It is possible to calculate the value of `DeltaH` for the reaction
`H_(2)(g) +Br_(2)(l) rarr 2HBr(g)` from the bond enthalpy data alone.

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules.What will be the enthalpy change for the following reaction. H_(2)(g) + Br_(2)(g) to 2HBr(g) Given that bond energy of H_(2), Br_(2) and HBr is 435 kJ mol^(-1) , 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively.

At standard conditions, if the change in the enthalpy for the following reaction is -109KJ mol^(-1) H_2(g) + Br_2(g) rarr 2HBr(g) Given that bond energy of H_2 and Br_2 is 435kj mol^(-1) and 192kj mol^(-1) , respectively. What is the bond energy of HBr ?

Using bond enthalpies (symbolized by epsilon ) an estimated value of DeltaH^(@) for the reaction H_(2)C=CH_(2)(g)+H_(2)(g)rarrCH_(3)-CH_(3)(g) would be

Use the bond enthalpies listed below to estimate the enthalpy change for the reaction H_(2)(g)+Br_(2)(g)rarr2HBr(g) Given: BE of H_(2), Br_(2) , and HBr is 435, 192 , and 368 kJ mol^(-1) , respectively.