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Bond enthalpy of bromine is 194 kJ mol^(...

Bond enthalpy of bromine is `194 kJ mol^(-)`. If enthalpy of vapourisation of `Br_2` is `+30 kJ mol^(-)`, electron gain enthalpy of `Br` is `-325 kJ mol^(-1)` and hydration enthalpy of bromide is `-339 kJ mol^(-1)` calculate the change in enthalpy for the reaction, `1/2Br_(2)(l) + e^(-) overset(aq)(rarr) Br^(-)(aq)`. 

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To calculate the change in enthalpy for the reaction: \[ \frac{1}{2} \text{Br}_2(l) + e^-(aq) \rightarrow \text{Br}^-(aq) \] we will use the provided data on bond enthalpy, enthalpy of vaporization, electron gain enthalpy, and hydration enthalpy of bromine. ...
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