Question

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3 = 192.5 J/mol K H2 = 130.6 J/mol K N2 = 191.5 J/mol K

Answer 1

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction
`(\Delta S^o)`.

`\Delta S^o=S_(product)-S_(reactant)`

`\Delta S^o=[n_(NH_3)* \Delta S^0_((NH_3))]-[n_(N_2)* \Delta S^0_((N_2))+n_(H_2)* \Delta S^0_((H_2))]`

where,

`\Delta S^o` = entropy of reaction = ?

n = number of moles

`\Delta S^0{(NH_3)}` = standard entropy of
`NH_3`

`\Delta S^0{(H_2)}` = standard entropy of
`H_2`

`\Delta S^0{(N_2)}` = standard entropy of
`N_2`

Now put all the given values in this expression, we get:

`\Delta S^o=[2mole* (192.5J/K.mole)]-[1mole* (191.5J/K.mole)+3mole* (130.6J/K.mole)]`

`\Delta S^o=-198.3J/K`

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K