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| 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)` | |