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Calculate the enthalpy change during the...

Calculate the enthalpy change during the reaction `:`
`H_(2(g))+Br_(2(g))rarr 2HBr_((g))`
Given, `e_(H-H)=435kJ mol^(-1),e_(Br-Br)=192kJ mol^(-1)` and `e_(H-Br)=368kJ mol^(-1).`

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`-109 kJ" "mol^(-1)`
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Calculate Delta G (in kJ) for the reaction at 300 K, H_(2)(g)+Cl_(2)(g) rarr 2HCl (g) Given at 300 K, BE_(H-H) = 435 kJ mol^(-1), BE_(Cl-Cl) = 240 kJ mol^(-1), BE_(HCl) = 430 kJ mol^(-1) Entropies of H_(2), Cl_(2) and HCl are 131, 223 and 187 JK^(-1) mol^(-1) respectively.

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.

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.

Calculate the entropy of Br_(2)(g) in the reaction H_(2)(g)+Br_(2)(g)rarr2HBr(g), Delta S^(@)=20.1 JK^(-1) given, entropy of H_(2) and HBr is 130.6 and 198.5 J mol^(-1)K^(-1) :-

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