Given the following information:

Energy of sublimation of Na(s) = 97 kJ/mol

Bond energy of HBr = 363 kJ/mol

Ionization energy of Na (g) = 496 kJ/mol

Electron affinity of Br (g) = ?324 kJ/mol

Lattice energy of NaBr (s) = ?781 kJ/mol

Bond energy of H, = 432 kJ/mol

Calculate the net change in energy for the following reaction:

Transcribed Image Text:

ΔΕ = 2Na(s) + 2HBr(g) → 2NaBr(s) + H₂(g) kJ ΔΕ = 2Na(s) + 2HBr(g) → 2NaBr(s) + H₂(g) kJ

Answer rating: 100% (QA)

2 Nas HBr g 2 NaBr s H 2 g From the data 2 Nas 2 Nag H 2 x 97 194 kJmole 2 Nag 2 Na g 2e H 2 x ... View the full answer