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

(A) The formation reaction for $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$.
$\Delta H_f = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$.
$-91 = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - BE(H-Cl)$.
$-91 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 242] - BE(H-Cl)$.
$-91 = [215 + 121] - BE(H-Cl)$.
$-91 = 336 - BE(H-Cl)$.
$BE(H-Cl) = 336 + 91 = 427 \, kJ \, mol^{-1}$.

(A) The heat of neutralization of a strong acid and a strong base is constant at $-57.1 \ kJ \ mol^{-1}$ because it involves the reaction $H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)$.
In the case of a weak acid,some energy is consumed to dissociate the weak acid molecules into ions,as they are not fully ionized in solution.
Therefore,the net enthalpy change is less than $-57.1 \ kJ \ mol^{-1}$ because the heat released is partially used to overcome the bond dissociation energy of the weak acid.
Thus,the correct reason is that energy has to be spent for the total dissociation of the weak acid.

(B) The heat of neutralization of a strong acid and a strong base is $-57.1 \ kJ \ mol^{-1}$.
Since $CH_3COOH$ is a weak acid,some energy is consumed in its dissociation.
Therefore,the heat of neutralization of $CH_3COOH$ and $NaOH$ is less than the standard value of $-57.1 \ kJ \ mol^{-1}$.
Given $\Delta H_f (H_2O) = X$,where $X$ is approximately $-285.8 \ kJ \ mol^{-1}$,the heat of neutralization is numerically less than the magnitude of $X$.

(C) The amount of heat produced in calories by the complete combustion of $1 \ g$ of a substance is defined as its calorific value. Therefore,for carbon,it is called the calorific value of carbon.

(B) The total enthalpy change of a chemical reaction is independent of the path taken,whether the reaction occurs in one step or several steps.
This principle is known as $Hess's \ \text{law}$.

(C) . Hess's law of constant heat summation is a direct consequence of the first law of thermodynamics,which is based on the principle of conservation of energy. Since enthalpy is a state function,the total enthalpy change for a reaction is the same whether it occurs in one step or several steps.

(D) 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 several steps.
This principle allows for the calculation of the heat of reaction,heat of formation,heat of transition,and other enthalpy changes by using known thermochemical data.
Therefore,it is applicable to all the given processes.

(B) The reaction $C(s) + O_2(g) \to CO_2(g)$ is a combustion reaction.
Combustion reactions are exothermic in nature,meaning heat is released during the process.
Therefore,the enthalpy change $\Delta H$ for this reaction is negative $(\Delta H < 0)$.

(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. It depends only on the initial state of the reactants and the final state of the products,regardless of the path taken.

(B) Hess's law states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction occurs in one step or in several steps.
It is a direct consequence of the first law of thermodynamics,which is the law of conservation of energy.
Since enthalpy is a state function,the change in enthalpy depends only on the initial and final states,not on the path taken.

(B) The given reaction is $2C + O_2 \to 2CO$ with $\Delta H = -220 \ kJ$.
$1$. The heat of combustion is defined for the complete combustion of $1 \ mole$ of a substance. Here,$2 \ moles$ of $C$ are reacting to form $CO$,not $CO_2$,so this is not the heat of combustion of carbon.
$2$. The negative value of $\Delta H$ $(-220 \ kJ)$ indicates that the reaction is exothermic,as heat is released.
$3$. Most chemical reactions,even if spontaneous,require an initial input of energy (activation energy) to proceed.
Therefore,only statement $B$ is correct.

(B) The heat of formation of methane is the enthalpy change for the reaction: $C(s) + 2H_2(g) \to CH_4(g)$.
Given equations:
$(I)$ $C + O_2 \to CO_2$,$\Delta H_1 = -94.2 \ kcal$
$(II)$ $H_2 + \frac{1}{2} O_2 \to H_2O$,$\Delta H_2 = -68.3 \ kcal$
$(III)$ $CH_4 + 2O_2 \to CO_2 + 2H_2O$,$\Delta H_3 = -210.8 \ kcal$
To obtain the formation reaction,perform: $\Delta H_f = \Delta H_1 + 2 \times \Delta H_2 - \Delta H_3$.
$\Delta H_f = -94.2 + 2(-68.3) - (-210.8)$
$\Delta H_f = -94.2 - 136.6 + 210.8$
$\Delta H_f = -230.8 + 210.8 = -20 \ kcal$.
Since the question asks for the heat of formation (which is typically defined as the energy released or absorbed),and the provided options match the magnitude,the correct value is $20 \ kcal$.

