Catalysis: Previous Year Questions with Solutions

Question

A catalyst speeds up a reaction but does not get consumed. Does it change (a) the equilibrium constant, (b) the activation energy, (c) the $\Delta G$ of the reaction, (d) the rate constant? Explain each.

Solution — Step by Step

This appeared in a previous year paper — let’s unpack it. Read what is asked and identify the concept — in this case, catalysis. Note down the given data and the unknown.

(a) No — the equilibrium constant depends only on thermodynamics ( $\Delta G^\circ$ ), which the catalyst does not change. (b) Yes — the catalyst provides an alternative path with lower activation energy. (c) No — $\Delta G$ is a state function and depends only on initial and final states. (d) Yes — lowering $E_a$ increases $k$ via the Arrhenius equation $k = Ae^{-E_a/RT}$ . Key insight: catalysts speed up both forward and reverse reactions equally, so equilibrium is reached faster but its position is unchanged.

Now we write the final answer in a single sentence so the examiner can find it at a glance. In board exams, bold or box the answer — it matters for marks.

Why This Works

This problem looks like a memory question but it is really a reasoning question. The key is to identify the physical or chemical principle at play and then apply it cleanly, without getting distracted by surface features. Students who only memorise facts get stuck on slight variations; students who reason from principles handle any variation.

Whenever you hit a chemistry question that mentions a trend or comparison, ask: ‘what property decides this?’ Then list the candidates — electronegativity, size, bond energy, polarity, resonance — and pick the one that varies in the question.

Alternative Method

A faster route to the same answer is to compare with a known benchmark. For instance, if you remember one clean example from class, you can often reason by analogy rather than rebuilding the theory from scratch. Toppers use this shortcut all the time during JEE Main, where every minute counts.

Common Mistake

Thinking a catalyst shifts equilibrium toward products. It does not. If a student sees ‘catalyst increases yield’, they should read carefully — it usually means yield per unit time, not maximum yield.

Exam Connection

This kind of question is a reliable marks-scorer in CBSE board papers and JEE Main. Examiners reuse the underlying concept with small cosmetic changes — different numbers, a renamed compound, a twist in the wording. If you understand the reasoning here, you will recognise the pattern in any future variant. That is the difference between solving one question and solving a question type.

Teachers often say, “practise till the method becomes boring.” What they mean is: work through enough variations that the decision tree inside your head becomes automatic. By the time you sit the exam, your pen should know the first three steps before your brain consciously thinks about them.

Practice Extension

Try these follow-up variations on your own and see if your method still works:

Change the starting quantity (e.g. 4.6 g becomes 9.2 g) and confirm that the final answer scales linearly.
Swap the reagent for a close analogue (e.g. $\text{KMnO}_4$ instead of $\text{K}_2\text{Cr}_2\text{O}_7$ ) and check which steps stay the same and which change.
Flip the question — given the product, work backwards to find the starting amount. This reverse drill tests whether you really understand the stoichiometry, or whether you were just plugging into a memorised chain.
Add an impurity (e.g. “the sample is only 80% pure”) and adjust the mole calculation. Purity questions are a classic twist in JEE Main.

In JEE Main, the cut-off between a 95 percentile and a 99 percentile is often just 3 to 4 questions of this exact difficulty. Practising these variations matters more than learning new topics during the last month of preparation.

Quick Recap

Identify the concept before touching numbers.
Write units on every line — they catch errors automatically.
Reach the final answer in a single bold or boxed expression so the examiner sees it instantly.
When stuck, ask: “what quantity is conserved here?” Conservation of mass, charge, and energy solve most chemistry problems.