1.

A certain reaction is at equilibrium at 82^(@)C and the enthalpy change for the reaction is 21.3 kJ. The value of Delta S (in JK mol^(-1) for the reaction is

Answer»

`55.0`
`60.0`
`68.5`
`120.0`

Solution :At equilibrium, `Delta G = 0`
`Delta G = Delta H - T Delta S = 0`
:. `Delta H = T Delta S`
T = 82 + 273 = 355 K
`Delta S = (Delta H)/(T) = (21.3 XX 1000)/(355K) J mol^(-1)`
`= 60 JK^(-1) mol^(-1)`


Discussion

No Comment Found

Related InterviewSolutions