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For a given reaction, `DeltaH = 35.5kJ mol^(-1)` and `DeltaS = 83.6 JK^(-1) mol^(-1)` .The reaction s spontaneous at `:` ( Assume at `DeltaH ` and `DeltaS` do not vary with temperature )A. `T lt425 K`B. `T gt 425 K`C. All temperaturesD. `T gt 298 K` |
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Answer» Correct Answer - B `DeltaS= DeltaH - T DeltaS` For the reaction to be spontaneous , `DeltaG = - ve` As`DeltaH` and`DeltaS` both are positive,`DeltaG` can be `-ve` only if`T DeltaS gt DeltaH` or`T gt ( DeltaH)/( DeltaS)` `i.e,T gt ( 35.5xx 1000J)/( 83.6JK^(-1)) ` or `T gt 425 K` |
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