Question

The standard enthalpies of formation of $ A\left( {N{H_3}} \right),\;B\left( {C{O_2}} \right),\;C\left( {HI} \right) $ and $ D\left( {S{O_2}} \right) $ are respectively -46.19, -393.4, +24.94 and -296.9 $ kJmo{l^{ - 1}} $. The increasing order of their stability is:
(a) $ B < D < A < C $
(b) $ C < A < D < B $
(c) $ D < B < C < A $
(d) $ A < C < D < B $

Answer 1

Hint: We know that when a reaction happens, there might be some energy changes. It might be absorbing energy or else evolution energy. That means there is always a change in enthalpy whenever a reaction takes place. This change is known as the enthalpy change.

Complete Step by step answer:
We have given four standard enthalpies of formation and have to find the increasing order of its stability. To answer this question first we will have to understand what is enthalpy of formation.
Enthalpy of formation can be defined as the heat change (it can be either evolved or absorbed) accompanying the formation of 1 mole of a substance from its elements under given condition of temperature and pressure. It is denoted by $ \vartriangle {H_f} $.

This is the definition of enthalpy of formation, but when we keep the values of temperature and pressure to its standard values we get standard enthalpy of formation. Standard enthalpy of formation is the enthalpy of formation under standard conditions of temperature and pressure i.e., 298K and 1 atmospheric pressure. It is represented by $ \vartriangle H_f^\circ $.

By stability means when a compound releases a maximum amount of energy it attains more stability, hence we can say that higher the negative value of enthalpy of formation, greater is the stability of the compound. The negative value represents the energy released.
Therefore, coming to question the highest negative value is for $ C{O_2} $ then $ S{O_2} $. $ HI $ has energy absorbed hence it is least stable.
Hence the order for stability is $ C{O_2} > S{O_2} > N{H_3} > HI $. i.e., $ C < A < D < B $.

Therefore, option (b) is correct.

Note: Standard enthalpy of formation of all elements and pure elemental compounds is taken as zero. In case of allotropes, the enthalpy of formation of the most stable allotrope is taken as zero. For example enthalpy of formation of graphite is zero, while that of diamond is not zero.