Example 1
Gas volume rises from 12 to 42mL between 10 and 25s. Find average rate.
Rate =25−1042−12=1530=2.0mLs−1.
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Year 12 · SACE Stage 2
Reaction rates, collisions and energy profiles
Read rates and apply collision and activation-energy rules.
Method and worked examples
Average rate is change in a measured amount divided by elapsed time. A tangent to an amount–time graph gives instantaneous rate; a steeper slope means a greater magnitude of rate. Reactant concentration normally decreases and product amount increases. State which quantity is being measured and use its units.
Successful collisions need sufficient energy and suitable orientation. Higher temperature increases the fraction above activation energy. Higher concentration, gas pressure or exposed solid surface usually increases collision frequency. Catalysts provide an alternative pathway with lower activation energy. They change rate, not reaction enthalpy or the equilibrium constant. Enzymes are catalysts whose activity also depends on structure and conditions.
Example 1
Example 2
Reactants have enthalpy 40, the peak 100, and products 10 kJ mol−1 on one energy scale. Find forward activation energy and ΔH.
Ea=100−40=60 kJ mol−1. ΔH=10−40=−30 kJ mol−1; the reaction is exothermic.
Watch out: It lowers activation energy and changes rate; the same reactants and products retain the same ΔH.