= Solution
For $\mathrm{C_6H_6(l)\rightarrow6C(graphite)+3H_2(g)}$, reversing the standard formation reaction gives $\Delta_rG^\circ=-124.3\,\mathrm{kJ\,mol^{-1}}$. Thus decomposition is thermodynamically favoured under the corresponding standard conditions. \b[<benzene> persists because it is kinetically stable on the laboratory timescale.] Breaking and reorganizing its bonding framework requires access to high-<activation energy> pathways, and suitable reactive intermediates or catalysts are absent under ordinary storage conditions. A negative <Gibbs free energy> change specifies the favoured thermodynamic direction, not a rapid <reaction rate>; thermal energy does not automatically overcome a large barrier.
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