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,

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

$\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]$

$\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]$

where,

n = number of moles

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

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

$Delta H=-105kJ$

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