Calculate the change in entropy for the following reaction
`2CO(g) +O_(2)(g) rarr 2CO_(2)(g)`
Given:
`S_(CO)^(Theta)(g)=197.6 J K^(-1)mol^(-1)`
`S_(O_(2))^(Theta)(g)=205.03 J K^(-1)mol^(-1)`
`S_(CO_(2))^(Theta)(g)=213.6 J K^(-1)mol^(-1)`

Explanation

a. `2CO(g) +O_(2)(g) rarr 2CO_(2)(g)`
`DeltaS^(Theta) = sum S^(Theta) ("products") -sum S^(Theta) ("reactants")`
`=2S^(Theta) [CO_(2)(g)] -{2S^(Theta)[CO(g)+S^(Theta)[O_(2)(g)]}`
`= 2(213.6)-{2(197.6)+(205.03)}`
`= 427.2 - 600.23`
`= -173.03 J K^(-1)`
b. `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(l)`
`DeltaS^(Theta) = {2S^(Theta)[H_(2)O(l)]} -{2S^(Theta)[H_(2)(g)] +S^(Theta)[O_(2)(g)]}`
`= 2(69.96) - {2(130.6)+205.03}`
`= 189.92 - 466.23`
`=- 326.32JK^(-1)`