2020 P1 Q12 - Compare Rate and Determine Volume of CO2 Formed

Let Chemistry Guru, Singapore's prestigious JC Chemistry tuition centre, guide you through 2020 A Levels H2 Chemistry Paper 1 Question 12.

We are required to deduce which time shows the reaction at the faster rate and the value of Z, the maximum volume of CO₂ produced at stp.

Let's do this part by part.

1. Compare Rate

Comparing rate at different times is fairly straight forward.

Since rate is the gradient of concentration time graph, we just need to compare the slope of the graph at both times t₁ and t₂.

From the graph we can see clearly that the gradient of the slope at t₂ is steeper, hence rate of reaction at t₂ is faster.

2. Calculate Max Volume of CO₂

Since the amounts of both reactants are given, we need to determine which of them is limiting.

Check out this video for a detailed discussion of deducing limiting reagent.

If 2.5 x 10^-4 moles of MnO₄^- were used up, amount of C₂O₄^2- required will be 6.25 x 10^-4 moles.

Since we only have 2.5 x 10^-4 moles of C₂O₄^2-, there is insufficient C₂O₄^2- to react with all the MnO₄^-.

Hence C₂O₄^2- is limiting.

We can now use the limiting reagent to determine amount and volume of carbon dioxide formed.

Hence 11.4 cm³ of CO₂ is formed at stp.

Let's run through the options and determine our answer.

Remember the faster rate is at t₂ and max volume of CO₂ is 11.4 cm³.

Therefore the answer to this question will be option C.

Topic: Kinetics and Mole Concept, Physical Chemistry, A Level Chemistry, Singapore

Back to list of questions for 2020 A Level H2 Chemistry Paper 1.

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