(D) The formation reaction of $H_2SO_4$ is: $H_2 + S + 2O_2 \to H_2SO_4$.
To obtain this,we add the given equations:
$(1) S + O_2 \to SO_2; \Delta H_1 = -298.2 \ kJ$
$(2) SO_2 + \frac{1}{2} O_2 \to SO_3; \Delta H_2 = -98.2 \ kJ$
$(3) SO_3 + H_2O \to H_2SO_4; \Delta H_3 = -130.2 \ kJ$
$(4) H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H_4 = -287.3 \ kJ$
Adding $(1) + (2) + (3) + (4)$:
$S + O_2 + SO_2 + \frac{1}{2} O_2 + SO_3 + H_2O + H_2 + \frac{1}{2} O_2 \to SO_2 + SO_3 + H_2SO_4 + H_2O$
Canceling common terms on both sides gives:
$H_2 + S + 2O_2 \to H_2SO_4$
$\Delta H_f = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4$
$\Delta H_f = (-298.2) + (-98.2) + (-130.2) + (-287.3) = -813.9 \ kJ$.

(B) The effect of temperature on the heat of reaction is given by Kirchhoff's equation,which is expressed as: $\Delta H_2 - \Delta H_1 = \Delta C_p (T_2 - T_1)$. Thus,the factor that affects the heat of reaction in this equation is temperature.

(C) The heat of neutralization for any strong acid and strong base is constant because the reaction essentially involves the formation of water from $H^+$ and $OH^-$ ions: $H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)$.
Since both $KOH$ and $HNO_3$ are strong electrolytes,and $NaOH$ and $HCl$ are also strong electrolytes,the enthalpy change for both reactions is approximately $-13.7 \, kcal/mol$ (or $-57.1 \, kJ/mol$).
Therefore,the molar neutralization heat is equal.

(B) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \, g/mol$.
According to the thermochemical equation,the combustion of $1 \, mol$ $(78 \, g)$ of benzene releases $3264.6 \, kJ$ of heat.
Therefore,the heat evolved for $39 \, g$ of benzene is calculated as:
$\text{Heat} = \frac{3264.6 \, kJ}{78 \, g} \times 39 \, g = \frac{3264.6}{2} = 1632.3 \, kJ$.
Thus,the correct option is $B$.

(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.
Let the given reactions be:
$(i) C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H_1 = -110.5 \ kJ$
$(ii) CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H_2 = -283.2 \ kJ$
Adding equation $(i)$ and $(ii)$:
$C(\text{graphite}) + \frac{1}{2} O_{2(g)} + CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{(g)} + CO_{2(g)}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$
The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 = -110.5 \ kJ + (-283.2 \ kJ) = -393.7 \ kJ$.

(C) The correct statement is $(C)$.
$1$. For an exothermic reaction,$\Delta H$ is negative $(-ve)$.
$2$. For an endothermic reaction,$\Delta H$ is positive $(+ve)$.
$3$. The enthalpy of neutralization for any strong acid and strong base is constant,approximately $-57.1 \ kJ \ mol^{-1}$,because it involves the same reaction: $H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)$.
$4$. The enthalpy of fusion is always positive $(+ve)$ because energy is required to overcome intermolecular forces during melting.

(A) The standard enthalpy of neutralisation for a strong acid and a strong base is $-57.33 \ kJ \ mol^{-1}$.
$MgO$ is a metal oxide that acts as a base,but it is not a strong soluble base like $NaOH$ or $KOH$.
When $MgO_{(s)}$ reacts with $HCl_{(aq)}$,energy is consumed to break the crystal lattice of $MgO$ and to convert it into ions.
Therefore,the net heat released is less than the standard value of $-57.33 \ kJ \ mol^{-1}$ (i.e.,it is less exothermic).
Thus,the enthalpy of neutralisation will be less than $-57.33 \ kJ \ mol^{-1}$.

(B) The standard enthalpy of formation,$\Delta H^o_f$,is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their most stable standard states at $298 \ K$ and $1 \ bar$ pressure.
For methanol $(CH_3OH_{(l)})$,the constituent elements are carbon (in its most stable form,graphite),hydrogen $(H_{2(g)})$,and oxygen $(O_{2(g)})$.
The balanced thermochemical equation is:
$C_{(graphite)} + 1/2 O_{2(g)} + 2H_{2(g)} \to CH_3OH_{(l)}$
Therefore,option $B$ correctly represents the standard enthalpy of formation of methanol.

