For the process Br₂(l)$\to$ 2Br(g), A) ($\Delta$H) is + and $\Delta$S is + for the reaction. B) ($\Delta$H) is + and $\Delta$S = 0 for the reaction. C) ($\Delta$H) is - and $\Delta$S is - for the reaction. D) ($\Delta$H) is - and $\Delta$S is + for the reaction. E) ($\Delta$H) is + and $\Delta$S is - for the reaction.

Question

Quizplus · Accepted Answer

The process involves the vaporization of liquid bromine to gaseous bromine atoms, which requires energy input, making ΔH positive. The increase in disorder from liquid to gas makes ΔS positive.

For the Process Br2(l) →\to→ 2Br(g), A) Δ\DeltaΔ H) Is + And

For the Process Br₂(l) $\to$ 2Br(g),
A) $\Delta$ H) Is + And