In: Chemistry
Ammonia is produced directly from nitrogen and hydrogen by using
the Haber process. The chemical reaction is
N2(g)+3H2(g)?2NH3(g) Answer is in kJ / 2moles
NH3
Part A.Use bond enthalpies to estimate the enthalpy change for the reaction. Express your answer using two significant figures.
Part C. Calculate the true enthalpy change as obtained using ?H?f values.
Express your answer using two significant figures.
Part D. Compare the enthalpy change you calculate in (a) to the true enthalpy change as obtained using ?H?f values. (essay question)
Part A. We will use the following standard bond energy values
Average Bond energy (kJ/mol)
N(triple bond)N (N2) 941
H-H
436
N-H 391
Let us first calculate the enthalpy change for the reaction per mole of NH3
0.5N2 + 1.5H2 ---------> NH3
Thus,
summing up bond energies of reactants (Bonds broken)
0.5 x N-N (triple): 0.5 * 941
1.5 x H-H: 1.5 * 436
Summing up bond energies of products (Bonds made)
3x N-H: 3 * 391
Enthalpy change of reaction = sum of bond energies of reactants - sum of bond energies of products
= (470.5 + 654) - (1173)
= -48.5 kJ/mol
Thus, the enthalpy change based on bond energies per mole is -48.5 kJ/mol
Therefore, the enthalpy change based on bond energies per mole is --97.00 kJ/2 mol of NH3
Part C. We will use the following standard values
delta Hof (kJ/mol)
N2(g) 0.0
H2(g) 0.0
NH3(g) -45.90
Let us calculate the true enthalpy change for the reaction
true delta Hof(reaction) = delta Hof(products) - delta Hof(reactants)
= 2 x -45.90 - (0.0 + 0.0)
= -91.8 kJ/2 moles of NH3
Thus, the true enthalpy change for the reaction is -91.8 kJ/2mole of NH3
Part D. The standard enthalpy change as calculated in Part A) is -97.00 kJ/2mole of NH3, whereas, the true enthalpy change calculated in Part C) is -91.8 kJ/2moles of NH3.
Thus, the amount of heat released during the reaction is 5.2 kJ/mol higher in case of Part A) when compared to Part B).