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If the enthalpy change for the transitio...

If the enthalpy change for the transition of liquid water to steam is `30 kJ mol^(-1)" at " 27^@C` the entropy change for the process would be

A

`1.0 J "mol"^(-1) K^(-1)`

B

`0.1 J "mol"^(-1) K^(-1)`

C

`100 J "mol"^(-1) K^(-1)`

D

`10 J "mol"^(-1) K^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaG^(@) = DeltaH^(@)-TDeltaS^(@)`
Given, `DeltaH_("vap")=30 KJ "mol"^(-1)`
T=27 + 273 =300 K
`DeltaG^(@)=0` at equilibrium
`DeltaS_("vap")=(DeltaH_("vap"))/(T)=(30xx10^(3) J "mol"^(-1))/(300 K)`
`=100 J "mol"^(-1) K^(-1)`
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