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The heat of formation of `NH_(3)(g)` is `-46 " kJ mol"^(-1)`. The `DeltaH` (in `" kJ mol"^(-1)`) of the reaction, `2NH_(3)(g)rarrN_(2)(g)+3H_(2)(g)` isA. 46B. `-46`C. 92D. `-92` |
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Answer» Correct Answer - C `(1)/(2) N_(2)(g) + (3)/(2) H_(2) (g) to NH_(3) (g)` `Delta H_(r) = - 46` kJ/mol ` 2NH_(3) (g) to N_(2) (g) + 3H_(2)(g)` `DeltaH_(r) = -2 DeltaH_(r)` = `-2(-46)` = 92 kJ |
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