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)` :-

Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) =- 206.8 J K^(-1) mol^(-1) .

Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) =- 206.8 J K^(-1) mol^(-1) .

Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) = 206.8 J K^(-1) mol^(-1) .

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) rarr 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.

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 372 kJ mol^(-1) , respectively.