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Question: Consider the following reactions: (i) \( {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) ...

Consider the following reactions:
(i) H+(aq)+OH−(aq)=H2O(l);ΔH=−X1KJmol−1{H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) = {H_2}O\left( l \right);\Delta H = - {X_1}KJmo{l^{ - 1}}
(ii) H2(g)+12O2(g)=H2O(l);ΔH=−X2KJmol−1{H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right) = {H_2}O\left( l \right);\Delta H = - {X_2}KJmo{l^{ - 1}}
(iii) CO2(g)+H2(g)=CO(g)+H2O(l);ΔH=+X3KJmol−1C{O_2}\left( g \right) + {H_2}\left( g \right) = CO\left( g \right) + {H_2}O\left( l \right);\Delta H = + {X_3}KJmo{l^{ - 1}}
(iv) C2H2(g)+52O2(g)=2CO(g)+H2O(l);ΔH=+X4KJmol−1{C_2}{H_2}\left( g \right) + \dfrac{5}{2}{O_2}\left( g \right) = 2CO(g) + {H_2}O(l);\Delta H = + {X_4}KJmo{l^{ - 1}}
Enthalpy of formation of is:
(A) +X1KJmol−1+ {X_1}KJmo{l^{ - 1}}
(B) −X2KJmol−1- {X_2}KJmo{l^{ - 1}}
(C) +X3KJmol−1+ {X_3}KJmo{l^{ - 1}}
(D) +X4KJmol−1+ {X_4}KJmo{l^{ - 1}}

Explanation

Solution

First we have to know, what is the enthalpy of formation, how the given data are useful for finding the solution and which reaction of the formation of the water in the liquid phase is suitable by the definition of enthalpy of formation.

Complete step-by-step answer
We start by defining what is enthalpy of formation:
Enthalpy of formation: The enthalpy of formation of any compound is the change in the enthalpy during the formation of the compound from its elements in standard conditions and elements must exist in nature.
Now, we have to choose which reaction is suitable:
H+(aq)+OH−(aq)=H2O(l){H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) = {H_2}O\left( l \right)
In the first, the water is formed from the H+{H^ + } ion and OH−O{H^ - } ion. Hence, the reactants do not exist in nature in this state. Hence, the reaction cannot be used for finding the enthalpy of formation of H2O{H_2}O .
H2(g)+12O2(g)=H2O(l){H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right) = {H_2}O\left( l \right)
Similarly, in the second, the water is formed from O2{O_2} and H2{H_2} . These reactants are present in nature and also they are in the standard form. Hence, this reaction can be used for finding the enthalpy of formation of H2O{H_2}O .
CO2(g)+H2(g)=CO(g)+H2O(l)C{O_2}\left( g \right) + {H_2}\left( g \right) = CO\left( g \right) + {H_2}O\left( l \right)
Similarly, in the third, the water is formed from H2{H_2} and CO2C{O_2} . These reactants are not only the elements which are required in the formation of H2O{H_2}O . Hence, this reaction cannot be used for finding the enthalpy of formation of H2O{H_2}O .
C2H2(g)+52O2(g)=2CO(g)+H2O(l){C_2}{H_2}\left( g \right) + \dfrac{5}{2}{O_2}\left( g \right) = 2CO(g) + {H_2}O(l)
Similarly, in the last, the water is formed from C2H2{C_2}{H_2} and O2{O_2} . These reactants are not only the elements which are required in the formation of H2O{H_2}O . Hence, this reaction cannot be used for finding the enthalpy of formation of H2O{H_2}O .
Hence, from the above discussion only from the second reaction is useful for making H2O{H_2}O . So, the enthalpy of formation of that reaction will be −X2KJmol−1- {X_2}KJmo{l^{ - 1}} .
Hence, the correct option is (B) −X2KJmol−1- {X_2}KJmo{l^{ - 1}} .

Note
The positive and negative sign in the enthalpy of formation has different significance. The negative sign shows that the heat is released from the reaction to the surroundings, that is the reaction is an exothermic reaction and the positive sign shows that the heat is absorbed from the surroundings, that is the reaction is endothermic.