Q5.3-62E - Chemistry (OpenStax)

Question

Which produces more heat?

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

or

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

for the phase change \({\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{s}} \right) \to {\bf{Os}}{{\bf{O}}_{\bf{4}}}\left( {\bf{g}} \right){\bf{                 \Delta H =  56}}{\bf{.4 kJ}}\)

Step-by-Step Solution

Answer

Enthalpy of formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( {\rm{g}} \right)\)is more than the enthalpy of formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( s \right)\). It means that the formation of \({\rm{Os}}{{\rm{O}}_{\rm{4}}}\left( {\rm{g}} \right)\)will produce more heat. 

Step 1: Enthalpy change for reaction 1

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:

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

ΔH_reaction = ΔH⁰_{formation(OsO4(s))} - [(2 x ΔH⁰_{formation(O2(g))}+ΔH⁰_{formation(Os(s))})]

ΔH_reaction = ΔH⁰_{formation(OsO4(s))................}(1)

Step 2: Enthalpy change for reaction 2

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:

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

ΔH_reaction = ΔH⁰_{formation(OsO4(s))} - [(2 x ΔH⁰_{formation(O2(g))}+ΔH⁰_{formation(Os(s))})]

ΔH_reaction = ΔH⁰_{formation(OsO4(s))................}(2)

Step 3: Comparison

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)      ΔH=56.4kJ

ΔH⁰_reaction=56.4kJ =ΔH⁰_OSO4 -ΔH⁰_OSO4 

ΔH⁰_OSO4=56.4kJ+ΔH⁰_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.