The industrial synthesis of methanol can be represented by the equation Selected…

CHEMISTRY · Multi-part · A Level · Chemistry · Energetics · Energetics

Problem

The industrial synthesis of methanol can be represented by the equation CO(g)+2 H2(g)→CH3OH(g)\mathrm{CO(g)} + 2\,\mathrm{H_2(g)} \to \mathrm{CH_3OH(g)} Selected standard enthalpies of formation at 298 K are given in the table below. Species ΔHf∘\Delta H_f^{\circ} (kJ mol−1kJ\,\text{mol}^{-1}) CO(g)\mathrm{CO(g)} -110.5 H2(g)\mathrm{H_2(g)} 0.0 CH3OH(g)\mathrm{CH_3OH(g)} -201.0 Kinetic studies indicate that, in the absence of a catalyst, the forward activation energy is 180 kJ mol−1180\,\text{kJ\,mol}^{-1}. Introduction of a Cu/ZnO catalyst lowers the forward activation energy by 45 kJ mol−145\,\text{kJ\,mol}^{-1} but does not alter the enthalpy change. Energy (kJ\text{kJ}) versus reaction progress is the coordinate system assumed for any diagrams you construct. (a) Calculate the standard enthalpy change ΔH∘\Delta H^{\circ} for the reaction as written. Show your work and report the final value with sign and units. (b) Determine the activation energy for the reverse (methanol decomposition) reaction in the absence of the catalyst. (c) On one set of fully labeled axes, sketch quantitatively scaled potential-energy diagrams for both the uncatalyzed and catalyzed pathways. Label numerical values for the energies of reactants, products, both transition states, EaE_a (forward and reverse) for each pathway, and ΔH∘\Delta H^{\circ}. (d) Using thermodynamic and kinetic reasoning based on your diagram, explain whether the catalyst has any effect on the equilibrium constant KK at 298 K.

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