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Given the bond energies `N=N` , `H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` isA. `93KJ`B. `102KJ`C. `90KJ`D. `105KJ` |
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Answer» Correct Answer - A `DeltaH=(945+3xx439)-(2xx3xx391)` `=2253-2346` `=-93KJ` |
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