From the given table answer the following question:
`{:(,CO(g),CO_(2)(g),H_(2)O(g),H_(2)(g)),(DeltaH_(298)^(@)(-“KCal”//”mole”),-26.42,-94.05,-57.8,0),(DeltaH_(298)^(@)(-“KCal”//”mole”),-32.79,-94.24,-54.64,0),(S_(298)^(@)(-“Cal”//”k mol”),47.3,51.1,?,31.2):}`
Reaction:`H_(2)O(g) + CO(g) hArr H_(2)(g)+CO_(2)(g)`
Calculate `S_(298)^(@) [H_(2)O(g)]`
`{:(,CO(g),CO_(2)(g),H_(2)O(g),H_(2)(g)),(DeltaH_(298)^(@)(-“KCal”//”mole”),-26.42,-94.05,-57.8,0),(DeltaH_(298)^(@)(-“KCal”//”mole”),-32.79,-94.24,-54.64,0),(S_(298)^(@)(-“Cal”//”k mol”),47.3,51.1,?,31.2):}`
Reaction:`H_(2)O(g) + CO(g) hArr H_(2)(g)+CO_(2)(g)`
Calculate `S_(298)^(@) [H_(2)O(g)]`
Correct Answer – (i) -9.83 kCal; (ii) -6.81 kCal, (iii) -10.13 Cal/K, (iv) -9.83 kCal, (v) +45.13 Cal/K
(i) `DeltaH=0-94.05-[-57.8-26.42]=-9.83` kCal
(ii) `DeltaG=0+(-94.24) -[-54.64-32.79]=-6.81` kCal
(iii) `DeltaS=(DeltaH-DeltaG)/(T)=-10.13` Cal
(iv) `DeltaH=DeltaU+Deltan_(g)RT” “rArr Deltan_(g)=0`
`DeltaU=-9.83` kCal
(v) `-10.13=31.2+51.1-[x+47.3]`
`x=45.13″ Cal/K”`