(A) The given thermochemical equations are:
$(i) \ H_2(g) + O_2(g) \to H_2O_2(l) \quad \Delta H_f^o = -188 \ kJ/mole$
$(ii) \ H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l) \quad \Delta H_f^o = -286 \ kJ/mole$
To obtain the reaction $2H_2O_2(l) \to 2H_2O(l) + O_2(g)$,we perform the operation: $2 \times (ii) - 2 \times (i)$.
$\Delta H_{rxn} = 2 \times (-286) - 2 \times (-188)$
$\Delta H_{rxn} = -572 + 376 = -196 \ kJ/mole$.

(B) We are given the following thermochemical equations:
$(1) \ C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = x \ kJ \ mol^{-1}$
$(2) \ C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H_2 = y \ kJ \ mol^{-1}$
To find the heat of transition for $C(\text{graphite}) \to C(\text{diamond})$,we subtract equation $(2)$ from equation $(1)$:
$C(\text{graphite}) - C(\text{diamond}) = x - y$
$C(\text{graphite}) \to C(\text{diamond}); \Delta H_t = (x - y) \ kJ \ mol^{-1}$

(C) According to Hess's Law,the enthalpy change for a process is the same whether it occurs in one step or multiple steps.
Sublimation is the direct conversion of a solid to a gas: $\text{Solid} \rightarrow \text{Gas} (\Delta H_{sub})$.
This process can also occur in two steps: $\text{Solid} \rightarrow \text{Liquid} (\Delta H_{fus})$ followed by $\text{Liquid} \rightarrow \text{Gas} (\Delta H_{vap})$.
Therefore,the total enthalpy change is the sum of the individual steps: $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$.

(C) An endothermic reaction is one in which heat is absorbed from the surroundings,characterized by a positive change in enthalpy ($\Delta H > 0$ or $\Delta E > 0$).
In the given options,the reaction $N_2 + O_2 \to 2NO$ has $\Delta E = +180.5 \ kJ$,which indicates that energy is absorbed.
Therefore,option $(C)$ is the correct example of an endothermic reaction.

(C) The standard enthalpy of formation $(\Delta_f H^\circ)$ is defined as the change in enthalpy when $1 \, \text{mole}$ of a compound is formed from its constituent elements in their most stable standard states.
In the given reaction,$1 \, \text{mole}$ of $HCl(g)$ is formed from its constituent elements $H_2(g)$ and $Cl_2(g)$ in their standard states.
Therefore,this enthalpy change is specifically called the enthalpy of formation.

(C) The enthalpy of neutralization for a strong acid and a strong base is constant at approximately $-57.3 \ kJ \ mol^{-1}$ (or $-13.7 \ kcal \ mol^{-1}$).
$HCl$ is a strong acid and $NaOH$ is a strong base.
Therefore,the neutralization of $HCl$ and $NaOH$ releases $57.3 \ kJ \ mol^{-1}$ of energy.
$HCl + NaOH \rightarrow NaCl + H_2O$.

(B) An endothermic reaction is defined as a process that absorbs heat from the surroundings.
For any endothermic reaction,the enthalpy of the products is greater than the enthalpy of the reactants.
Therefore,the enthalpy change $\Delta H = H_{products} - H_{reactants}$ is positive $(+ve)$.

(B) The heat of neutralisation of a strong acid and a strong base is always $13.7 \, kcal \, mol^{-1}$.
Here,$NaOH$ is a strong base and $CH_3COOH$ is a weak acid.
Some energy is consumed in the dissociation of the weak acid $(CH_3COOH)$,which is an endothermic process.
Therefore,the net heat released during the neutralisation of a weak acid and a strong base is always less than $13.7 \, kcal \, mol^{-1}$.

(B) The target reaction for the combustion of carbon monoxide is: $CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}$
Using the formula for the heat of reaction: $\Delta H_{comb} = \Delta H_f^o(CO_2) - [\Delta H_f^o(CO) + \frac{1}{2}\Delta H_f^o(O_2)]$
Since the heat of formation of an element in its standard state is zero,$\Delta H_f^o(O_2) = 0$.
Substituting the given values: $\Delta H_{comb} = -94.0 \ kcal - (-26.4 \ kcal) = -67.6 \ kcal$.

(A) The combustion reactions are:
$(i) \ H_2(g) + 1/2 O_2(g) \to H_2O(l), \Delta H_1 = -241 \ kJ/mol$
$(ii) \ C_6H_{10}(l) + 17/2 O_2(g) \to 6CO_2(g) + 5H_2O(l), \Delta H_2 = -3800 \ kJ/mol$
$(iii) \ C_6H_{12}(l) + 9O_2(g) \to 6CO_2(g) + 6H_2O(l), \Delta H_3 = -3920 \ kJ/mol$
The hydrogenation reaction is:
$C_6H_{10}(l) + H_2(g) \to C_6H_{12}(l)$
This can be obtained by: $(i) + (ii) - (iii)$
$\Delta H_{hydrogenation} = \Delta H_1 + \Delta H_2 - \Delta H_3$
$\Delta H_{hydrogenation} = -241 + (-3800) - (-3920)$
$\Delta H_{hydrogenation} = -4041 + 3920 = -121 \ kJ/mol$.

