Question

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ΔH° for the reaction written below, using the bond energies given. N2(g) + 3H2(g) → 2NH3(g) Bond: N≡N H–H N–H Bond energy (kJ/mol): 945 432 391

Answer 1

Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* B.E(reactant)]-\sum [n* B.E(product)]`

`\Delta H=[(n_(N_2)* B.E_(N_2))+(n_(H_2)* B.E_(H_2)) ]-[(n_(NH_3)* B.E_(NH_3))]`

`\Delta H=[(n_(N_2)* B.E_(N\equiv N))+(n_(H_2)* B.E_(H-H)) ]-[(n_(NH_3)* 3* B.E_(N-H))]`

where,

n = number of moles

Now put all the given values in this expression, we get

`\Delta H=[(1* 945)+(3* 432)]-[(2* 3* 391)]`

`Delta H=-105kJ`

Therefore, the enthalpy change for this reaction is, -105 kJ