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

(B) The chemical equation for the formation of methane is: $C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
The molar mass of graphite $(C)$ is $12.0 \ g \ mol^{-1}$.
Given that $6.0 \ g$ of graphite releases $37.4 \ kJ$ of heat,the enthalpy change $(\Delta H)$ for this amount is $-37.4 \ kJ$.
To find the standard enthalpy of formation $(\Delta H_f^{\circ})$,we calculate the heat released for $1 \ mol$ $(12.0 \ g)$ of graphite:
$\Delta H_f^{\circ} = \frac{-37.4 \ kJ}{6.0 \ g} \times 12.0 \ g \ mol^{-1} = -74.8 \ kJ \ mol^{-1}$.

(A) The chemical formula of acetaldehyde is $CH_3CHO$. The molar mass of $CH_3CHO = (2 \times 12) + (4 \times 1) + 16 = 44 \ g \ mol^{-1}$.
Given that the heat of combustion for $1 \ mol$ $(44 \ g)$ of $CH_3CHO$ is $-1172 \ kJ \ mol^{-1}$.
For $66 \ g$ of $CH_3CHO$,the number of moles $n = \frac{66 \ g}{44 \ g \ mol^{-1}} = 1.5 \ mol$.
The heat liberated = $n \times \Delta H = 1.5 \ mol \times 1172 \ kJ \ mol^{-1} = 1758 \ kJ$.
Therefore,the amount of heat liberated is $1758 \ kJ$.

(B) To find the enthalpy of formation of $CO_{(g)}$,we need the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.
Given equations:

$(i) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$	$\Delta H = -X$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$	$\Delta H = -Y$

Subtracting equation $(ii)$ from equation $(i)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} - CO_{(g)} \rightarrow 0$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
The enthalpy change is $\Delta_f H = (-X) - (-Y) = Y - X$.

(C) The combustion reaction is: $C(s) + O_{2}(g) \longrightarrow CO_{2}(g)$
The enthalpy change for the formation of $1 \ mol$ $(44 \ g)$ of $CO_{2}$ is $\Delta H = -393.5 \ kJ / mol$.
Heat released for $44 \ g$ of $CO_{2} = 393.5 \ kJ$.
Heat released for $1 \ g$ of $CO_{2} = \frac{393.5}{44} \ kJ$.
Heat released for $35.2 \ g$ of $CO_{2} = \frac{393.5}{44} \times 35.2 \ kJ = 314.8 \ kJ \approx 315 \ kJ$.
Since heat is released,the value is $-315 \ kJ$.

(B) According to Hess's law,the total enthalpy change of a reaction is the same whether the reaction occurs in one step or in several steps. Therefore,the heat of reaction depends only on the initial and final states of the reactants and products,not on the path taken.

(B) The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is represented as: $H^+ + OH^- \rightarrow H_2O$; $\Delta H = -57.0 \ kJ \ mol^{-1}$.
Since $0.2 \ mol$ of $KOH$ is the limiting reagent,it will react with $0.2 \ mol$ of $HNO_3$ to produce $0.2 \ mol$ of $H_2O$.
The heat evolved is calculated as: $\text{Heat} = \Delta H \times \text{moles of water formed} = 57.0 \ kJ \ mol^{-1} \times 0.2 \ mol = 11.4 \ kJ$.

(C) The reaction for the formation of water is as follows:
$H_2(g) + \frac{1}{2} O_2(g) \longrightarrow H_2O(l); \quad \Delta H = -260 \ kJ$
For the decomposition of water,the reaction is reversed:
$H_2O(l) \longrightarrow H_2(g) + \frac{1}{2} O_2(g); \quad \Delta H = +260 \ kJ$
This means that $260 \ kJ$ of heat is required to decompose $1 \ mol$ of $H_2O$.
Therefore,the amount of $H_2O$ decomposed by $130 \ kJ$ of heat is:
$\text{Moles of } H_2O = \frac{1 \ mol}{260 \ kJ} \times 130 \ kJ = 0.5 \ mol$

(B) Given equations:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ ...$(I)$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$ ...$(II)$
We need to find the heat of formation of $CO$,which corresponds to the reaction:
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)}; \Delta H = ?$
Subtracting equation $(II)$ from equation $(I)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \longrightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} - CO_{(g)} \longrightarrow 0$
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)}$
The enthalpy change for this reaction is $\Delta H = r - s$.

