Problem 26
Question
At \(298 \mathrm{K},\) for the reaction \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+\) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{Br}_{2}(1)+2 \mathrm{HNO}_{2}(\mathrm{aq}), \Delta H^{\circ}=-61.6 \mathrm{kJ}\) and the standard molar entropies are \(\mathrm{H}^{+}(\mathrm{aq}), 0 \mathrm{JK}^{-1}\) \(\mathrm{Br}^{-}(\mathrm{aq}), 82.4 \mathrm{JK}^{-1} ; \mathrm{NO}_{2}(\mathrm{g}), 240.1 \mathrm{JK}^{-1} ; \mathrm{Br}_{2}(1), 152.2\) \(\mathrm{J} \mathrm{K}^{-1} ; \mathrm{HNO}_{2}(\mathrm{aq}), 135.6 \mathrm{JK}^{-1} .\) Determine (a) \(\Delta G^{\circ}\) at 298 K and (b) whether the reaction proceeds spontaneously in the forward or the reverse direction when reactants and products are in their standard states.
Step-by-Step Solution
VerifiedKey Concepts
Entropy
- For reactants, we consider the entropy contributions of \(2\ \mathrm{H}^{+}, 2\ \mathrm{Br}^{-}, \) and \(2\ \mathrm{NO}_{2}.\)
- For products, we use \(\mathrm{Br}_{2}\) and \(2\ \mathrm{HNO}_{2}.\)
Enthalpy
- A negative enthalpy (\(\Delta H^{\circ} < 0\)) means the reaction is exothermic, releasing heat to the surroundings.
- Conversely, a positive value would mean absorption of heat (endothermic).
Spontaneity
- When \(\Delta G^{\circ} < 0,\) the reaction occurs spontaneously in the forward direction.
- If \(\Delta G^{\circ} > 0,\) it favors the reverse direction.
Thermodynamics
- First Law (Conservation of Energy): Energy can neither be created nor destroyed.
- Second Law: Entropy of an isolated system always increases in spontaneous processes.