2P(s) 3CI2(g) 2PCI3 ∆H= 151.8 kJ PCl3+ CI2(g) , PCI5(g); ∆H = -32.8 kJFrom the following data, the heat of formation of PCI5 comes out to be :
Correct answer: D. +119 kJ
- A. -217.4 kJ
- B. -119 kJ
- C. +217.4 kJ
- D. +119 kJ
Explanation
To find the heat of formation of PCl5, we need to add the enthalpy changes of the given reactions. Using Hess's Law, we have:2P(s) + 3Cl2(g) → 2PCl3(g); ΔH = 151.8 kJPCl3(g) + Cl2(g) → PCl5(g); ΔH = -32.8 kJBy combining these reactions, we obtain the enthalpy for the formation of 2 moles of PCl5. The total enthalpy change is 151.8 kJ + (-32.8 kJ) = 119 kJ. Since we are considering the formation of 1 mole of PCl5, the heat of formation is +119 kJ.The other options result from incorrect calculations or sign errors.
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About Thermochemistry and Energetics of Chemical Reactions
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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