1.

Calculate the enthalpy change for the reaction: H_(2) (g) + Br_(2) (g) rarr 2HBr(g) Given the bond enthalpies H_(2), Br_(2) and HBr are 435 kJ mol^(-1), 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively.

Answer»

SOLUTION :`-109 KJ MOL^(-1)`


Discussion

No Comment Found