Bond dissociation enthalpies of H2(g) and N2(g) are 436.0 kJ mol-1 and 941.8 kJ mol-1, respectively, and enthalpy of formation of NH3(g) is -46 kJ mol-1. What are the enthalpy of atomisation of NH3(g) and the average bond enthalpy of N — H bond respectively (in kJ mol-1) ?

Question

Bond dissociation enthalpies of H2(g) and N2(g) are 436.0 kJ mol-1 and 941.8 kJ mol-1, respectively, and enthalpy of formation of NH3(g) is -46 kJ mol-1. What are the enthalpy of atomisation of NH3(g) and the average bond enthalpy of N — H bond respectively (in kJ mol-1) ? (A) 1170.9, 390.3 (B) 117, 300 (C) 300, 200 (D) 2000, 1975

Tardigrade Admin · Accepted Answer

N2(g)+3H2(g) → 2NH3(g) ; Δ H = -2 × 46 kJ mol-1 Δ H = Sigma(BE)reactants - Sigma (BE)products = (941.8 + 3 × 436) - (6x) = - 2 × 46 (here x = BE of N — H bonds) x = 390.3 kJ mol-1 NH3 → N + 3(H) Heat of atomisation = 3 × 390.3 = 1170.9 kJ mol-1

$1170.9$ , $390.3$

$117$ , $300$

$300$ , $200$

$2000$ , $1975$

Q. Bond dissociation enthalpies of H2(g)​ and N2(g)​ are 436.0kJmol−1 and 941.8kJmol−1, respectively, and enthalpy of formation of NH3(g)​ is −46kJmol−1. What are the enthalpy of atomisation of NH3(g)​ and the average bond enthalpy of N—H bond respectively (in kJmol−1) ?

1170.9, 390.3

117, 300

300, 200

2000, 1975

N2(g)​+3H2(g)​→2NH3(g)​ ; ΔH=−2×46kJmol−1 ΔH=Σ(BE)reactants​−Σ(BE)products​ =(941.8+3×436)−(6x)=−2×46 (here x=BE of N—H bonds) x=390.3kJmol−1 NH3​→N+3(H) Heat of atomisation =3×390.3=1170.9kJmol−1

$1170.9$ , $390.3$

$117$ , $300$

$300$ , $200$

$2000$ , $1975$