Sequential and Parallel Reactions

Transcript

0:00Introduction

Hello everyone! Before we start, one small puzzle. I take sixteen grams of sulphur. It becomes sulphur dioxide, that becomes sulphur trioxide, and that becomes sulphuric acid. Three reactions, one after another. How much acid do I get? Will you balance all three equations and walk through every middle step? Relax, there is a one line way, and today you will learn it. And a second puzzle. Ten grams of a mixture of magnesium and aluminium, with excess hydrochloric acid, gives half a mole of hydrogen. How much of each metal? Two metals, both making hydrogen at the same time. Look, chains and splits like these come in JEE again and again. So, let's begin.

0:48Linking equations: a chain or a split

So, sometimes one equation is not enough. Two or three reactions are linked together. Now think. In what ways can two reactions be linked? Yes! Either one after the other, like a chain, or side by side, sharing the same reactant. Two ways, and today we learn both. First, sequential. The product of one step becomes the reactant of the next step, like a chain, okay? Our sulphur puzzle is exactly this: sulphur, then sulphur dioxide, then sulphur trioxide, then the acid. Second, parallel. One reactant splits, and part of it goes into one reaction, part into another, side by side. Our metal puzzle? The ten grams of mixture splits between two reactions, and both of them make hydrogen. Clear?

1:39Sequential: follow the chain

Alright, first the chain. P gives Q, and then Q gives R, okay? Q is made in the first step and used up in the second. We call it the intermediate. Now suppose I give you the moles of P, and I ask for the moles of R, with me? How will you get there? Through Q, exactly! There is no other road. So, way one: multiply the mole ratios along the chain. Each equation gives you one ratio, okay? Moles of R per mole of P equals moles of Q per mole of P, from the first equation, times moles of R per mole of Q, from the second. Two ratios, multiplied, and you have jumped from P straight to R. Way two, and this is the topper's way. POAC, the principle of atom conservation. Atoms are conserved, okay? So if one element is carried through the whole chain, just follow that element from the start to the end. And notice, the intermediates simply drop out. When one element is carried from the starting reactant all the way to the final product, you skip everything in the middle. That's it, one line. Got it?

2:48Parallel: one reactant, two paths

Now the split. One reactant, and two reactions sharing it at the same time, okay? So what is the difficulty here? What exactly do we not know? How much goes each way, exactly! The question never tells you the share of each reaction. So, make those amounts the unknowns. Call one of them x, and if the total is given, the other one is simply the total minus x. Next, write one equation for every piece of data the question gives you. The total mass, the total amount of gas formed: each one becomes an equation, right? Then solve them together, and the split is found. Alright, enough theory. Let's solve both puzzles.

3:32Example 1: sulphuric acid from 16 g of sulphur

Now, remember our first puzzle? Let's finish it. Sulphur is converted to sulphuric acid: sulphur to sulphur dioxide, to sulphur trioxide, to sulphuric acid. What mass of acid can be made from sixteen grams of sulphur? Pause the video, and try it first. Done? Okay, let's check. Some students say: sixteen grams of sulphur goes in, so sixteen grams of acid comes out. Right? No, no, careful! This is the JEE trap. Yes, all that sulphur ends up inside the acid. But the acid also carries hydrogen and oxygen, doesn't it? So never follow grams along a chain; follow atoms. And look at the chain: each step keeps one sulphur per molecule. So, POAC on sulphur. Moles of sulphuric acid equal moles of sulphur. See, sulphur dioxide and sulphur trioxide have simply dropped out. We never even touched the middle steps! Moles of sulphur: sixteen grams over thirty two. Zero point five moles. So, zero point five moles of sulphuric acid. Its molar mass is ninety eight, so zero point five times ninety eight is forty nine grams! Puzzle solved, in one line, just like I promised. Not sixteen grams, forty nine.

4:54Example 2: Mg and Al mixture composition from H2 evolved

Now a harder one, JEE level, and it is our second puzzle. Ten grams of a mixture of magnesium and aluminium gives zero point five moles of hydrogen with excess hydrochloric acid. Find the composition. The atomic masses are twenty four for magnesium and twenty seven for aluminium. Pause and try. Hint: let the mass of magnesium be x grams. Done? Okay. Two metals, two paths to hydrogen, so this is a parallel problem. Let magnesium be x grams. The first piece of data, the total of ten grams, tells us that aluminium is ten minus x grams. Now, how much hydrogen does each metal give? One mole of magnesium gives one mole of hydrogen. So aluminium also gives one to one, right? No, no, careful! This is exactly where students lose marks. Magnesium to hydrogen is one to one. But two aluminium give three hydrogen, okay? So the hydrogen from aluminium is three by two times the moles of aluminium. Now the second piece of data, the hydrogen. x over twenty four, plus three by two, times ten minus x over twenty seven, equals zero point five. Moles of hydrogen from magnesium, plus moles of hydrogen from aluminium, equals the total. One equation, one unknown. Now just solve it. Why two hundred and sixteen? Because it is the LCM of twenty four and twenty seven, so multiplying everything by it clears all the fractions, right? Nine x, plus one hundred and twenty, minus twelve x, equals one hundred and eight. Nine x minus twelve x is minus three x, and one hundred and twenty minus one hundred and eight is twelve. So, three x equals twelve, which gives x equals four. Magnesium is four grams, and aluminium is ten minus four, that is six grams. Quick check: four by twenty four is one sixth of a mole of hydrogen. Six by twenty seven, times three by two, is one third. One sixth plus one third is one half. Spot on!

7:11Quick revision: chain ratios, POAC and parallel unknowns

So, quick revision. Sequential reactions: multiply the mole ratios along the chain, or use POAC on the element carried through. Parallel reactions: make the amount going each way an unknown, and write one equation per piece of data. And always follow atoms, not grams, and check every mole ratio. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!