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Given the bond energies `N-=N`, H-H and N-H bonds are 945,436 and 391 kJ `"mole"^(-1)` respectively, the enthalpy of the following reaction `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` isA. `-93` kJB. 102 kJC. 90 kJD. 105 kJ |
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Answer» Correct Answer - A `underset("Energy absorbed")ubrace(underset(945)(N-=)Nunderset(+3xx436)(+3H-H))rarrunderset("Energy released")undersetubrace(2xx(3xx391)=2346)(2underset(H)underset(|)overset(H)overset(|)N-H)` Net. Energy released = 2346-2253=93 kJ i.e. `DeltaH=-93 kJ`. |
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