Method of Initial Rates

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a reaction where A plus B gives products: one A and one B in the balanced equation. So the rate must be k, times the concentration of A, times the concentration of B, right? But wait, we learnt that orders come only from experiment. But which experiment? Nobody can watch molecules colliding. Today we learn the method of initial rates: a few simple runs in the lab, and the whole rate law comes out. At the end, we will test that guess with real numbers. Questions like this come up often in JEE papers. So, let's begin.

0:41Change one concentration at a time

So, here is the rate law. Small r is the rate, small k is the rate constant, and the square brackets mean concentration, in moles per litre. So r equals k, times the concentration of A to the power m, times the concentration of B to the power n. The powers m and n are the orders, and our job is to find m, n and k. Okay? Now think. Suppose we change both concentrations together, and the rate goes up. Which one caused it, A or B? We cannot tell, can we? So the rule is simple: change only one concentration at a time, and keep the rest fixed. Then any change in the rate comes from that one reactant. Clear? We do a few such runs. In each run, we measure the rate right at the start. That is called the initial rate, and it gives the method its name. Got it?

1:38Reading the order from the rate

Now, why does this work? We keep B fixed, and we double A, just as the first line says. In the rate law, A becomes two times A, in round brackets, to the power m. The two to the power m comes out in front, and the rest is just the old rate. So the new rate equals two to the power m, times r. Okay? So doubling A multiplies the rate by two to the power m. Measure the factor, and you have the order. With me? Suppose doubling A makes the rate four times; the cross on screen means times. So the order in A is four, right? Oh no, careful here! Four is the factor, not the order. Two to the power what gives four? Two, so the order is two. The same way, if the rate does not change at all, the order is zero, since two to the power zero is one. And if the rate doubles, the order is one. Clear? And it is not only doubling. Multiply A by any factor x, and the rate is multiplied by x to the power m. So tripling A multiplies the rate by three to the power m. Right? Once the orders are known, only k is left. Put any one run into the rate law, and solve for k. That's it! Got it?

2:56Gases: squeeze the vessel

Now one more idea, for gases. We squeeze the gas mixture to one over x of its volume. For example, x equals two means half the volume. Okay? The amount of gas stays the same, but the volume is x times smaller. So what happens to each concentration? Yes, exactly! Concentration is moles per litre, and the litres went down x times. So every concentration is multiplied by x. Right? And the rate? Each concentration brings its own power: x to the power m from A, times x to the power n from B, okay? That is x to the power m plus n, and m plus n is the overall order. So the rate is multiplied by x to the power of the overall order. With me?

3:45Example 1: 2NO + Cl2, volume halved, rate becomes 8 times

Now our first example. Two N O plus C l two gives two NOC l. The small two in C l two counts the chlorine atoms, and the arrow means gives. The rate equals k, times the concentration of N O squared, times the concentration of C l two. How does the rate change if the volume of the vessel is halved? Pause the video and try it first. Done? Okay, let's check. Half the volume means one over two, so x equals two. So the concentrations of N O and of C l two both double. Okay? Both concentrations double, so the rate simply doubles too, right? No, no, careful here! Each concentration carries its own power. N O has order two, and C l two has order one, so the overall order is two plus one, three. Got it? So the rate is multiplied by two to the power three, which is eight. Halve the volume, and this reaction goes eight times faster, not two times. Clear?

4:52Example 2: three-run table, orders in A and B, then k

Now a harder one, and it is our puzzle from the start. For A plus B gives products, here are three runs. Capital M means molar, moles per litre, and M s to the power minus one means molar per second. Run one: A is zero point one zero, B is zero point one zero, and the rate is two point zero times ten to the power minus three. Run two: A is zero point two zero, B is zero point one zero, rate eight point zero times ten to the power minus three. Run three: A is zero point one zero, B is zero point three zero, rate six point zero times ten to the power minus three. Find the rate law and k. Pause the video and try it first. Done? Okay, let's check. First, the order in A: which two runs should we compare? Runs one and two, exactly! B stays at zero point one zero, and A doubles. The rate goes from two to eight, times ten to the power minus three: four times. And four is two squared, so the order in A is two. Okay? For B, compare runs one and three, where A stays the same. B is tripled, and the rate goes from two to six, also three times. Three is three to the power one, so the order in B is one. And runs two and three? Both concentrations change there, so on their own they tell us nothing. Right? Now the rate constant, from run one. Two point zero times ten to the power minus three equals k, times zero point one zero squared, times zero point one zero. Zero point one zero squared, times zero point one zero, is one times ten to the power minus three, so k equals two point zero. The overall order is three, so the units are litre squared, per mole squared, per second; capital L is litre and m o l is mole. Careful here: many students forget to square the zero point one zero, so watch for that. So r equals two point zero, times the concentration of A squared, times the concentration of B to the power one. Now, our puzzle. From the equation, we guessed k times A times B, with power one on each. That guess was wrong, wasn't it? The experiment says the order in A is two. Orders come from experiment, not from the balanced equation, unless the reaction is a single elementary step. Got it?

7:22Quick revision: one change at a time, gas volume, k

So, quick revision. Change one concentration at a time, and keep the rest fixed. If multiplying A by x multiplies the rate by x to the power m, the order in A is m. For gases, squeezing to one over x of the volume multiplies the rate by x to the power of the overall order, and k comes from any one run. Now go and try the practice problems on this page. All the best!