Question

the bond enthalpy of h2 is 436 kj mol-1 and the n2 is 941.3 kl mol-1. calculate the average bond enthalpy of an N-H bond in ammonia ​

Answer 1

Answer:

390.3 kJmol−1

Explanation:

N2(g)+3H2(g)→2NH3(g);  

ΔH=−2 x 46 kJmol−1

ΔH=∑(BE)reactants−∑(BE)products

                  =(941.8+3×436)−(6x)=−2×46

(Here x= BE of N−H bonds)

x= 390.3 kJmol−1