Reading the Question(s) in IB Chemistry

By Daniel Tritton

(Biology, Chemistry tutor at The Edge Learning Center)

All too often I encounter students with innate ability in Chemistry making unnecessary mistakes and losing easy marks in their mock exams for IB Chemistry. Apart from having the knowledge and necessary problem solving skills, students should be adept at READING the question in the exam; as the question gives the student all the information they need to gain full marks, all they need do is apply understanding and some Chemistry know how!

 

 

Let’s consider the example below, taken from a previous Paper 1 exam:

 

 

 

 

 

 

 

This question concerns the topic of “Chemical Kinetics” and can be easily answered with a little knowledge and closely READING the question. Firstly, they are asking which option the “rate-determining step” could be, which is the slowest step, by definition and the step upon which the overall rate of reaction depends. Therefore, we CANNOT select a step which will occur quickly or spontaneously! After reading the question carefully, one can spot that the question has already given us the answer in the first sentence: “Sodium thiosulfate, NaS2O3(aq), and hydrochloric acid, HCl(aq), react spontaneously to produce solid sulfur, S(s), according to the equation below . . .”. Hence the answer cannot be A or B since thiosulfate is reacting with H+ (from hydrochloric acid) and it cannot be C, since one of the products is sulfur solid. Thus, it must be D!

Let’s look at another example question from a previous paper 2 exam:

 

This question covers the topic of “Redox Processes” and, once again can be scored with full marks by reading carefully and applying understanding to solve the problem. For (i) we need to know what the “oxidation number” is, which is simply a number assigned to an element/ compound (involved in a redox reaction) to indicate if said element/ compound is being “oxidised” i.e. losing electrons or “reduced” i.e. gaining electrons. It is important to note that oxidation and reduction may also be the gain of oxygen/loss of hydrogen and gain of hydrogen/ loss of oxygen, respectively. Hence, it is important to know that a specific term used in Chemistry may have several meanings. Nevertheless, if the oxidation number rises, the concerned species is being oxidised and vice versa for reduction. Paying attention to the equation in step 1, we can clearly see that Cu is changing to Cu2+ and thus the oxidation number is changing from 0 to +2. For nitrogen, it is a slightly different case as it is part of a compound, nitric acid (HNO3) in this case. So HNO3 changes to NO2 and given that the charges usually associated with H and O are 1+ and 2-, respectively then the oxidation number of nitrogen is changing from +5 to +4.

 

For (ii), we need to notice two things: (1) what an “oxidising agent” is and (2) pay attention to the word agent. An oxidising agent oxidises other species (unsurprisingly!) and thus becomes reduced itself. We know from (i) that nitrogen is being reduced to allow copper to oxidise, however nitrogen is part of the agent HNO3 and therefore the oxidising agent is HNO3 or nitric acid (both the name and chemical formula would be acceptable answers).

 

In conclusion, remember to READ the question to ensure you are getting all the marks you deserve, granted you are answering correctly!

P.S. Chemistry allows us to study matter and manipulate it to make new materials for humankind, which is cool!

 

 


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