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At 1 atm and 298K DeltaH^(0) value of th...

At 1 atm and 298K `DeltaH^(0)` value of the reaction `2H_(2)(g)+O_(2)(g)to2H_(2)O(l)` is -572 kJ. Calculate the change in entropy of the system and surroundings for this reaction. Is this reaction spontaneous at that temperature and pressure? Given: Standard molar entropies of `H_(2)(g),O_(2)(g)&H_(2)O(l)` at 298K are 130.6, 205.0 and 69.90`J*K^(-1)*mol^(-1)` respectively.

Text Solution

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The change in entropy of the given reaction
`DeltaS_(sys)^(0)=[2S^(0)(H_(2)O,l)]-[2xxS^(0)(H_(2),g)+S^(0)(O_(2),g)]`
`=[2xx69.9-(2xx130.6+205)]J*K^(-1)`
`=-326.4J*K^(-1)`
Change in entropy of the surroundings in the reaction,
`DeltaS_("surr")=-(DeltaH^(0))/(298)=(572xx10^(3))/(298)J*K^(-1)=+1919.4J*K^(-1)`
`therefore`Total change in entropy,
`DeltaS_("univ")=DeltaS_("sys")+DeltaS_("surr")=-326.4+1919.4`
`=+1593J*K^(-1)`
Since `DeltaS_("univ")gt0`, then the reaction will occur spontaneously at 298K and 1 atm.
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