Nitrogen and hydrogen combine to form ammonia in the Haber process.

Calculate (in kJ) the standard enthalpy change image from custom entry toolH for the reaction written below, using the bond energies given.

N2(g) + 3H2(g) ---> 2NH3 (g)

Bond energy (kJ/mol): Nimage from custom entry toolN 945; H - ---- H 432; N ----- H 391

Respuesta :

Answer:

-105 kJ

Explanation:

The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.

The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:

Reactants: 945 + 3*432 = 2241 kJ

Products: 6*(-391) = -2346 kJ

ΔH = 2241 - 2346

ΔH = -105 kJ