91Ó°ÊÓ

To comment on the question, we have to calculate the enthalpy change for each chemical reaction separately.

Reaction 1:

\({\bf{Os}}\left( {\bf{s}} \right) \to {\bf{2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right)\)

Chemical reaction:

\({\bf{Os}}\left( {\bf{s}} \right) \to {\bf{2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right)\)

Δ±á_reaction= Δ±á⁰_{formation(OsO4(s))}- [(2 x Δ±á⁰_{formation(O2(g))}+Δ±á⁰_{formation(Os(s))})]

Δ±á_reaction= Δ±á⁰_{formation(OsO4(s))................}(1)

Now, we will calculate the enthalpy change for the next reaction separately.

Reaction 2:

\({\bf{Os}}\left( {\bf{s}} \right) \to {\bf{2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right)\)

Chemical reaction:

\({\bf{Os}}\left( {\bf{s}} \right) \to {\bf{2}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right)\)

Δ±á_reaction= Δ±á⁰_{formation(OsO4(s))}- [(2 x Δ±á⁰_{formation(O2(g))}+Δ±á⁰_{formation(Os(s))})]

Δ±á_reaction= Δ±á⁰_{formation(OsO4(s))................}(2)

Now, we have to compare the change in enthalpy of both reactions.

For doing so, we are going to use the third equation given there.

O_SO₄(s)→ O_SO₄(g) Δ±á=56.4kJ

Δ±á⁰_reaction=56.4kJ =Δ±á⁰_OSO4-Δ±á⁰_OSO4

Δ±á⁰_OSO4=56.4kJ+Δ±á⁰_OSO4

Hence, we can easily conclude that the enthalpy of formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( {\rm{g}} \right)\)is more than the enthalpy of formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( s \right)\).This means that the formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( {\rm{g}} \right)\)will produce more heat.