Respuesta :

ΔH° and ΔS° for the reaction at 298 K is 90.7kJ and 220.7J/K

For the reaction

CH₃OH(g) ⇄ CO(g) + 2H₂(g)

At 298K,

ΔH° = ∑mΔH° (products) -  ∑mΔH° (reactants)

= [1 mol *ΔH° CO(g)+ 2 mol *ΔH°H2(g) ]-[1 mol *ΔH°CH3OH(g)]

= [1 mol * (-110.5kJ)/mol +2 mol*0 kJ/mol] - [ 1 mol * -201.2kJ/mol]

= 90.7kJ

At 298 K,

ΔS° = ∑mS° (products) -  ∑mS° (reactants)

= [1 mol *ΔS° CO(g)+ 2 mol *ΔS°H2(g) ]-[1 mol *ΔS°CH3OH(g)]

= [1 mol * (-197.5kJ)/mol +2 mol*130.6 kJ/mol] - [ 1 mol * -238kJ/mol]

=220.7J/K

Learn more about ΔS here:

https://brainly.com/question/17087340

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