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In today's T-Level Chemistry lesson, we are going to study A twelve point four, Catalysts in Industrial Reactions.

By the end of this lesson, you should be able to explain how a catalyst increases industrial rate, how that shortens turnaround time and can reduce costs, and why delta H does not change.

A catalyst increases reaction rate without being permanently chemically changed overall. It provides an alternative pathway with lower activation energy.

Activation energy is the minimum energy required for a reaction to occur.

Specification point A twelve point four point one links increased rate to faster turnaround and reduced costs.

The industrial chain is: catalyst, increased reaction rate, shorter reaction stage, faster turnaround. That can allow more production cycles in the same period, which can reduce operating costs or increase capacity.

Do not claim that every catalyst makes every part of a process cheaper. Real cost still includes catalyst purchase, recovery or replacement, energy, maintenance and separation.

Case study from the lesson. Without a catalyst the reaction stage takes eight point zero hours. With a catalyst it takes three point zero hours.

The reduction is five point zero hours. The percentage time saved is eight minus three, divided by eight, times one hundred, which is sixty two point five percent less time on the reaction stage.

An examination example uses ten hours reduced to four hours. The time reduction is six hours.

Process X runs twelve batches a week without the catalyst and eighteen batches a week with it. That is six extra batches each week. The reaction-stage cost falls from nine hundred pounds to seven hundred pounds per batch, a saving of two hundred pounds per batch.

Specification point A twelve point four point two: catalysts reduce activation energy.

Watch the energy profile. Reactant enthalpy is the same. Product enthalpy is the same. The uncatalysed peak is higher. The catalysed peak is lower. Delta H is the same on both pathways.

At the same temperature, the lower barrier means a greater proportion of collisions can react. Successful collisions occur more frequently. The catalyst does not give particles extra kinetic energy.

Correct the misconceptions. A catalyst does not heat the mixture. It does not change delta H. It is not consumed overall. And a faster reaction does not mean zero cost.

Examination-style question.

Part a. Give one industrial advantage of increasing rate using a catalyst. Three marks.

Rate increases. The stage completes sooner, so turnaround is faster. Costs can fall, or more production cycles can be completed.

Part b. Explain how a catalyst increases rate in terms of activation energy. Three marks.

It provides an alternative pathway. Activation energy is lower. More collisions overcome the barrier, so more successful collisions occur per unit time.

Part c. A stage falls from ten hours to four hours. State the time reduction. One mark.

Six hours.

Part d. Why does a catalyst not change delta H? Two marks.

Reactant enthalpy is unchanged. Product enthalpy is unchanged, so delta H is unchanged.

Before we finish, complete the model. A catalyst lowers E a and increases rate. Faster rate shortens the reaction stage and can reduce costs through faster turnaround. Delta H stays the same because the enthalpy levels of reactants and products do not change.

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