(B) Hess's law states that the total enthalpy change for a reaction is the same,whether it occurs in one step or several steps.
This is a direct consequence of the law of conservation of energy,which is also known as the first law of thermodynamics.
Therefore,both $B$ and $C$ are technically correct,but since the law is fundamentally a statement of the conservation of energy,$B$ is the most appropriate answer.

(A) The given reaction is $2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_{2}O_{(g)}$ with $\Delta H^{\circ} = -573.2 \ kJ$.
This is the heat of formation for $2 \text{ moles}$ of water.
The decomposition reaction is the reverse of the formation reaction: $H_{2}O_{(g)} \longrightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$.
For this reaction,the enthalpy change is $\Delta H = -(\frac{-573.2 \ kJ}{2}) = +286.6 \ kJ/mol$.

(C) The enthalpy change $(\Delta H)$ of a reaction is defined as the difference between the potential energy of the products $(H_P)$ and the potential energy of the reactants $(H_R)$.
$\Delta H = H_P - H_R$
From the given graph:
Potential energy of reactants $(H_R)$ = $150 \ kJ$
Potential energy of products $(H_P)$ = $100 \ kJ$
Therefore,$\Delta H = 100 \ kJ - 150 \ kJ = -50 \ kJ$.
Thus,the correct option is $C$.

$\textbf{(C) According to Hess's Law, the enthalpy change for a cyclic process is zero.}$
$\text{For the cycle } A \rightarrow 2B \rightarrow C \rightarrow A, \text{ we have:}$
$\Delta H_{A \to 2B} + \Delta H_{2B \to C} + \Delta H_{C \to A} = 0$
$\text{Given } \Delta H_{A \to 2B} = +10 \text{ J and } \Delta H_{2B \to C} = +25 \text{ J}.$
$\text{Substituting these values:}$
$10 \text{ J} + 25 \text{ J} + \Delta H_{C \to A} = 0$
$35 \text{ J} + \Delta H_{C \to A} = 0$
$\Delta H_{C \to A} = -35 \text{ J}$

(D) For the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
$\Delta H^{\circ}_{reaction} = \Sigma \Delta H_{\text{bonds broken}} - \Sigma \Delta H_{\text{bonds formed}}$
$\Delta H^{\circ}_{reaction} = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$
$\Delta H^{\circ}_{reaction} = (433 + 192) - (2 \times 364) \ kJ \ mol^{-1}$
$\Delta H^{\circ}_{reaction} = 625 - 728 \ kJ \ mol^{-1}$
$\Delta H^{\circ}_{reaction} = -103 \ kJ \ mol^{-1}$

(C) Given equations:
$1) \ S_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}; \Delta H_{1} = 2x \ kJ$
$2) \ SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}; \Delta H_{2} = y \ kJ$
We need to find the heat of formation of $SO_{2}$,which is the enthalpy change for the reaction:
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}; \Delta H_{3} = ?$
Reverse equation $(2)$:
$SO_{3(g)} \rightarrow SO_{2(g)} + \frac{1}{2} O_{2(g)}; \Delta H_{4} = -y \ kJ$ $(3)$
Add equation $(1)$ and equation $(3)$:
$(S_{(s)} + \frac{3}{2} O_{2(g)}) + (SO_{3(g)}) \rightarrow (SO_{3(g)}) + (SO_{2(g)} + \frac{1}{2} O_{2(g)})$
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
The enthalpy change is $\Delta H_{3} = \Delta H_{1} + \Delta H_{4} = 2x - y \ kJ$.

(D) The given thermochemical equation is:
$2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$; $\Delta H = -571.6 \ kJ$
The decomposition of water is represented by the reverse reaction:
$H_2O_{(l)} \rightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$
To obtain this,we reverse the given equation and divide it by $2$:
$\Delta H_{\text{decomposition}} = -(\frac{\Delta H}{2}) = -(\frac{-571.6 \ kJ}{2}) = +285.8 \ kJ$
Thus,the heat of decomposition of water is $+285.8 \ kJ$.

