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H(2)O(g)toH(2)O(l),DeltaH=-40.4kJ*mol^(-...

`H_(2)O(g)toH_(2)O(l),DeltaH=-40.4kJ*mol^(-1) and DeltaS= - 108.3J*K^(-1)*mol^(-1)`. At which temperature the process will be spontaneous at constant 1 atm?

Text Solution

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In this process `DeltaH lt 0 and DeltaS lt 0`. For such type of process, the temperature t which the reaction will occur spontaneously, `Tlt(DeltaH)/(DeltaS)`.
given: `DeltaH=-40.4xx10^(3)J*mol^(-1),DeltaS=-108.3J*K^(-1)*mol^(-1)`
`therefore T lt (DeltaH)/(DeltaS) or, T lt (40.4xx10^(3))/(108.3) or, T lt 373K`
`therefore`The process will be spontaneous below 373K `(100^(@)C)`.
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