Ammonia is produced directly from nitrogen and hydrogen by using the Haber proce

Question

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)

Explanation / Answer

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).

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