(C) The heat of neutralisation for a strong acid and a strong base is always $13.7 \ kcal \ mol^{-1}$.
Since $NH_4OH$ is a weak base,some heat is consumed in the dissociation of the weak base into its ions.
Therefore,the net heat of neutralisation is less than $13.7 \ kcal \ mol^{-1}$.

(A) The combustion reaction for methane is: $CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}$.
The heat of combustion $(\Delta H_c)$ is calculated using the formula: $\Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$.
$\Delta H_c = [\Delta H_f(CO_{2(g)}) + 2 \times \Delta H_f(H_2O_{(l)})] - [\Delta H_f(CH_{4(g)}) + 2 \times \Delta H_f(O_{2(g)})]$.
Given $\Delta H_f(O_{2(g)}) = 0 \ kcal/mol$.
$\Delta H_c = [-94.0 + 2 \times (-68.4)] - [-17.9 + 2 \times 0]$.
$\Delta H_c = [-94.0 - 136.8] - [-17.9]$.
$\Delta H_c = -230.8 + 17.9 = -212.9 \ kcal$.

(D) The heat of reaction is a state function. According to $Hess's \ Law$,the total enthalpy change for a chemical reaction is the same,regardless of whether the reaction is completed in one step or several steps. Therefore,it does not depend upon the path or the method by which the final products are obtained from the reactants.

(B) The heat of neutralisation is the enthalpy change when $1 \ gram \ equivalent$ of an acid is neutralised by $1 \ gram \ equivalent$ of a base.
When both the acid and the base are strong,they completely dissociate into ions in an aqueous solution.
The reaction essentially involves the combination of $H^+$ ions from the acid and $OH^-$ ions from the base to form water: $H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$.
The enthalpy change for this reaction is constant at approximately $-57.1 \ kJ \ mol^{-1}$.
If either the acid or the base is weak,some energy is consumed in the dissociation of the weak electrolyte,resulting in a lower heat of neutralisation value.

(A) An endothermic reaction is one that absorbs heat from the surroundings.
Decomposition reactions,such as the thermal decomposition of calcium carbonate $(CaCO_3)$,require the input of heat energy to break chemical bonds.
Therefore,the reaction $CaCO_3 \to CaO + CO_2$ is endothermic.
Options $B$,$C$,and $D$ represent synthesis,neutralization,and combustion reactions,respectively,which are typically exothermic.

(D) The enthalpy change $(\Delta H)$ associated with a chemical reaction,where reactants are converted into products as per the stoichiometric coefficients,is defined as the Heat of reaction.
Therefore,for the given reaction,$\Delta H = 80 \ kJ$ represents the Heat of reaction.
Thus,the correct option is $(D)$.

(B) The heat of combustion is defined as the enthalpy change when $1 \ mole$ of a substance is completely burnt in the presence of oxygen. Since combustion is an exothermic process,the enthalpy change $(\Delta H)$ is always negative.

(B) The reaction $H_2 + \frac{1}{2} O_2 \to H_2 O$ represents the formation of $1 \ mol$ of water from its constituent elements in their standard states.
By definition,the enthalpy change associated with the formation of $1 \ mol$ of a compound from its constituent elements is known as the standard enthalpy of formation or heat of formation.
Therefore,the correct option is $B$.

(C) The given reaction is $2CO + O_2 \to 2CO_2$.
Dividing by $2$,we get $CO + \frac{1}{2}O_2 \to CO_2$.
Since $1 \ mole$ of $CO$ is completely oxidized in the presence of oxygen,the heat change associated with this reaction is specifically known as the heat of combustion of $CO$.

(D) The combustion reactions are given as follows:
$S_{\text{(rhombic)}} + O_2 \to SO_2$,$\Delta H_1 = -70960 \ cal$ ... $(I)$
$S_{\text{(monoclinic)}} + O_2 \to SO_2$,$\Delta H_2 = -71030 \ cal$ ... $(II)$
To find the heat of conversion for $S_{\text{(rhombic)}} \to S_{\text{(monoclinic)}}$,we subtract equation $(II)$ from equation $(I)$:
$S_{\text{(rhombic)}} - S_{\text{(monoclinic)}} = -70960 - (-71030)$
$S_{\text{(rhombic)}} \to S_{\text{(monoclinic)}}$,$\Delta H = -70960 + 71030 = +70 \ cal$
Thus,the heat of conversion is $+70 \ cal$.