(C) The reaction is $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$.
For any reaction,$\Delta H_r = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$.
Since the elements $N_2$ and $H_2$ are in their standard states,their enthalpy of formation is $0$.
$\Delta H_r = [1 \times \Delta H_f^{\circ}(N_2) + 3 \times \Delta H_f^{\circ}(H_2)] - [2 \times \Delta H_f^{\circ}(NH_3)]$.
$\Delta H_r = [0 + 0] - [2 \times (-46 \ kJ \ mol^{-1})]$.
$\Delta H_r = +92 \ kJ$.

(C) Given,$(BE)_{A_{2}} : (BE)_{B_{2}} : (BE)_{AB} = 2 : 1 : 2$,where $BE$ is bond enthalpy.
Let $(BE)_{A_{2}} = 2x$,$(BE)_{B_{2}} = x$,and $(BE)_{AB} = 2x$.
The reaction for the formation of $AB$ is: $\frac{1}{2} A_{2}(g) + \frac{1}{2} B_{2}(g) \rightarrow AB(g)$.
The enthalpy of formation is given by: $\Delta H^{\circ}_{f} = \Sigma (BE)_{\text{reactants}} - \Sigma (BE)_{\text{products}}$.
Substituting the values: $-100 = [\frac{1}{2}(2x) + \frac{1}{2}(x)] - 2x$.
$-100 = [x + 0.5x] - 2x$.
$-100 = 1.5x - 2x$.
$-100 = -0.5x$.
$x = \frac{100}{0.5} = 200 \ kJ \ mol^{-1}$.
Thus,the bond enthalpy of $B_{2}$ is $200 \ kJ \ mol^{-1}$.

(A) The enthalpy of formation of a substance is defined as the change in enthalpy when $1 \ mol$ of a substance is formed from its constituent elements in their most stable states at standard conditions ($298 \ K$ and $1 \ bar$ pressure).
Since graphite is the most stable allotrope of carbon at standard conditions,its standard enthalpy of formation is defined as zero.
Therefore,the Assertion $(A)$ is correct because it states the standard enthalpy of formation of graphite is zero.
The Reason $(R)$ is also correct because the stability of graphite as the most stable allotrope is the fundamental reason why its enthalpy of formation is taken as zero.
Thus,$(R)$ is the correct explanation of $(A)$.

(A) The balanced chemical equation for the combustion of ethanol is:
$C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of reaction $\Delta_rH^{\ominus}$ is calculated using the formula:
$\Delta_rH^{\ominus} = \sum \Delta_fH^{\ominus}(\text{products}) - \sum \Delta_fH^{\ominus}(\text{reactants})$
Substituting the given values:
$\Delta_rH^{\ominus} = [2 \times \Delta_fH^{\ominus}(CO_{2(g)}) + 3 \times \Delta_fH^{\ominus}(H_2O_{(l)})] - [\Delta_fH^{\ominus}(C_2H_5OH_{(l)}) + 3 \times \Delta_fH^{\ominus}(O_{2(g)})]$
Since $\Delta_fH^{\ominus}$ for an element in its standard state $(O_{2(g)})$ is $0$:
$\Delta_rH^{\ominus} = [2y + 3z] - [x + 3(0)] = 2y + 3z - x \ kJ \ mol^{-1}$

(D) The combustion reaction for ethanol is: $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of reaction is calculated using the formula: $\Delta_r H^{\ominus} = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})$
$\Delta_r H^{\ominus} = [2 \times \Delta_f H^{\ominus}(CO_{2(g)}) + 3 \times \Delta_f H^{\ominus}(H_2O_{(l)})] - [1 \times \Delta_f H^{\ominus}(C_2H_5OH_{(l)}) + 3 \times \Delta_f H^{\ominus}(O_{2(g)})]$
Given $\Delta_f H^{\ominus}(O_{2(g)}) = 0 \ kJ \ mol^{-1}$ (standard state of an element).
Substituting the values: $\Delta_r H^{\ominus} = [2 \times (-393) + 3 \times (-286)] - [-277 + 3 \times 0]$
$\Delta_r H^{\ominus} = [-786 - 858] - [-277]$
$\Delta_r H^{\ominus} = -1644 + 277 = -1367 \ kJ \ mol^{-1}$.

