Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Suppose a first-order reaction is half done in ten minutes. When will it be seventy five percent done? The first fifty percent took ten minutes, so the next twenty five percent should take only five more. Fifteen minutes in all, right? We will check that carefully today. Today: Set one of the mixed problems on chemical kinetics, short numericals of the kind that come up often in JEE Main. So, let's begin.

0:33Set 1: nine short problems, three themes

So, Set one has nine short solved problems, from three themes. First, the rate law and stoichiometry: changing concentrations in a rate law, the rates of different species from the balanced equation, and the order from the units of k, the rate constant. Remember, the order itself comes from experiment, not from the balanced equation. Three problems. Okay? Second, integrated laws and half-lives, for zero, first and second order. Four problems. Right? And third, temperature: the Arrhenius equation, and the temperature coefficient. Two problems. Today's two examples come from the last two lines: a half-life problem first, then an Arrhenius one. With me?

1:18Half-lives: first order is special

Now, the tools. Look at this graph from the course. Along the bottom is time, t, marked in half-lives, so two means two half-lives. Up the side is the concentration of A over its starting value: square brackets mean concentration, and the small zero means at the start. All three curves, zero, first and second order, start at one and cross the dashed line, one half, together. Okay? First order. The half-life, t half, written t with a small one over two below it, equals zero point six nine three over k. Right? Now, if you double the starting concentration, A naught, does the first-order half-life change? It does not change at all. Correct! There is no A naught in the formula, so it is independent of the starting concentration. The second half-life takes exactly as long as the first, even though less A is left. Clear? So, after one half-life, half is left. After two half-lives, how much is left? One quarter, yes! It is half of a half. In general, after n half-lives, one half to the power n remains. On the graph, at two, the dark blue first-order curve is at one quarter. Got it? Now, a trap the course warns about. x percent complete means one hundred minus x percent remains. So seventy five percent complete means twenty five percent is left, not seventy five. Read the question slowly, okay? Zero order is different. Its half-life is A naught over two k, so it grows with A naught. And it is complete at A naught over k, two half-lives. On the graph, the zero-order line hits zero at two. Right? And second order: t half equals one over k times A naught. So for zero and second order, the half-life depends on the starting concentration; for first order, it does not. With me?

3:19Temperature: Arrhenius at two temperatures

Next, temperature. The Arrhenius equation: k equals capital A times e to the power minus E a over R T, as on screen. Capital A is the frequency factor; E a, capital E with a small a, is the activation energy; R is the gas constant; and T is the temperature. Okay? On the right, the course plots l n k, the natural log of k, against one over the temperature. It is a straight line with slope minus E a over R, and it cuts the vertical axis at natural log of capital A, the intercept. Now, if you know k at just two temperatures, can you find E a? Yes, you can! Two points fix the slope of a straight line. Natural log of k two over k one equals E a over R, times one over T one, minus one over T two. The small one and two below k and T mark the first and the second temperature. Clear? Two rules here. T must be in kelvin, never in degrees Celsius. And in this chapter, take R as eight point three one four joules per mole per kelvin. Right? One more trap. The formula has l n, the natural log. If you use log to the base ten, multiply it by two point three zero three. Use one or the other, never a mix. Got it? Last, the temperature coefficient: k at T plus ten, over k at T, the factor for a ten degree rise. Near room temperature, it is usually two to three. With me?

4:55Example 1: 75% completion of a first order reaction, 20 min

Now, Example one, an easy one from Set one. Here, a first-order reaction has k equals zero point zero six nine three per minute. How long does seventy five percent completion take? Pause the video, and try it first. Done? Okay, let's check. First order, so t half equals zero point six nine three over k. Zero point six nine three over zero point zero six nine three: exactly ten minutes. So this is our puzzle reaction, half done in ten minutes! Now, the puzzle answer. Fifty percent took ten minutes, so seventy five percent takes fifteen minutes, right? No, no, wait, that is exactly the trap! First order does not remove equal amounts in equal times. Seventy five percent complete means twenty five percent remains, one quarter; the double arrow means so. Okay? And one quarter is one half times one half, one half squared. So, two half-lives. The second half-life takes just as long as the first, even though it removes only twenty five percent. Clear? So t equals two times ten; the cross means times. Twenty minutes, not fifteen. And that solves our puzzle! For first order, seventy five percent completion always takes two half-lives. Very good!

6:19Example 2: k quadruples from 300 K to 320 K, find Ea

And now, Example two, a step harder. Here, a rate constant quadruples, it becomes four times, when the temperature goes from three hundred to three hundred and twenty kelvin. Find E a, the activation energy. Pause the video and try it. Hint: use the two-temperature form. Done? Okay, let's check. k two over k one is four. T one is three hundred kelvin and T two is three hundred and twenty kelvin, with capital K for kelvin. Already in kelvin, so no conversion is needed. Okay? Now the bracket: one over three hundred, minus one over three hundred and twenty. That is twenty over ninety six thousand, which is exactly one over four thousand eight hundred, per kelvin. Positive, because T one is the lower temperature. Right? So, natural log of four equals E a over eight point three one four, times one over four thousand eight hundred. Clear? Now, which key do you press on the calculator here: l n, or log? The natural log, l n! Oh no, careful with the log key: log to the base ten would make the answer two point three zero three times too small. Natural log of four is one point three eight six. So, E a equals eight point three one four, times one point three eight six, times four thousand eight hundred. Got it? That is about fifty five thousand three hundred joules per mole; the wavy equals sign means approximately. Divide by one thousand: fifty five point three kilojoules per mole. See? Just a twenty kelvin rise made the reaction four times faster. Perfect!

8:08Quick revision: 0.693/k, 75% done = 2 half-lives, Arrhenius

So, quick revision. First order: t half equals zero point six nine three over k, the same for any starting concentration. x percent complete leaves one hundred minus x percent, and for first order, seventy five percent complete is two half-lives. And Arrhenius at two temperatures: natural log of k two over k one equals E a over R, times one over T one, minus one over T two, with T in kelvin. Now go and try the practice problems on this page. All the best!