1.

Calculate the entropy change for the following reaction `H_(2)(g) +CI_(2)(g) rarr 2HCI (g) at 298 K` Given `S^(Theta)H_(2) = 131 J K^(-1) mol^(-1), S^(Theta)CI_(2) = 233 J K^(-1) mol^(-1)`, and `S^(Theta) HCI = 187 J K^(-1) mol^(-1)`

Answer» `DeltaS^(Theta) = sum S^(Theta)(Products) - sum S^(Theta) (Reactant)`
`= 2xx187 -(131 +223)`
`= 374 - 131 - 223 = 20 J K^(-1) mol^(-1)`


Discussion

No Comment Found

Related InterviewSolutions