(D) $1$. For diatomic molecules like $H_2$,$Cl_2$,and $F_2$,the enthalpy of atomization is equal to the bond dissociation enthalpy because breaking the bond results in the formation of two atoms.
$2$. For polyatomic molecules like $CH_4$,the enthalpy of atomization is the energy required to break all $C-H$ bonds to form gaseous atoms $(CH_4(g) \rightarrow C(g) + 4H(g))$.
$3$. The bond dissociation enthalpy for $CH_4$ refers to the energy required to break a single $C-H$ bond,which varies for each step of dissociation.
$4$. Therefore,for $CH_4$,the enthalpy of atomization is the sum of four different $C-H$ bond dissociation enthalpies,making it not equal to the bond dissociation enthalpy of a single bond.

(A) The given reaction is: $2 A_{2(g)} + B_{2(g)} \rightarrow 2 A_2 B_{(g)} + 600 \ kJ$.
This indicates that the formation of $2 \ mol$ of $A_2 B$ releases $600 \ kJ$ of energy,meaning the enthalpy change for the reaction is $\Delta H = -600 \ kJ$.
By definition,the standard enthalpy of formation $(\Delta_f H^{\circ})$ is the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
$\Delta_f H^{\circ}(A_2 B) = \frac{\Delta H}{2} = \frac{-600 \ kJ}{2} = -300 \ kJ \ mol^{-1}$.

(A) Given:
$\Delta H_f(A) = 9.7 \ kJ \ mol^{-1}$
$\Delta H_f(B) = -110 \ kJ \ mol^{-1}$
$\Delta H_f(C) = 81 \ kJ \ mol^{-1}$
$\Delta H_f(D) = -393 \ kJ \ mol^{-1}$
Reaction: $A_{(g)} + 3B_{(g)} \longrightarrow C_{(g)} + 3D_{(g)}$
$\Delta_r H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$\Delta_r H = [\Delta H_f(C) + 3 \times \Delta H_f(D)] - [\Delta H_f(A) + 3 \times \Delta H_f(B)]$
$\Delta_r H = [81 + 3 \times (-393)] - [9.7 + 3 \times (-110)]$
$\Delta_r H = [81 - 1179] - [9.7 - 330]$
$\Delta_r H = -1098 - (-320.3)$
$\Delta_r H = -1098 + 320.3$
$\Delta_r H = -777.7 \ kJ \ mol^{-1}$

(A) The standard enthalpy of formation $\Delta_{f} H^\theta$ 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 $H_{2(g)} + Br_{2(l)} \rightarrow 2 HBr_{(g)}$,the enthalpy change $\Delta_{r} H^\theta$ is given as $-72.8 \ kJ$ for the production of $2 \ mol$ of $HBr_{(g)}$.
The relationship between reaction enthalpy and enthalpy of formation is:
$\Delta_{r} H^\theta = \sum \Delta_{f} H^\theta \text{(products)} - \sum \Delta_{f} H^\theta \text{(reactants)}$
Since $\Delta_{f} H^\theta$ for elements in their standard states ($H_{2(g)}$ and $Br_{2(l)}$) is $0$,we have:
$-72.8 \ kJ = 2 \times \Delta_{f} H^\theta (HBr_{(g)}) - (0 + 0)$
$\Delta_{f} H^\theta (HBr_{(g)}) = \frac{-72.8 \ kJ}{2 \ mol} = -36.4 \ kJ \ mol^{-1}$.

(B) The reaction is $CCl_{4(g)} \rightarrow C_{(g)} + 4Cl_{(g)}$.
First,we find the enthalpy of formation of $CCl_{4(g)}$ using the enthalpy of vaporization:
$\Delta_{f} H^{\theta} (CCl_{4(g)}) = \Delta_{f} H^{\theta} (CCl_{4(l)}) + \Delta_{vap} H^{\theta} (CCl_{4(l)}) = -136.0 + 30.0 = -106.0 \ kJ \ mol^{-1}$.
The enthalpy change for the reaction is given by:
$\Delta_{r} H^{\theta} = \Delta_{a} H^{\theta} (C) + 4 \times \Delta_{a} H^{\theta} (Cl) - \Delta_{f} H^{\theta} (CCl_{4(g)})$.
Note that $\Delta_{a} H^{\theta} (Cl) = \frac{1}{2} \Delta_{a} H^{\theta} (Cl_2) = \frac{242.0}{2} = 121.0 \ kJ \ mol^{-1}$.
$\Delta_{r} H^{\theta} = 714.0 + 4(121.0) - (-106.0) = 714.0 + 484.0 + 106.0 = 1304.0 \ kJ \ mol^{-1}$.
The mean bond enthalpy of $C-Cl$ is $\frac{\Delta_{r} H^{\theta}}{4} = \frac{1304.0}{4} = 326 \ kJ \ mol^{-1}$.

