At some temperature, Kp = 1.3 x 105 for the reaction: H2(g) +Cl2(g) <---> 2 HCl(g) If the equilibrium partial pressures ofCl2 and HCl are 0.0027 bar and 0.89 ...

A mixture of 0.47 mole of $\mathrm{H}_{2}$ and 3.59 moles of $\mathrm{HCl}$ is heated to $2800^{\circ} \mathrm{C}$. Calculate the equilibrium partial pressures of $\mathrm{H}_{2}, \mathrm{Cl}_{2},$ and $\mathrm{HCl}$ if the total pressure is 2.00 atm. For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) $$ $K_{P}$ is 193 at $2800^{\circ} \mathrm{C}$

So this question gives us the chemical equation s o to seal to gas reverse. Italy reacts to form S O to gas plus C l to gas and tells us that r k p is two point nine one times ten to the third And let's get rid of that and make that a little more clear. Two point nine one times ten to the third. OK, we're also told that the partial pressure of eso tu is one thirty seven tour and the partial pressure of C L too is to eighty five tour and so we can use the equation for KP, which is equal to the pressure of the products. So P s o two times PCL too over the pressure of the reactant eso tu si l too. And this pressure down here, pressure off s O to see l two is what we're trying to find. Yes, we can rearrange this equation to solve for that. So P s o to Seattle to is equal to P s o two times pcl too divided by K p. And so I'll go to a new page here and plug in those values So P s O to see L too is equal to P s O two, which is one thirty seven times PCL too, which is two eighty five, divided by the K P two point nine one times ten to the third. And this equals thirteen point four tour. And that is the partial pressure of S O to seal too.

So in this problem, the pressure of the hydrogen arctic liam has to be found when B he goes to 0.1 to 80 M. And be hcl equals to 0.0 when the radio. So we'll, K P is equal to 1.37 to 10. Initially power 21. So this is the ingredient equation. We c w cl six gas, yes, plus three H two gases, gifts W solids, six HL Yes, Yes. So the first step is to write the equilibrium concern for the balanced chemical equation. This is the current situation chemical equation and the Cleveland coast. It will be ph cl power six to P W C L B. So you re Pwc and not readable field. It's pW six Power six P H two. Our three. So now the exponents are determined by the coefficients in the reaction. Substituting the values of the pressures in the constant will get K. P equals two. We'll get K P equals two. We have developed for K P. That is 1.37 into 10 X. To the power 21. Now we'll have to take the value of would take out develop for pH two. That on the equation that on the calculation comes out to be 3.93 to 10 H to the power minus 9 80 M. So the pressure of hydrogen at the equilibrium is pH two, which is equal to 3.93 to 10 Initiative Afghanistan Native Thanks

We're trying to find the direction of spontaneous spontaneity of the following reaction di atomic hydrogen gas and die atomic, Korean gas Go to make two moles of hcl gas and we've been given thes things down here Been given that the temperatures to 98 Calvin we confined from our appendix d inner textbook that Delta RG not of just hcl gas is negative. 92.31 killed joules per mole. And we know the partial pressures of all of these gases that are problem 0.5 bar. So we're gonna first find Delta RG not for the entire reaction. And we can do that using this guy right here. We're essentially going to set up a, um, problem where we can find this by taking the reacted minus the are the products minister react. It's, um And since we have these as our reactant, they both have Delta are gene on values of zero. Then we're going to use this value right here. We'll split by two because there's two moles of problems gonna find that that is two times that value. Negative too negative. 92.321 So are delta RG not for the entire problem is negative. 1 84 0.62 coaches purple. So that's our first up Done. The next step we're gonna find que you relate that back to Delta RG later. Who could do this? Using these partial pressures right here we're gonna do is, uh, ratio of parts of a reactive. Um and essentially, we're gonna have 0.5 square on top because we've got two moles of HCL and then teach to conceal to each have a partial pressure of 0.5 and they are the reactions. They go on bottom. So this is essentially going to be que of one nice and easy. It's the same on top and bottom was one. So that is simple. And now we're gonna put all together that and in this following reaction equation, we're gonna find dot org and that's equal to Delta RG not plus rt lnk. Ellen que I'm sorry. And we know of these things. You found this? We know this. This is given. And then we just found this. So we're gonna plug away adult RG again is equal to that value refund. Earlier 1 84 wait six to negative. That is plus dear quits here. 0831 That's always constant. You threw him for times to get Calvin our problem. I many times natural log of one and easy thing about this, the natural log of one zero. So that actually makes this whole term you call this year. Oh, plus negative 1 84 You doing that? Sorry. 0.62 plus zero. So that means that are dealt RG for This problem is negative. 1 84.62 just like our Delta RG not since this is negative. We calculated it. Um, with this as the product, when we set up our cue and when we set up, our adults already know it and these as the reactant. Then we could be certain that this reaction is spontaneous in the forward direction because we set up our equations all like that. And since we got such a origin, negative tell top, she value