(D) The process of vaporization is $H_2O_{(l)} \rightarrow H_2O_{(g)}$,which is an endothermic process.
The enthalpy of vaporization $\Delta H_{vap}$ is approximately $+9.7 \ kcal/mol$.
The heat of formation of $H_2O_{(g)}$ is given by $\Delta H_f(H_2O_{(g)}) = \Delta H_f(H_2O_{(l)}) + \Delta H_{vap}$.
Therefore,$\Delta H_f(H_2O_{(g)}) = -68.0 + 9.7 = -58.3 \ kcal$.
Thus,the correct option is $D$.

(B) The calorific value is defined as the amount of heat energy produced by the complete combustion of $1 \ kg$ of a fuel.
Among the given options,the approximate calorific values are:
$1$. Charcoal: $33,000 \ kJ/kg$
$2$. Kerosene: $45,000 \ kJ/kg$
$3$. Wood: $17,000 \ kJ/kg$
$4$. Dung: $6,000 - 8,000 \ kJ/kg$
Therefore,kerosene has the highest calorific value among the given options.

(B) The combustion reaction is: $C_{(s)} + O_{2(g)} \to CO_{2(g)}$
For this reaction,the change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 1 = 0$.
Using the relation $\Delta H = \Delta E + \Delta n_g RT$,since $\Delta n_g = 0$,we have $\Delta H = \Delta E$.
Given $\Delta H = -94 \ kcal$,therefore $\Delta E = -94 \ kcal$.
Since $\Delta E = E_{CO_2} - (E_C + E_{O_2})$ and the internal energy of elements in their standard state is taken as zero $(E_C = 0, E_{O_2} = 0)$,we get $E_{CO_2} = \Delta E = -94 \ kcal$.

(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 $0.2 \, mole$ of $KOH$ is the limiting reagent,it will react with $0.2 \, mole$ of $HNO_3$ to produce $0.2 \, mole$ of $H_2O$.
The heat released is calculated as: $\text{Heat} = \Delta H \times \text{moles of water formed} = 57.0 \, kJ \, mol^{-1} \times 0.2 \, mol = 11.4 \, kJ$.

(B) The enthalpy of a compound is defined as the heat of formation of that compound from its constituent elements in their standard states. Therefore,the enthalpy of a compound is equal to its $Heat \ of \ formation$.

(D) The heat of neutralization is defined as the amount of heat released when $1 \ gram$ equivalent of an acid is neutralized by $1 \ gram$ equivalent of a base in a dilute solution at a stated temperature.

(B) The heat of formation $(\Delta H_f)$ represents the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
If $\Delta H_f < 0$ (negative),the formation process releases energy,which characterizes an exothermic compound.
Therefore,the correct option is $(B)$.

(B) The heat of neutralization for the reaction between a strong acid and a strong base is constant at $13.7 \ kcal$ per mole of $H_2O$ formed.
$H_2SO_4$ is a strong diprotic acid,meaning it provides $2 \ moles$ of $H^{+}$ ions per mole of $H_2SO_4$.
For complete neutralization of $1 \ mole$ of $H_2SO_4$ by a strong base,$2 \ moles$ of $H_2O$ are produced.
Therefore,the total heat of neutralization $= 2 \times 13.7 \ kcal = 27.4 \ kcal$.

(B) The heat of neutralization for a strong acid and a strong base is constant at $-57.1 \ kJ \ mol^{-1}$.
When a weak acid or a weak base is involved,some energy is consumed in the dissociation of the weak electrolyte.
Since both $CH_3COOH$ (weak acid) and $NH_4OH$ (weak base) are weak electrolytes,the net heat of neutralization is the lowest because a significant amount of energy is required for their ionization.
Therefore,the correct option is $(B)$.

(C) The heat of neutralization is defined as the heat released when $1 \ mol$ of $H^+$ ions reacts with $1 \ mol$ of $OH^-$ ions to form $1 \ mol$ of $H_2O$.
Given,$\Delta H_{neut} = -57.1 \ kJ \ mol^{-1}$ for $1 \ mol$ of reaction.
For $0.25 \ mol$ of $NaOH$ reacting with $0.25 \ mol$ of $HCl$,the amount of heat released is calculated as:
$\text{Heat released} = \Delta H_{neut} \times \text{moles of reaction}$
$\text{Heat released} = 57.1 \ kJ \ mol^{-1} \times 0.25 \ mol = 14.275 \ kJ \approx 14.3 \ kJ$.