Topic 9: Introduction to Kinetics and Equilibria

When comparing one factor, keep the others constant: use the same amounts and temperature for a surface-area experiment, or the same gas temperature when changing pressure.

Reactant bonds or electron arrangements must be disturbed before product bonds can form. Collisions below $E_a$ separate without reaction; collisions at or above it can access the reaction pathway.

On a reaction-profile diagram, forward $E_a$ is the vertical energy difference from the reactant level to the top of the energy barrier. Reverse $E_a$ is measured from the product level to that same barrier.

For initial rate, draw a tangent at $t=0$. For rate at a stated time, draw a tangent touching the curve at that point. Choose two well-separated points on the tangent—not necessarily data points—form a large triangle and calculate rise/run.

A product graph has a positive gradient; a reactant-amount graph has a negative gradient. Reaction rate is normally quoted as the positive magnitude unless a signed species rate is specifically requested.

Write units from the graph axes and retain scale factors. A steeper tangent has a larger rate; a plateau has gradient zero.

A Maxwell–Boltzmann graph plots number of molecules against molecular energy. The curve starts at the origin, rises to a peak and approaches the energy axis asymptotically; total area represents the fixed number of molecules.

The increase in the high-energy tail is much more important than the modest increase in collision frequency. More particles have $E\geq E_a$, so successful collisions per unit time rise and the reaction is faster.

At the same temperature, the molecular energy distribution is unchanged. Lowering $E_a$ means a larger existing fraction of collisions has enough energy, so successful collisions occur more frequently.

Plot enthalpy on the vertical axis and reaction progress on the horizontal axis. Put reactants and products at the same levels for both paths; their vertical difference is the unchanged $\Delta H$.

Label both activation energies with upward arrows from the relevant starting level to a peak, label the catalyst intermediate at the valley, and label $\Delta H$ directly between reactant and product levels.

Benefits must be compared with catalyst manufacture, toxicity, scarcity, cost, lifetime, recovery and recycling. A valuable catalyst may still be economical because it is regenerated and used for many cycles.

For one fixed balanced reaction, a catalyst does not change its stoichiometric atom economy. Higher atom economy arises only when catalysis makes a different, more selective reaction route possible.

At one fixed temperature, draw one Maxwell–Boltzmann curve with axes number of molecules and energy. Mark the uncatalysed $E_a$ line and a catalysed $E_a$ line farther left.

The area to the right of a threshold represents molecules with $E\geq E_a$. Because the catalysed line is lower, its right-hand area is larger, so a greater proportion of collisions is energetic enough to react.

The alternative pathway therefore increases successful-collision frequency and rate. The curve itself, its peak and its total area remain the same because temperature and molecule count have not changed.

Before equilibrium, concentrations change as the two rates approach equality. At equilibrium, concentration–time curves become horizontal, but their values need not be equal.

Write the equilibrium equation, mark the forward reaction as exothermic or endothermic, count gaseous coefficients only for pressure, identify the imposed change, then state both shift direction and the specific reason.

If gaseous mole totals are equal, pressure has no effect on position. A catalyst speeds both directions and does not shift equilibrium. Pure solid amounts do not determine a homogeneous gas/solution equilibrium position in this qualitative model.

Use the supplied enthalpy sign, gas mole ratio, rate/yield data and cost information. Select a moderate temperature and pressure where extra yield or speed still justifies marginal energy, equipment and safety costs.

An industrial optimum maximises viable output per time and cost, not equilibrium percentage alone. Catalyst choice, feedstock conversion, separation and recycle can change the best compromise.