Question

Given:

N2 + 3H2 → 2NH3


Bond Bond Energy (kJ/mol)

N≡N 942

H–H 432

N–H 386


Use the bond energies to calculate the change in enthalpy for the reaction.


The enthalpy change for the reaction is ____ kilojoules. Express answer in 2 sig figs

Answer 1

Answer:-78 kilojoules

Step-by-step explanation:

Answer 2

ΔH = -78 kj/mol

Step-by-step explanation:

Given data:

Bond energies:

N≡N = 942 kJ/mol

H-H = 432 kj/mol

N-H = 386 kj/mol

Change in enthalpy of reaction = ?

Solution:

N₂ + 3H₂ → 2NH₃

942 + 3×432 = 2(386×3)

942+1296 = 2316

2238 = 2316

ΔH = Bond broken - bond formed

ΔH = 2238 kj/mol - 2316 kj/mol

ΔH = -78 kj/mol