1.

For the reactions A rarr B, DeltaH = + 24 kJ//mol and B rarr C, DeltaH = - 18kJ//mol, the decreasing order of enthalpy of A, B, C follows the order

Answer»

A, B, C
B, C, A
C, B, A
C, A, B

Solution :`ArarrB, DeltaH=+24 kJ//mol`
`implies H_(B)-H_(A)=+24 "...(i)"`
`BrarrC, DeltaH=-18 kJ//mol`
`implies H_(C)-H_(B)=-18`
`implies H_(B)-H_(C)=+18"...(II)"`
From EQS.(i) and (ii), we have
`H_(C)-H_(A)=6`
`thereforeH_(B)gtH_(C)gtH_(A)`.


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