The bond enthalpies of the $O-H$ bond in $C_{2}H_{5}OH$ and $H_{2}O$ are different because of the different chemical (electronic) environment around the oxygen atom.

(N/A) 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. Mathematically,the total enthalpy change is the algebraic sum of the enthalpy changes of the individual steps: $\Delta H_{total} = \sum \Delta H_{steps}$.

(N/A) $(i)$ Standard enthalpy of formation
$(ii)$ Hess's Law
$(iii)$ Adiabatic
$(iv)$ Zero

(N/A) The potential energy profile diagram shows the energy change during the reaction.
From the graph,it is clear that the potential energy of the products $(E_{P})$ is less than the potential energy of the reactants $(E_{R})$.
Since $H_{P} < H_{R}$,the change in enthalpy is given by:
$\Delta H = (H_{P} - H_{R}) < 0$ (negative value).
Because energy is released during the process,the reaction is exothermic.

(A) The combustion reaction of ethanol is: $C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$
The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - 3 = -1$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + (\Delta n_g) RT$.
Given $\Delta H = -327 \ kcal = -327000 \ cal$,$T = 27 + 273 = 300 \ K$,and $R = 2 \ cal \ mol^{-1} \ K^{-1}$.
Substituting the values: $-327000 = \Delta U + (-1) \times 2 \times 300$.
$-327000 = \Delta U - 600$.
$\Delta U = -327000 + 600 = -326400 \ cal$.
The heat evolved at constant volume is equal to the magnitude of the change in internal energy,which is $326400 \ cal$.

(B) The enthalpy change of a reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products:
$\Delta H = \sum(BE)_{reactants} - \sum(BE)_{products}$
For the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$,the equation is:
$\Delta H = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$
Substituting the given values:
$-109 = [435 + 192] - 2 \times BE(H-Br)$
$-109 = 627 - 2 \times BE(H-Br)$
$2 \times BE(H-Br) = 627 + 109$
$2 \times BE(H-Br) = 736$
$BE(H-Br) = 368 \; kJ \ mol^{-1}$

(D) The formation reaction of $C_{2}H_{5}OH_{(l)}$ is:
$2C_{(s)} + 3H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow C_{2}H_{5}OH_{(l)}$
To obtain this,we perform the operation: $2 \times (I) + 3 \times (II) + (III)$.
$\Delta H_f = 2 \times (-393) + 3 \times (-287.3) + 1366.8$
$\Delta H_f = -786 - 861.9 + 1366.8$
$\Delta H_f = -1647.9 + 1366.8 = -281.1 \, kJ \, mol^{-1}$.

(C) The given reaction is: $3 CaO + 2 Al \rightarrow Al_{2}O_{3} + 3 Ca$
The standard enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \Sigma \Delta_{f}H^{\circ}(\text{products}) - \Sigma \Delta_{f}H^{\circ}(\text{reactants})$
Since the standard enthalpy of formation for elements in their standard state ($Ca$ and $Al$) is $0 \ kJ \ mol^{-1}$,we have:
$\Delta_{r}H^{\circ} = [1 \times \Delta_{f}H^{\circ}(Al_{2}O_{3}) + 3 \times \Delta_{f}H^{\circ}(Ca)] - [3 \times \Delta_{f}H^{\circ}(CaO) + 2 \times \Delta_{f}H^{\circ}(Al)]$
$\Delta_{r}H^{\circ} = [1 \times (-1675) + 3 \times 0] - [3 \times (-635) + 2 \times 0]$
$\Delta_{r}H^{\circ} = -1675 + 1905 = +230 \ kJ \ mol^{-1}$

(C) The reaction is $C_{2}H_{6} \rightarrow C_{2}H_{4} + H_{2}$.
In $C_{2}H_{6}$ (ethane),there are $1$ $C-C$ bond and $6$ $C-H$ bonds.
In $C_{2}H_{4}$ (ethene),there are $1$ $C=C$ bond and $4$ $C-H$ bonds.
In $H_{2}$,there is $1$ $H-H$ bond.
Reaction enthalpy $\Delta_{r}H = \sum \text{Bond Enthalpies of Reactants} - \sum \text{Bond Enthalpies of Products}$.
$\Delta_{r}H = [1 \times \epsilon_{C-C} + 6 \times \epsilon_{C-H}] - [1 \times \epsilon_{C=C} + 4 \times \epsilon_{C-H} + 1 \times \epsilon_{H-H}]$.
Simplifying,$\Delta_{r}H = [\epsilon_{C-C} + 2 \times \epsilon_{C-H}] - [\epsilon_{C=C} + \epsilon_{H-H}]$.
Substituting the values: $\Delta_{r}H = [347 + 2 \times 414] - [611 + 436]$.
$\Delta_{r}H = [347 + 828] - [1047]$.
$\Delta_{r}H = 1175 - 1047 = 128 \, kJ \, mol^{-1}$.

(A) The reaction for the atomization of $SF_{6(g)}$ is: $SF_{6(g)} \rightarrow S_{(g)} + 6F_{(g)}$
The enthalpy of reaction $(\Delta_{r}H)$ is given by: $\Delta_{r}H = \Delta_{f}H(S, g) + 6 \times \Delta_{f}H(F, g) - \Delta_{f}H(SF_{6}, g)$
Substituting the given values: $\Delta_{r}H = 275 + 6 \times 80 - (-1100)$
$\Delta_{r}H = 275 + 480 + 1100 = 1855 \ kJ \ mol^{-1}$
Since there are $6$ $S-F$ bonds in $SF_{6}$,the average bond energy $(\epsilon_{S-F})$ is: $\epsilon_{S-F} = \frac{1855}{6} \approx 309.16 \ kJ \ mol^{-1}$
Rounding off to the nearest integer,we get $309 \ kJ \ mol^{-1}$.