The Graphical and Differential Methods

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Two students plot the same readings of the same first order reaction: a log of the concentration against time. Both get a straight line. But one finds the slope is minus k, and the other finds a slope of only minus k over two point three zero three! So who made the mistake? Maybe nobody made a mistake, right? Today we find the order of a reaction from graphs and rates, and solve this mystery on the way. These graphs come up often in JEE questions. So, let's begin.

0:36Which plot is a straight line?

So, how do we find the order? Remember, order comes from experiment, not from the balanced equation, unless the step is elementary. So we measure the concentration at different times, and draw graphs. Now think. In zero order, the concentration falls by the same amount every minute. Which graph is a straight line? Concentration against time itself, yes! The square brackets around A mean the concentration of A, and v s means versus, plotted against time t. The line goes down with slope minus k, where k is the rate constant. Okay? First order: equal fractions are used up in equal times, so the concentration curves. But its natural log, written l n, falls in a straight line, again with slope minus k. Right? Second order: we plot one over the concentration against time. As the reactant is used up, does this line go up or down? It goes up, exactly! The concentration falls, so one over it rises, and the slope is plus k. Clear? So from one table of readings, draw all three graphs. Whichever comes out straight gives the order. With me?

1:53The trap: log to the base ten

Now the trap. Many questions use log to the base ten, not the natural log. It is still a straight line, so the slope is minus k again, right? No, no, careful here! The natural log of x is two point three zero three times the log to the base ten of x; the small ten below the log is the base. Okay? So this plot has slope minus k over two point three zero three, not minus k. And that solves our puzzle! One used the natural log, the other log to the base ten, and both are correct. Got it? So k equals minus two point three zero three times the slope, which comes out positive. Before using any slope, always check the axis: natural log, or log to the base ten? Clear?

2:40Rate against concentration

Now a different graph: rate against concentration. For zero order, does the rate change as the reactant is used up? Not at all! Rate equals k, a constant, so we get a flat, horizontal line. Okay? First order: rate equals k times the concentration of A, so double the concentration, double the rate. That is a straight line through the origin, with slope k. Right? Second order: rate equals k times the concentration of A to the power two, which curves upward into a parabola. With me?

3:15Quick tests on a table

Sometimes you do not even need a graph: two quick tests on a table. Equal amounts used up in equal times: zero order. Okay? Equal fractions used up in equal times, like half gone in every equal interval: first order. Right? And the half-lives. t half, written t with a small one over two, is the time for half the reactant to go. Zero order: A naught over two k, where A naught, with the small zero, is the starting concentration. First order: zero point six nine three over k. Second order: one over k times A naught. Only the first order half-life does not depend on the starting concentration. Got it?

3:58The differential (van't Hoff) method

Now the differential method, also called van't Hoff's method. Here we measure the rate, small r, at different concentrations. Rate equals k times the concentration of A to the power n, where n is the order. Take the log of both sides. Okay? Log r equals log k, plus n times log of the concentration; the power n comes down in front. Right? This has the same shape as y equals m x plus c. So what should we plot against what, to get a straight line? Yes, exactly! Log r against log of the concentration. The slope is n, the order itself, and the intercept is log k. Nice, right? And with just two readings, no graph is needed. n equals log of r two over r one, all over log of a two over a one, where a one and a two are the two concentrations; the small one and two are labels. Clear?

4:53Example 1: zero order from a table, k and half-life 12.5 min

Now let's try one. The concentration of A is zero point five zero, zero point four zero, zero point three zero and zero point two zero molar, at zero, five, ten and fifteen minutes. Capital M means molar, moles per litre, and m i n means minutes. Find the order, k, and the half-life. Pause the video and try it first. Done? Okay, let's check. Every five minutes, the concentration drops by zero point one zero molar. Equal amounts in equal times: zero order. Very good! So k is just the fall per minute: zero point one zero molar over five minutes, that is zero point zero two moles per litre per minute. The minus one power means per. Okay? Now the half-life. Many students write zero point six nine three over k straight away. Is that right here? Oh no, be careful here, because that formula is only for first order. For zero order, t half equals A naught over two k: zero point five zero over two times zero point zero two, which is zero point five zero over zero point zero four. Twelve point five minutes. Perfect!

6:02Example 2: differential method, rate ratio 9 gives order 2

Now a harder one, with the differential method. The rate is two point zero times ten to the power minus three moles per litre per second at zero point one zero molar, and one point eight times ten to the power minus two moles per litre per second at zero point three zero molar. On screen, s to the power minus one means per second. Find the order, and k. Pause the video and try it first. Done? Okay, let's check. r two over r one: one point eight times ten to the power minus two, over two point zero times ten to the power minus three, that is nine. And a two over a one: zero point three zero over zero point one zero, that is three. Okay? So n equals log nine over log three. Nine is three squared, so log nine is two log three, and n equals two. Concentration three times, rate nine times: second order. Right? Now k. Some students divide the rate by zero point one zero and get zero point zero two. Is that right? No, no, wait here! The order is two, so rate equals k times the concentration squared. Divide by zero point one zero squared, which is zero point zero one. Clear? That gives k equals zero point two zero litres per mole per second; capital L is litres, and mol to the power minus one means per mole. Check with the second reading: one point eight times ten to the power minus two, over zero point three zero squared, also gives zero point two zero. Perfect!

7:40Quick revision: straight plots, log10 slope, log r vs log A

So, quick revision. Against time, a straight line for the concentration means zero order, for its natural log, first order, and for one over it, second order. With log to the base ten, the slope is minus k over two point three zero three. And log r against log concentration has slope n and intercept log k. Now go and try the practice problems on this page. All the best!