1.

If the activation energy for the forward reaction is 150 kJ mol^(-1) and that of the reverse reaction is 260 kJ mol^(-1) , what is the enthalpy change for the reaction

Answer»

410 kJ `mol^(-1)`
`-110kJ mol^(-1)`
110 kJ `mol^(-1)`
`-410 kJ mol^(-1)`

Solution :For a reversible reaction ,
`DELTA H = E_(a)` (forward) `- E_(a)` (backward)
`DeltaH = 150 - 260 = -110 kJ mol^(-1)`.


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