Question

which statement is true for the reaction below?
$2\ce{o_3}(g) \
ightarrow 3\ce{o_2}(g)$
$\delta h = -427\\ \text{kj/mol}$
the reaction will occur at all temperatures.
the reaction will occur only at high temperatures.
the reaction will occur only at low temperatures.
the reaction will never occur.

Explanation:

To determine the temperature dependence of the reaction \( 2\text{O}_3(\text{g})
ightarrow 3\text{O}_2(\text{g}) \) with \( \Delta H = -427 \, \text{kJ/mol} \), we analyze the enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) changes:

1. Enthalpy Change (\( \Delta H \)): The negative \( \Delta H \) (\( -427 \, \text{kJ/mol} \)) indicates the reaction is exothermic (releases heat).
2. Entropy Change (\( \Delta S \)): The number of moles of gas increases (2 moles of \( \text{O}_3 \) produce 3 moles of \( \text{O}_2 \)), so \( \Delta S \) is positive (disorder increases).

For spontaneity, we use the Gibbs free energy equation: \( \Delta G = \Delta H - T\Delta S \). Since \( \Delta H < 0 \) (exothermic) and \( \Delta S > 0 \) (entropy increase), the term \( -T\Delta S \) will also be negative (because \( T \) is always positive in Kelvin). Thus, \( \Delta G \) will be negative (\( \Delta G < 0 \)) at all temperatures, meaning the reaction is spontaneous (will occur) regardless of temperature.

Answer:

The reaction will occur at all temperatures.