Calculate the energy change for the formation of MgBr₂(s) from its elements in their standard states: Mg(s) + Br₂(l) → MgBr₂(s) ?
Mg(s) → Mg(g) +147.1 kJ/mol
Mg(g) → Mg⁺(g) + e^- +737.8 kJ/mol
Mg⁺(g) → Mg²⁺(g) + e^- +1451 kJ/mol
Br₂(l) → Br₂(g) +30.9 kJ/mol
1/2 Br₂(g) → Br(g) +112.0 kJ/mol
Br(g) + e^- → Br^-(g) -325.0 kJ/mol
MgBr₂(s) → Mg²⁺(g) + 2Br^-(g) +2440 kJ/mol

Question

Quizplus · Accepted Answer

To calculate the energy change for the formation of MgBr₂(s) from its elements in their standard states, we need to add up the energy changes for each step of the formation reaction. Mg(s) → Mg(g) +147.1 kJ/mol (sublimation) Mg(g) → Mg⁺(g)+ e^- +737.8 kJ/mol (ionization energy) Mg⁺(g) → Mg²⁺(g)+ e^- +1451 kJ/mol (second ionization energy) Br₂(l) → Br₂(g) +30.9 kJ/mol (vaporization) 1/2 Br₂(g) → Br(g) +112.0 kJ/mol (bond dissociation) Br(g) + e^- → Br^-(g)-325.0 kJ/mol (electron affinity) Mg⁺(g) + 2 Br^-(g) → MgBr₂(s) +2440 kJ/mol (formation reaction) Adding all these energy changes together, we get: ΔH = 147.1 + 737.8 + 1451 + 30.9 + 112.0 - 325.0 + 2440 = 3494.8 kJ/mol However, we need the energy change for the formation of MgBr₂(s) from its elements in their standard states, so we need to reverse the signs for Mg(s) → Mg(g) and Br₂(l) → Br₂(g): ΔH = -147.1 - 737.8 - 1451 - 30.9 - 112.0 + 325.0 + 2440 = -499.2 kJ/mol Therefore, the energy change for the formation of MgBr₂(s) from its elements in their standard states is -499.2 kJ/mol, which corresponds to choice C.

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