(B) The decomposition reaction is: $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$
$\Delta H_{r}^{\circ} = [\Delta_{f} H^{\circ}(CaO) + \Delta_{f} H^{\circ}(CO_2)] - [\Delta_{f} H^{\circ}(CaCO_3)]$
$\Delta H_{r}^{\circ} = [(-634) + (-393)] - (-1210)$
$\Delta H_{r}^{\circ} = -1027 + 1210 = +183 \ kJ \ mol^{-1}$

(D) The standard enthalpy of atomization $(\Delta_{a}H^{\circ})$ is the sum of the bond dissociation enthalpies of all bonds in the molecule.
For ethane $(C_2H_6)$,there are $6$ $C-H$ bonds and $1$ $C-C$ bond.
Thus,$\Delta_{a}H^{\circ} = (6 \times \Delta_{C-H}H^{\circ}) \Delta_{C-C}H^{\circ}$.
Given: $\Delta_{a}H^{\circ} = 622 \ kJ \ mol^{-1}$ and $\Delta_{C-H}H^{\circ} = 90 \ kJ \ mol^{-1}$.
Substituting the values: $622 = (6 \times 90) \Delta_{C-C}H^{\circ}$.
$622 = 540 \Delta_{C-C}H^{\circ}$.
$\Delta_{C-C}H^{\circ} = 622 - 540 = 82 \ kJ \ mol^{-1}$.

(A) The formation reaction is: $2C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \longrightarrow CH_3CHO(g)$
The enthalpy of formation $\Delta H_f$ is calculated using the formula: $\Delta H_f = \Sigma \Delta H_{\text{sublimation/atomization}} - \Sigma BE_{\text{products}}$
$\Delta H_f = [2 \times \Delta H_f(C) + 2 \times \Delta H_f(H) + \frac{1}{2} \times \Delta H_f(O_2)] - [3 \times BE(C-H) + 1 \times BE(C-C) + 1 \times BE(C=O)]$
Note: $\Delta H_f(O_2) = 2 \times 250 = 500 \ kJ \ mol^{-1}$
$\Delta H_f = [(2 \times 700) + (2 \times 200) + (\frac{1}{2} \times 500)] - [(3 \times 400) + (1 \times 350) + (1 \times 700)]$
$\Delta H_f = [1400 + 400 + 250] - [1200 + 350 + 700]$
$\Delta H_f = 2050 - 2250 = -200 \ kJ \ mol^{-1}$

(C) To find $\Delta_{r} H^0$ for $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$,we manipulate the given equations:
$1. H_2O_{(g)} \rightarrow H_{2(g)} + 1/2 O_{2(g)} ; \Delta H = +242 \ kJ \ mol^{-1}$ (Reverse of equation $2$)
$2. H_{2(g)} \rightarrow 2 H_{(g)} ; \Delta H = +436 \ kJ \ mol^{-1}$ (Equation $3$)
$3. 1/2 O_{2(g)} \rightarrow O_{(g)} ; \Delta H = 496 / 2 = +248 \ kJ \ mol^{-1}$ (Half of equation $4$)
Adding these equations:
$H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$
$\Delta_{r} H^0 = 242 + 436 + 248 = 926 \ kJ \ mol^{-1}$

(D) The enthalpy of reaction $\Delta_r H$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$.
Given values are: $\Delta_f H(CO) = -110 \ kJ \ mol^{-1}$,$\Delta_f H(CO_2) = -393 \ kJ \ mol^{-1}$,$\Delta_f H(N_2O) = 81 \ kJ \ mol^{-1}$,and $\Delta_f H(N_2O_4) = 9.7 \ kJ \ mol^{-1}$.
For the reaction $N_2O_{4(g)} + 3 CO_{(g)} \longrightarrow N_2O_{(g)} + 3 CO_{2(g)}$:
$\Delta_r H = [\Delta_f H(N_2O) + 3 \times \Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3 \times \Delta_f H(CO)]$.
Substituting the values:
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)]$.
$\Delta_r H = [81 - 1179] - [9.7 - 330]$.
$\Delta_r H = (-1098) - (-320.3)$.
$\Delta_r H = -1098 + 320.3 = -777.7 \ kJ \ mol^{-1}$.
Rounding off,we get $\Delta_r H \approx -778 \ kJ \ mol^{-1}$.

