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Why is adsorption always exothermic ? |
Answer» Correct Answer - Adsorption is accompanied by decrease of randomness, i.e. this factor opposes the process, i.e. `DeltaS` is -ve. For the process to be spontaneous, `DeltaG` must be -ve. Hence, according to eqn, `DeltaG=DeltaH-TDeltaS, DeltaG` can be -ve only if `DeltaH` is -ve. (a) `3I^(-)+S_(2)O_(8)^(2-) rarr I_(3)^(-)+2SO_(4)^(2-)` `(-d[S_(2)O_(8)^(2-)])/(dt)=1.5xx10^(-3)" "(-1)/(3)(Delta[I^(-)])/(Deltat)=(-Delta[S_(2)O_(8)^(2-)])/(Deltat)` `-1/3(Delta[I^(-)])/(Deltat)=1.5xx10^(-3)" "-(Delta[I^(-)])/(Deltat)=4.5xx10^(-3)` (b) `-(d[S_(2)O_(8)^(2-)])/(dt)=1/2(d[SO_(4)^(2-)])/(dt)=1.5xx10^(-3)=1/2(d[SO_(4)^(2-)])/(dt)` `(d[SO_(4)^(2-)])/(dt)=3xx10^(-3)` |
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