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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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Answer» Solution :`{:((1)/(2)Br_(2)(l)rarr(1)/(2)Br_(2)(g),,Delta=+15kJmol^(-1)),((1)/(2)Br_(2)(g)rarrBr(g),,DeltaH=+97kJmol^(-1)),(Br(g)rarrBr^(-)(g),,DeltaH=-325kJmol^(-1)),(Br^(-)(g)+AQ rarr Br^(-)(aq),,DeltaH=-339kJ MOL^(-1)):}` Adding these EQUATIONS we get, `(1)/(2)Br_(2)(l)+e^(-)overset(aq)rarrBr^(-)(aq)),DeltaH=-552kJ mol^(-1)` |
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