(C) The enthalpy of reaction $(\Delta_r H^0)$ is calculated using the formula:
$\Delta_r H^0 = \sum \Delta_f H^0 (\text{products}) - \sum \Delta_f H^0 (\text{reactants})$
For the reaction: $C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
$\Delta_r H^0 = [2 \times \Delta_f H^0 (CO_2) + 3 \times \Delta_f H^0 (H_2O)] - [\Delta_f H^0 (C_2H_5OH) + \frac{7}{2} \Delta_f H^0 (O_2)]$
Given: $\Delta_f H^0 (O_2) = 0 \ kJ \ mol^{-1}$ (standard state element).
$\Delta_r H^0 = [2(-400) + 3(-290)] - [-280 + 0]$
$\Delta_r H^0 = [-800 - 870] + 280$
$\Delta_r H^0 = -1670 + 280 = -1390 \ kJ \ mol^{-1}$

(B) The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$.
$CH_{4(g)} \longrightarrow C_{(g)} + 4H_{(g)} ; \Delta_a H = 1660 \ kJ \ mol^{-1}$
Mean $C-H$ bond enthalpy $= \frac{1660}{4} = 415 \ kJ \ mol^{-1}$.
Let the bond enthalpies of the four $C-H$ bonds be $E_1, E_2, E_3, E_4$.
Given: $E_1 = 415 + 15 = 430 \ kJ \ mol^{-1}$,$E_2 = 415 + 30 = 445 \ kJ \ mol^{-1}$,$E_3 = 415 + 45 = 460 \ kJ \ mol^{-1}$.
The sum of all bond enthalpies equals the atomisation enthalpy:
$E_1 + E_2 + E_3 + E_4 = 1660$
$430 + 445 + 460 + E_4 = 1660$
$1335 + E_4 = 1660$
$E_4 = 1660 - 1335 = 325 \ kJ \ mol^{-1}$.

(A) The enthalpy of neutralisation for a strong acid and strong base is given by: $H^{+} + OH^{-} \longrightarrow H_2O; \Delta_r H^{\circ} = -55.84 \ kJ \ mol^{-1}$.
For the weak acid $CH_3COOH$,the neutralisation reaction is: $CH_3COOH + OH^{-} \longrightarrow CH_3COO^{-} + H_2O; \Delta_r H = -49.86 \ kJ \ mol^{-1}$.
The enthalpy of ionisation $(\Delta H_i)$ is the energy required to dissociate the weak acid: $CH_3COOH \longrightarrow CH_3COO^{-} + H^{+}; \Delta H_i = ?$.
This can be calculated as: $\Delta H_i = \Delta H_{\text{neutralisation(weak)}} - \Delta H_{\text{neutralisation(strong)}}$.
$\Delta H_i = -49.86 - (-55.84) = 5.98 \ kJ \ mol^{-1}$.

(A) The standard molar enthalpy of formation,$\Delta_f H^{\circ}$,is a measure of the stability of a compound relative to its constituent elements in their standard states. $A$ more negative value indicates greater stability.
For the given alkali metal halides,the lattice energy is the primary factor determining the stability.
According to the Born-Landé equation,lattice energy is inversely proportional to the interionic distance $(r_+ + r_-)$.
Since the size of the halide ion increases in the order $F^- < Cl^- < Br^- < I^-$,the interionic distance increases in the order $NaF < NaCl < NaBr < NaI$.
Consequently,the lattice energy decreases in the order $NaF > NaCl > NaBr > NaI$.
Therefore,$NaF_{(s)}$ has the most negative (highest magnitude) standard molar enthalpy of formation.

