Question -
Given the bond energies N≡N, H−H and N−H bonds are 945,436 and 391kJmole−1 respectively, the enthalpy of the following reaction N2(g)+3H2(g)→2NH3(g) is
Options:
A .  -93 kJ
B .  102 kJ
C .  90 kJ
D .  105 kJ
Answer: Option A : A N≡N+3H−H→2N|H|H−H 945+3x4362×(3×391)=2346 EnergyabsorbedEnergyreleased Net. Energy released = 2346−2253=93kJ i.e., ΔH=−93kJ
Submit Your Solution hear: