### Related Formula Delta H = E_a(f) - E_a(b)$\Delta H = E_{a(f)} - E_{a(b)}$ ### Core Logic A catalyst accelerates both forward and backward path steps symmetrically by carving a lower activation energy profile route. * Uncatalyzed values: Delta H = 180 - 200 = -20mathrm~kJ~mol^-1$\Delta H = 180 - 200 = -20\mathrm{~kJ~mol}^{-1}$. * Catalyzed values: E_a(f)' = 80mathrm~kJ~mol^-1$E_{a(f)}' = 80\mathrm{~kJ~mol}^{-1}$ and E_a(b)' = 100mathrm~kJ~mol^-1$E_{a(b)}' = 100\mathrm{~kJ~mol}^{-1}$, leading to Delta H' = 80 - 100 = -20mathrm~kJ~mol^-1$\Delta H' = 80 - 100 = -20\mathrm{~kJ~mol}^{-1}$. Thermodynamic parameters (Delta H$\Delta H$, Delta G$\Delta G$, Delta S$\Delta S$) depend strictly on the initial and final energy states of reactants and products, meaning they are completely unaltered by the presence of a catalyst. ### Pattern Recognition Catalysts alter only kinetic properties (rate, activation barriers). They have zero impact on equilibrium positions or thermodynamic state parameters like Delta G$\Delta G$ or Delta H$\Delta H$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics