Chalk−1

Chemistry · General chemistry II · Worked example

Read an energy diagram with and without a catalyst

A reaction has ΔH = −40 kJ/mol and Ea = 50 kJ/mol. A catalyst lowers the forward barrier to 35 kJ/mol. Find the reverse barriers with and without the catalyst, say what happens to ΔH and K, and estimate how much faster the catalyzed reaction is at 298 K.

Δ⁢H=−40 kJ/mol,Ea=50 kJ/mol,Ea,cat=35 kJ/mol

Place the levels

ΔH is negative, so the products sit 40 kJ/mol below the reactants: the reaction is exothermic. The uncatalyzed transition state is 50 kJ/mol above the reactants; the catalyzed one is 35 kJ/mol above.

Reverse barriers

Each reverse barrier is measured from the product level up to the same transition state.

Ea,rev=50−(−40)=90 kJ/mol,Ea,rev,cat=35−(−40)=75 kJ/mol

What the catalyst leaves alone

Both barriers fall by 15 kJ/mol, so the forward and reverse rates rise by the same factor. The reactant and product levels do not move: ΔH is still −40 kJ/mol and K is unchanged.

How much faster

Assuming the same frequency factor A, the rate constants differ by the Arrhenius factor for the 15 kJ/mol drop.

kcatk=e15000⁢/⁢(8.314×298)=e6.05≈4.3×102

Result

Reverse barriers: 90 kJ/mol uncatalyzed and 75 kJ/mol catalyzed. ΔH (−40 kJ/mol) and K are unchanged. Both directions run about 430 times faster at 298 K.

Your turn

An endothermic reaction has ΔH = +30 kJ/mol and Ea = 80 kJ/mol. What is the reverse activation energy, and could a catalyst lower the forward Ea to 25 kJ/mol?

Show the answer and explanation

50 kJ/mol; no.

Ea,rev = 80 − 30 = 50 kJ/mol. The transition state must lie above the products, which are 30 kJ/mol above the reactants, so no pathway can have a forward barrier below 30 kJ/mol.

Ea,rev=80−30=50 kJ/mol

Keep exploring

Open the diagram in Energy diagrams, then make ΔH positive and see which barrier the transition state must clear.

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Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.