Question

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ/mole) the standard enthalpy change DH° for the reaction written below, using the bond energies given. Just enter a number (no units).
N2(g) + 3H2(g) -> 2NH3(g)Bond: N≡N H-H N-HBond energy: 945 432 391

Answer 1

Answer: 104 KJ mol

Each NH3 has 3 N-H bonds

Explanation:

Enthalpy change(DHo)= sum t enthalpy of products - sum of enthalpy of Reactants

DHo= 2x3 dHo (N-H) - {dHo(N2)+3dHo(H-H)}

= 6(391)-(945+3(432))

=2346-(945+1296)

=2346-2241

DHo= 105 KJ/mole

Answer 2

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