Question

Consider the following reactions:
(i) $ {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) = {H_2}O\left( l \right);\Delta H = - {X_1}KJmo{l^{ - 1}} $
(ii) $ {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) $ C{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) $ {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) $ + {X_1}KJmo{l^{ - 1}} $
(B) $ - {X_2}KJmo{l^{ - 1}} $
(C) $ + {X_3}KJmo{l^{ - 1}} $
(D) $ + {X_4}KJmo{l^{ - 1}} $

Answer 1

Hint: 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^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) = {H_2}O\left( l \right) $
In the first, the water is formed from the $ {H^ + } $ ion and $ 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 $ {H_2}O $ .
 $ {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 $ {O_2} $ and $ {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 $ {H_2}O $ .
 $ 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 $ {H_2} $ and $ C{O_2} $ . These reactants are not only the elements which are required in the formation of $ {H_2}O $ . Hence, this reaction cannot be used for finding the enthalpy of formation of $ {H_2}O $ .
 $ {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 $ {C_2}{H_2} $ and $ {O_2} $ . These reactants are not only the elements which are required in the formation of $ {H_2}O $ . Hence, this reaction cannot be used for finding the enthalpy of formation of $ {H_2}O $ .
Hence, from the above discussion only from the second reaction is useful for making $ {H_2}O $ . So, the enthalpy of formation of that reaction will be $ - {X_2}KJmo{l^{ - 1}} $ .
Hence, the correct option is (B) $ - {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.