K_p and its Relation to K_c

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Two students study the same equilibrium, nitrogen and hydrogen making ammonia, at seven hundred and seventy three kelvin. One reports the equilibrium constant as zero point zero five eight. The other reports one point four four times ten to the power minus five, about four thousand times smaller! So who made the mistake? Hold that thought; by the end of this class, you will settle it yourself. And this kind of conversion comes up often in JEE equilibrium questions. So, let's begin.

0:36Kp: the equilibrium constant in pressures

So, you know K c, written with concentrations. But for gases, there is another natural measure: the partial pressure, small p. Take small a moles of capital A, plus small b moles of capital B, giving small c moles of capital C, plus small d moles of capital D, at equilibrium; that is what the two half arrows mean. With me? So we write the same fraction with partial pressures. K p, capital K with a small p below for pressure, equals p of C to the power small c, times p of D to the power small d, over p of A to the power small a, times p of B to the power small b. Products over reactants, exactly like K c. Okay? Now, K p and K c describe the same equilibrium, so they must be connected. Think of an ideal gas. How is its pressure linked to its concentration? The ideal gas equation, yes, exactly! p equals n over capital V, times R T, where R is the gas constant and capital T is the temperature in kelvin. And n over V is just the concentration; the square brackets with a dot inside stand for it. So the pressure equals the concentration times R T, right? Now put this into K p. Every pressure brings one factor of R T: c plus d of them on top, and a plus b below. So, K p equals K c, times R T to the power delta n g. The little triangle is the capital Greek letter delta, meaning change, and the small g below means gas. Clear? Now, delta n g is a change in moles, so we count every species in the equation, solids and liquids too, right? Oh no, careful! Delta n g equals c plus d, minus the sum a plus b, counting gaseous species only: moles of gaseous products minus moles of gaseous reactants. Got it?

2:28Using Kp = Kc(RT)^Δn: which R, when Kp = Kc

Alright, let's use it. Pressures in atmospheres, concentrations in moles per litre: which value of R do you take? Pressures in atm with concentrations in moles per litre: R equals zero point zero eight two one, litre atmosphere per kelvin per mole. The minus one powers simply mean per kelvin and per mole. Okay? And with pressures in bar, again with moles per litre, use zero point zero eight three one, litre bar per kelvin per mole. Right? Now, a quick question. If delta n g is zero, how are K p and K c related? Equal, yes! Very good! R T to the power zero is one, so K p equals K c. Clear? If delta n g is positive, the greater than sign tells you K p is greater than K c, as long as R T is greater than one. With me? And if delta n g is negative, the less than sign: K p is less than K c, again for R T greater than one. These three cases let you check an answer in a second.

3:31Mole fractions and Kx

Okay? One more form. Each gas has a mole fraction, small x with a small i below, for gas number i, and capital P is the total pressure. Then the partial pressure, p i, equals x i times capital P, right? Put that into K p, and the same algebra gives K p equals K x, times capital P to the power delta n g, where K x is the same fraction written with mole fractions. Clear? Now, a trap. At a fixed temperature, I squeeze the mixture, and the total pressure doubles. So K p changes too, right? No, no, wait! It is K x that changes with the total pressure, because capital P to the power delta n g sits in that formula; unless delta n g is zero, of course. Okay? K p and K c do not; they change only with temperature. The mole fractions readjust, and K p stays exactly the same. Got it?

4:25Example 1: Kp for PCl5 from Kc at 500 K

Now, example one, from the course. Phosphorus pentachloride, P C l five, splits into phosphorus trichloride, P C l three, and chlorine; the little numbers below are subscripts, counting atoms. K c is zero point zero four zero moles per litre at five hundred kelvin, written with a capital K on screen. Find K p. Pause the video and try it first. Done? Okay, let's check. Two moles of gas on the product side, one on the reactant side, so delta n g is two minus one, plus one. So K p equals K c times R T, to the power one. Right? K c is in moles per litre and we want K p in atmospheres, so R is zero point zero eight two one. Zero point zero four zero, times zero point zero eight two one, times five hundred. With me? Before you multiply: what will the units of K p be? Atmospheres, correct! Moles per litre, times litre atmosphere per kelvin per mole, times kelvin: only atmosphere survives. And the number is one point six four, so K p equals one point six four atm. Delta n g is positive, so K p is bigger than K c, as expected.

5:42Example 2: Kc for ammonia synthesis from Kp at 773 K

Okay? Now a harder one, and it is our puzzle reaction. Nitrogen plus three hydrogen gives two ammonia. K p is one point four four times ten to the power minus five, per atmosphere squared, the atm to the minus two on screen, at seven hundred and seventy three kelvin. Find K c. Pause the video and try it. Done? Okay, let's check. Two moles of ammonia, minus the sum one plus three, that is, four moles of reactant gases. So delta n g is minus two. Clear? So, K c equals K p times R T to the power minus two, right? No, no, careful! This is the sign trap. The formula says K p equals K c times R T to the power minus two. Make K c the subject, and the double arrow shows the result: K c equals K p times R T squared. Moving R T across flips the sign of its power. Got it? Now, R T is zero point zero eight two one times seven hundred and seventy three; squared, that is about four thousand and twenty seven. The wavy equals sign means approximately. Okay? So, K c is one point four four times ten to the power minus five, times four thousand and twenty seven: about zero point zero five eight, litre squared per mole squared. Now, our puzzle! Zero point zero five eight was K c, and the tiny number was K p. Same equilibrium, and both students were right! Delta n g is negative, so K p is the smaller one, as expected. Right?

7:20Quick revision: Kp, Kc and counting gaseous moles

So, quick revision. K p is written with partial pressures in place of concentrations. K p equals K c times R T to the power delta n g, counting gaseous moles only. And K x changes with the total pressure, unless delta n g is zero, while K p and K c change only with temperature. Now go and try the practice problems on this page. Count delta n g carefully, and you will find them easy. All the best!