Question

Given: N2 + 3H2 → 2NH3 Bond Bond Energy (kJ/mol) N≡N 942 H–H 432 N–H 386 Use the bond energies to calculate the change in enthalpy for the reaction. The enthalpy change for the reaction is kilojoules.

Answer 1

Answer: -78 KJ.

Explanation:

• The reaction is between gaseous molecules, so we can use the bond energy to calculate the change in enthalpy.

• The change in enthalpy for a reaction can be calculated by the sum of all of the bond energies of reactants and products.

• For reactants, the bonds are broken and the energy is required for this. While, for products, the bonds are formed and the energy released from this.

• Thus, bond breaking is endothermic and the bond energy is positive.

• Also, bond formation is exothermic and the bond energy is negative.

• For N₂; there is only one triple bond and the bond energy (N≡N) = 942 KJ/mol.

• For H₂, there is only one single bond and the bond energy (H₋H) = 432 KJ/mol and there are 3 molecules are included in this reaction. So, the bond energy will by multiplied by 3.

• For NH₃, there are 3 single bonds between N and H and the bond energy (N₋H) = -386 KJ/mol (negative sign because the bond here is formed) and there are 2 molecules will be formed from this reaction. So, the bond energy will be multiplied by 6.

• ΔH = [bond energy (N≡N)] + 3 [bond energy (H₋H)] + 6 [bond energy (N₋H)]

• ΔH = [942] + 3 [432] + 6 [-386] = -78 KJ.

• This means that the reaction 78 KJ of energy will be released from this reaction (Exothermic reaction).

Answer 2

The change in enthalpy for the reaction N₂ + 3H₂ → 2NH₃, calculated from the values of bond energies N≡N 942, H–H 432, and N–H 386 (all in kJ/mol) is -78 kJ.  

The balanced reaction is:

N₂ + 3H₂ → 2NH₃    (1)

The change in enthalpy for reaction (1) is given by:

$\Delta H^{0} = \Sigma \Delta H_{r} - \Sigma \Delta H_{p}$   (2)

Where r is for reactants and p for products

The energy bonds of the compounds are:

• N₂: 1 mol of 1 bond N≡N → 1 mol*942 kJ/mol

• 3H₂: 3 moles of 1 bond H-H → 3 moles*432 kJ/mol  

• 2NH₃: 2 moles of 3 bonds N-H → 2 moles*3*386 kJ/mol    

By entering the above values into equation (2), we have:

$\Delta H^{0} = \Delta H_{N_{2}} + 3*\Delta H_{H_{2}} - 2*\Delta H_{NH_{3}}$

$\Delta H^{0} = \Delta H_{N\equiv N} + 3*\Delta H_{H-H} - 2*3*\Delta H_{N-H}$

$\Delta H^{0} = (942 + 3*432 - 2*3*386) kJ = -78 kJ$  

Therefore, the enthalpy change for the reaction is -78 kJ.

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I hope it helps you!