If bond dissociation energies of N≡N, H-H and N-H are x1, x2 and x3 respectively, the enthalpy of formation of NH3 (g) is:
Correct answer: C. ΔHf = x1/2 + 3/2x2 - 3x3
- A. ΔHf = x1+ 3x2 - 6x3
- B. ΔHf = 3x3 - 1/2x1 - 3/2 x2
- C. ΔHf = x1/2 + 3/2x2 - 3x3
- D. ΔHf = 6x3 - x1 - 3x2
Explanation
To find the enthalpy of formation (ΔHf) of NH3, consider the bond dissociation energies: The formation of NH3 from N2 and H2 involves breaking one N≡N bond and three H-H bonds, and forming six N-H bonds. Thus, the enthalpy change is calculated as: ΔHf = x1 + 3x2 - 6x3, where x1 is the energy to break N≡N, x2 is the energy to break H-H, and x3 is the energy released in forming each N-H bond. The correct option is C because it correctly accounts for the energy changes involved in breaking and forming the relevant bonds.
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Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
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