(D) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H_{rxn}^{\circ} = \sum \Delta H_f^{\circ} (\text{products}) - \sum \Delta H_f^{\circ} (\text{reactants})$.
For the reaction $3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,the expression is: $\Delta H_{rxn}^{\circ} = [1 \times \Delta H_f^{\circ}(C_6 H_6)] - [3 \times \Delta H_f^{\circ}(C_2 H_2)]$.
Substituting the given values: $\Delta H_{rxn}^{\circ} = [1 \times 90 \ kJ \ mol^{-1}] - [3 \times 250 \ kJ \ mol^{-1}]$.
$\Delta H_{rxn}^{\circ} = 90 \ kJ \ mol^{-1} - 750 \ kJ \ mol^{-1} = -660 \ kJ \ mol^{-1}$.

(A) The formation reaction for $CH_3OH_{(l)}$ is:
$C_{(graphite)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CH_3OH_{(l)}$
We can obtain this by combining the given equations:
$(iii) + 2 \times (ii) - (i)$
Substituting the values:
$\Delta_fH^{\circ} = (-393) + 2 \times (-286) - (-726)$
$\Delta_fH^{\circ} = -393 - 572 + 726$
$\Delta_fH^{\circ} = -965 + 726 = -239 \ kJ \ mol^{-1}$

(A) The combustion reaction is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$,$\Delta H = -394.0 \ kJ \ mol^{-1}$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g \ mol^{-1}$.
Since the formation of $44 \ g$ of $CO_2$ releases $394.0 \ kJ$ of heat,the enthalpy change for the formation of $17.6 \ g$ of $CO_2$ is calculated as:
$\Delta H = \frac{-394.0 \ kJ \ mol^{-1} \times 17.6 \ g}{44 \ g \ mol^{-1}} = -157.6 \ kJ$.

(D) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
The standard enthalpy of combustion $(\Delta_{c} H^{\circ})$ is calculated using the formula: $\Delta_{c} H^{\circ} = [\sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})]$.
Substituting the given values: $\Delta_{c} H^{\circ} = [\Delta_{f} H^{\circ}(CO_2) + 2 \times \Delta_{f} H^{\circ}(H_2O)] - [\Delta_{f} H^{\circ}(CH_4) + 2 \times \Delta_{f} H^{\circ}(O_2)]$.
Since $\Delta_{f} H^{\circ}(O_2) = 0$,we have: $\Delta_{c} H^{\circ} = [-393 + 2 \times (-286)] - [-74.0]$.
$\Delta_{c} H^{\circ} = [-393 - 572] + 74.0$.
$\Delta_{c} H^{\circ} = -965 + 74.0 = -891 \ kJ \ mol^{-1}$.

(D) The chemical equation for the oxidation of methanol is: $CH_3OH(l) + \frac{1}{2} O_2(g) \longrightarrow HCHO(g) + H_2O(l)$
The enthalpy change of the reaction $(\Delta_r H)$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$
Given values:
$\Delta_f H^{\circ}(CH_3OH) = -239 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(HCHO) = -116 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state)
Substituting the values:
$\Delta_r H = [(-116) + (-286)] - [-239 + 0]$
$\Delta_r H = -402 - (-239)$
$\Delta_r H = -402 + 239 = -163 \ kJ \ mol^{-1}$

(B) The enthalpy of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$.
Given equations:
$(i) \ C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$
$(ii) \ 2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$
Divide equation $(ii)$ by $2$:
$CO_{(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -\frac{y}{2} \ kJ \ mol^{-1} \ (iii)$
Subtract equation $(iii)$ from equation $(i)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2}O_{2(g)}) \longrightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$
$\Delta H^{\circ}_{f} = -x - (-\frac{y}{2}) = \frac{y}{2} - x = \frac{y-2x}{2} \ kJ \ mol^{-1}$.

(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.
Given equations:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
To get the target reaction,perform the operation: $2 \times (A) + \frac{1}{2} \times (C) - (B)$.
$2 \times [Na_{(s)} + H_2O_{(l)}$ $\longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)}] \implies 2Na_{(s)} + 2H_2O_{(l)}$ $\longrightarrow 2NaOH_{(s)} + H_{2(g)} \quad \Delta H^{\circ} = 2 \times (-146) = -292 \ kJ$
$\frac{1}{2} \times [2Na_2O_{(s)} + 2H_{2(g)}$ $\longrightarrow 4Na_{(s)} + 2H_2O_{(l)}] \implies Na_2O_{(s)} + H_{2(g)}$ $\longrightarrow 2Na_{(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = \frac{259}{2} = 129.5 \ kJ$
$-(B) \implies 2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = -418 \ kJ$
Adding these: $(2Na_{(s)} + 2H_2O_{(l)} + Na_2O_{(s)} + H_{2(g)} + 2NaOH_{(s)} + SO_{3(g)})$ $\longrightarrow (2NaOH_{(s)} + H_{2(g)} + 2Na_{(s)} + H_2O_{(l)} + Na_2SO_{4(s)} + H_2O_{(l)})$
Simplifying gives: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.
$\Delta H^{\circ} = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

