To solve the problem, we need to calculate the total energy involved in breaking the bonds of the reactants and the total energy involved in forming the bonds of the products, then find the change in enthalpy (ΔH).
1. Determine the bonds broken in the reactants:
- For [tex]\(N_2\)[/tex], there is one [tex]\(N \equiv N\)[/tex] triple bond.
- For [tex]\(3H_2\)[/tex], there are three [tex]\(H - H\)[/tex] bonds.
2. Calculate the total bond energy of the reactants:
- Energy to break [tex]\(1 x N \equiv N\)[/tex] bond: [tex]\(1 \times 942 \, kJ/mol = 942 \, kJ\)[/tex]
- Energy to break [tex]\(3 x H - H\)[/tex] bonds: [tex]\(3 \times 432 \, kJ/mol = 1296 \, kJ\)[/tex]
- Total energy for reactants: [tex]\(942 \, kJ + 1296 \, kJ = 2238 \, kJ\)[/tex]
3. Determine the bonds formed in the products:
- For [tex]\(2NH_3\)[/tex], there are six [tex]\(N - H\)[/tex] bonds because each [tex]\(NH_3\)[/tex] molecule has three [tex]\(N - H\)[/tex] bonds and there are two [tex]\(NH_3\)[/tex] molecules.
4. Calculate the total bond energy of the products:
- Energy to form [tex]\(6 x N - H\)[/tex] bonds: [tex]\(6 \times 386 \, kJ/mol = 2316 \, kJ\)[/tex]
5. Calculate the change in enthalpy (ΔH):
- ΔH = Total bond energy of products - Total bond energy of reactants
- ΔH = [tex]\(2316 \, kJ - 2238 \, kJ = 78 \, kJ\)[/tex]
6. Express the enthalpy change to two significant figures:
- ΔH is [tex]\(78 \, kJ\)[/tex]
Therefore, the enthalpy change for the reaction is [tex]\( \boxed{78} \)[/tex] kilojoules.
