Consider the reaction: I2(g) + Cl2(g) ⇌ 2 ICl(g) K_p = 81.9 at 25 °C Calculate ΔG rxn for the reaction at 25 °C under each of the following conditions: c. PICl = 2.55 atm; PI2 = 0.325 atm; PCl2 = 0.221 atm

Video transcript

Welcome back everyone. Our next problem says, consider the reaction I two in the gas form plus CL two in the gas form, irreversible reaction forms to IC L in a gas form with KP equaling 81.9 at 25 °C. Calculate delta G of the reaction for the reaction at 25 °C under the following conditions. P IC L 2.55 atmospheres, P I 20.325 atmospheres, PC L 20.221 atmosphere. A 0.3 kilojoules, B 0.5 kilojoules, C 1.3 kilojoules or D 1.8 kilojoules. So we are looking for delta G of the reactions, the change in free energy at a specific set of conditions, 25 °C and then these constant partial pressures of the different gaseous reactants and products. So the equation we need to use is delta G of reaction equals delta G and standard conditions. So delta G with that little degree symbol plus RT and then the natural log of Q. So delta G zero or delta G standard would be that delta G at equilibrium conditions standard conditions, which means everything equilibrium. And then RT LNQ Q being the reaction constant or excuse me, the reaction quotient cue, which would be of course equal to KP at equilibrium. So let's think about these amounts. Here R is our gas constant T is our temperature, which is given to us, we need to calculate delta G, understand our conditions. And Q to be able to solve this. A quick note on temperature will need the units in Kelvin. So we'll just briefly convert that. So we have 25 °C plus 273.15. It is the Kelvin amount. So our temperatures going to equal 298.15 Kelvin. And a quick note that since this is an addition problem, our last significant figure is the one's place. So we'll just list that as 298 Kelvin, we've got our temperature in the correct units. Let's work on Delta G standard. The equation for that is negative RT times the natural log of KP. Note that we have this negative sign here. I'm going to put brackets around this expression because when we have a negative sign right next to an equal sign, it's all too easy to lose track of it. And that will really mess you up down the line. So let's plug in our numbers here. So we have negative and then our R value we're working with gasses. So we want to for R, use the, the value of R with the units liters of atmospheres, liters atmospheres per mole Kelvin. Except that note what we're actually solving for. We're solving for delta G, which is an energy. So we actually want to use the other value of R with jewels in it since we want to end up with units of jewels or kilojoules. So put 8.314 jewels for more. Kelvin. Now note that all my answers are in kilojoules. So I'll go ahead and throw in a conversion factor to end up with a value in kilojoules here. So we'll end up writing one kill a jewel. So now we're multiplying by our conversion factor one kilojoule divided by 1000 joules. Now we need to keep multiplying here. So we will multiply it by the temperature which is 298 Kelvin. And finally, the natural log of KP which is given to us. So 81.9. So let's solve that there. I'll go ahead and note here the natural log so that you can just double check and make sure you're getting the correct answer in your calculations. I have for the natural log of 81.9 it equals 4.4055. So that we calculate that delta G standard is equal to. After we plug all those numbers into our calculator, we get negative. Don't forget that negative there 10.9 kilojoules and that's again after rounding off to get our three significant figures. So we have our delta G zero or delta G standard. We just need to calculate Q our reaction quotient. We know that that equals at a particular point, the concentration of the products divided by the concentration of the reactants. And remembering that the coefficients in equation will be exponents in our calculations. So in the numer reader, we have the concentration of the product, there's just one note that we have a coefficient of two. Now we're working with gasses. So our concentrations were using the partial pressures. So the partial pressure of it's going to be the partial pressure of CL or excuse me IC L these I's and L's get a little confusing and that will be squared because it has that coefficient of two. And that's going to be divided by the partial pressure of I two multiplied by the partial pressure of CL. So when we plug in those numbers, partial pressure of IC was given to us in our problem 2.55 we're losing our units again because like equilibrium constants, this is expressed unit list. Since the numbers are actually ratios of the given amount to one atmosphere, the condition under standard conditions. So we have 2.55 squared. And in the denominator, we have those given P I two which is 0.325 and PC which is 0.221 and when we plug those numbers in the value we get for Q is nine 0.533. So now we've got the two things we need to plug into our reaction. So we have delta G plug into our equation for the delta G reaction is equal to delta G understand conditions which is negative 10.9 kilojoules plus RT LNQ. So R once again, 8.314 jules, Primo Kelvin, and then we'll throw in our conversion factor to kilojoules. And I'm just going to move my answer to Q out of the way because I'm going to run into it. Otherwise put that up here equals 90.533. See if we can squeeze that in and we can't, and then we'll convert our 8.314 to kilojoules. So we add the conversion factor of one kilojoule divided by 1000 joules multiplied by our temperature of 298 Kelvin and then multiplied by the natural log of our calculated Q value. So 90.533. Now when I calculate that natural log, I'm going to get the answer of 4.5057. If I give that to you, so that you can just double check yourself before you enter these numbers into your calculator. Let me get delta G. Every action is equal to negative 10.9 kilojoules. Oops, there are 10.9 kilojoules plus 11.169 kilojoules. Now remember I'm going to get my answer here equals 0.2488 kilojoules. But when I look at my significant figures, this is an addition problem. So in terms of our city event figures, we can only go to one decimal point to the 10th place. So round it off, we're going to answer zero point three kilojoules for our answer, we go back to where the answer choices. We see that choice A is 0.3 kilojoules. One really important note here not to get mixed up. Note that when we solve that equation for delta G understanding conditions, we have that negative sign negative RT times the natural log of KP. But when we plug that all into our delta G reaction, we do not have a negative in front of the RT LNQ. So really important to keep that straight um where you have a negative and where you don't. But again, we work out two, our delta G of our reaction with the given conditions is 0.3 kilojoules. See you in the next video.