Something odd happened today, something I’d not seen before. It relates to CIE questions found here…

**Paper 11, Summer 2012, Question 9** and **Paper 12, Summer 2013, Question 7**

In paper 1, you’ve got roughly one minute thirty seconds to answer all the questions. Because there is so little time, I thought that if there is a quick method to answer a question, then it would good to use it. I proposed the method of just randomly creating your own simple stoichiometries (molar ratio’s) for the reaction equation, but keeping the ratio numbers small Then, carrying through that assumption, working out the algebraic result for the total number of moles in the eqm mixture.. The hope was that you may get lucky and hit upon a right answer straight away, but if you didn’t get the answer straight away, then hopefully you could spot what adjustment in your proposed equation was necessary to get the correct answer.

Well, for **11/M/J/2012 question 9**, this method and the equation, I chose **1P <–> 1Q + 1R**, I happened to use for the purposes of class discussion just happened to give the same answer as in the mark scheme – the answer was **2+x**, but the particular P Q R equation I dreamt up wasn’t one of the answers offered in the paper. I had never seen this before (yes, I’m still not as familiar with CIE papers as I’d like to be) and I couldn’t understand why it happened. There was no calculation error. My answer was valid but at the time I started thinking “*how can this be? – surely a different equation would give a different total number of total moles present at equilibrium*” **[note: I was wrong, but I got locked into thinking this. Actually you can still get the same algebraic number of moles at equilibrium, even with a different chemical equation, as I will demonstrate later]** My method completely failed to help to identify which of the correct answers presented was the correct one, and just as bad, if not worse, because I got stuck on my wrong idea, I began thinking I cannot change the 1,1,1 ratio’s as if I do then I’ll no longer get the 2+x answer.

😦

After all classes, I decided to start afresh, and map out my chem equation and pair it alongside the answer the mark scheme gave, to compare them to each other step by step, so see exactly what was going on…scientific approach 😉 and it worked.

See if you can spot what was going on from the picture below… Once you think you know, then scroll down to read my comment on it. DO TRY AND PROPOSE AN ANSWER FIRST BEFORE YOU LOOK FOR MY COMMENT

(you may need to click the image to expand it)

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

*The (1), (2) and (3) in the picture simply refer to sequence of giving the algebraic terms. There is nothing special about following that order, but sometimes it helps illustrate a way on how to proceed.*

.

OK, P and Q both have the mole ratio of 2 in them relative to R.

So the relative increase in Q will be matched by the relative decrease in P. Adding up an increase to a decrease, when they are exactly the same magnitude, e.g. -3 added to +3, will give you zero. So, IN THIS PARTICULAR QUESTION, any chemical equation we proposed in which the ratio of P and Q are equal, would have given the same 2-x answer.

Try and work out the answer for 3P <–> 3Q + R

where x is the number of moles of R present at equilibrium, and an only initial amount of 2 moles of P was present and allowed to decompose.

You should get 2-(3x) + 3x + x for the amount of moles at eqm.

The -(3x) will cancel with the +3x, leaving 2+x as the total number of moles at the end of the reaction.

It’s east to imagine my method could do something similar for different equations, hence until I can try and figure out an improved method (if any!) then it looks like simply doing each question to see which of the given equations gives the right answer, is the best way to do this kind of question.

Actually while typing all this out, I’ve just realized that as a matter of fact, using the equations given to you **is** actually the smartest way after all. Because: you have to work through an equation anyway, and by using one of the equations given, then straight away you have a 25% chance of having chosen the right answer, and if not, then because you’ve worked through it, you can still spot what adjustment was necessary, but if you just dream up your own ratios – as I did – then really you I stand a chance of NEVER getting the correct answer, and as this post shows, you may even hit across an answer that confuses you – exactly what happened to me in class… LOL.

Oh well. I’m learning too.

The working for each question given for the **paper 12, Summer 2013, Question 7** is given in the graphic below…

(click to expand if necessary)

Of course for THIS TYPE OF QUESTION, when you do it in the exams, just write down the relevant x terms under each species on the test paper, and add them up at the end.