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A certain reaction is at equilibrium at ...

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

A

`55.0`

B

`60.0`

C

`68.5`

D

`120.0`

Text Solution

Verified by Experts

The correct Answer is:
C
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)`
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