Question

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia given the following mean bond enthalpies (kJmol-1): N≡N 942; H-H 436; N-H 386

Answer 1

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!