(C) The formation reaction of ethylene is: $2C_{(graphite)} + 2H_{2(g)} \rightarrow C_2H_{4(g)}$
Using Hess's Law,the enthalpy of formation is given by: $\Delta H_f = 2 \times \Delta H_I + 2 \times \Delta H_{II} - \Delta H_{III}$
Substituting the given values:
$\Delta H_f = 2(-393.5 \ kJ) + 2(-286.2 \ kJ) - (-1410.8 \ kJ)$
$\Delta H_f = -787.0 \ kJ - 572.4 \ kJ + 1410.8 \ kJ$
$\Delta H_f = 51.4 \ kJ$

(C) The molecule $CH_4$ contains four $C-H$ bonds.
To break one mole of $CH_4$ into gaseous carbon and hydrogen atoms,all four $C-H$ bonds must be broken.
The total energy required is $4 \times 416 \ kJ \ mol^{-1} = 1664 \ kJ \ mol^{-1}$.
Therefore,the correct thermochemical equation is $CH_{4(g)} + 1664 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$.

(C) The standard enthalpy of formation $\Delta_{f} H^{\circ}$ for the most stable allotrope of an element is defined as $0 \ kJ/mol$.
Graphite $(II)$ is the most stable allotrope of carbon,so its $\Delta_{f} H^{\circ} = 0 \ kJ/mol$.
Diamond $(I)$ is less stable than graphite,with $\Delta_{f} H^{\circ} \approx 1.90 \ kJ/mol$.
Fullerene $(III)$ is the least stable among these,with $\Delta_{f} H^{\circ} \approx 38.1 \ kJ/mol$.
Therefore,the correct order of $\Delta_{f} H^{\circ}$ values is $III > I > II$.

(B) To find the enthalpy change for the reaction $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$,we manipulate the given equations:
Equation $(3): CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$
Reverse Equation $(2): CO_{2(g)} \rightarrow C_{(s)} + O_{2(g)} ; \Delta H^{\ominus} = +394 \ kJ$
Adding these two equations:
$CH_{4(g)} + 2O_{2(g)} + CO_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} + C_{(s)} + O_{2(g)}$
Simplifying,we get:
$CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$
The enthalpy change is the sum of the enthalpy changes of the steps:
$\Delta H = -890 \ kJ + 394 \ kJ = -496 \ kJ$.

(C) The standard enthalpy of reaction $(\Delta_r H^{\circ})$ is calculated using the formula: $\Delta_r H^{\circ} = \Sigma \Delta_f H^{\circ} (\text{products}) - \Sigma \Delta_f H^{\circ} (\text{reactants})$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the expression is: $\Delta_r H^{\circ} = [2 \times \Delta_f H^{\circ}(NH_3)] - [\Delta_f H^{\circ}(N_2) + 3 \times \Delta_f H^{\circ}(H_2)]$.
Since $N_2$ and $H_2$ are elements in their standard states,their standard enthalpy of formation is $0 \ kJ \ mol^{-1}$.
Substituting the values: $\Delta_r H^{\circ} = [2 \times (-46.2 \ kJ \ mol^{-1})] - [0 + 3 \times 0] = -92.4 \ kJ \ mol^{-1}$.

(B) The formation reaction for methanol is: $C(graphite) + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(l)}$
Given combustion reactions:
$(i)$ $C(graphite) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_1 = -393 \ kJ \ mol^{-1}$
(ii) $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}$,$\Delta H_2 = -286 \ kJ \ mol^{-1}$
(iii) $CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$,$\Delta H_3 = -726 \ kJ \ mol^{-1}$
To get the formation reaction,we perform: $(i) + 2 \times (ii) - (iii)$
$\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$
$\Delta H_f = -393 + 2(-286) - (-726)$
$\Delta H_f = -393 - 572 + 726 = -239 \ kJ \ mol^{-1}$