Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 4 / a / v / Solution

For ${\mathrm{C}}_6{\mathrm{H}}_6(\mathrm{l})\to 6C(\mathrm{graphite})+3{\mathrm{H}}_2(\mathrm{g})$, reversing the standard formation reaction gives ${\Delta }_r{G}^{\circ }=-124.3\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Thus decomposition is thermodynamically favoured under the corresponding standard conditions. 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.