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Consider the reactionCO(g) + Cl2(g)COCl2(g)Using standard thermodynamic data at 298K, calculate the entropychange for the surroundings when 2.46 moles of CO(g) reac...

Question

Consider the reactionCO(g) + Cl2(g)COCl2(g)Using standard thermodynamic data at 298K, calculate the entropychange for the surroundings when 2.46 moles of CO(g) react atstandard conditions. S°surroundings =

Consider the reaction CO(g) + Cl2(g)COCl2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.46 moles of CO(g) react at standard conditions. S°surroundings =



Answers

Calculate the entropy change in surroundings when $1.00 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is formed under standard conditions. $\Delta_{f} H^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Of. So for this question, you want to know if Delta age of reaction is minus 112 killed, Jules and one temperature will dealt SF system. The Don't ask that the system equal deltas have surroundings and were given that Delta s of the system is 354. Jules Perth. Calvin, save us some trouble. Later on, I'm going to go ahead and cover this Jules to killer jewels. So this is equivalent Thio by the 19. About 1000. This is equivalent to three points. 0.354 killed Jules for heaven. And so how are these two related? Well, looking after a textbook you can find that Delta H is equal to you're right by the delta s of surroundings is equal to minus dealt it h over tea. And so what you need to do is just treat delta surroundings is equal to Delta s subsistence. So if we said but to move into each other and plug in actual values, we get this. Zero point 35 four killed. Jules for Calvin equals minus negative 112 killed Jules ever t It's out. It's positive. And from here, you just do algebra. So I see your point 354 t equals 112. Divide both sides by that and you get the answer. So 112 ever see her? Uh, for because t they've killed Jules over kill Jules. Per Kelvin. So this is just making sure that units cancel out. And so you get 316. Kelvin. Um, 300 in 16.38 to be exact. Says the temperature. We're adults s of systems equals delta s of surroundings.

This question is about our reaction has dealt a hate our reaction to be 1123 to do more and down to heads to be 35 for Jill Carroll. So the question is asking mouse about the temperature. Ah, Well, uh, Jinyan Entra people, the reaction is because to continue in interview police around like they already know that Gotta hess around the hex. You rivers is the question. I got a hair. So the system loss down that's ahead of us. Okay, Now, for the reaction we can calculate, which is a system calculations temperature. So we know that the hands, of course to Dr Hate or team and here were given the doubt has to be 354 Joe's for camping is across two. Gotta hedges minus one wants you in here. We can just, uh you know, the sigh. You can see we are not interested in Yeah, kind off reaction, which is, uh, except chemical and attendants. And we're just going to say it's won't want you. This is killer Joe. So to convert your job because this is a job much applied by turns to part three. Joe Calvin. Yes, sir. This is just pure adjusted out job divided by temperature, so temperature will be Won't want you. Time stands 53 Joe's divided by 354 Joe Carroll and distribute three waas 6.2 hits So this is the temperature.

Just before we start this problem in the original problem. Uh, this right here in front of hydrogen is a 1/2. And if we were to make this equation balance, this would have to be three halves. If not, then this would just be a nature end. That's not what that says. So the goal of this problem is to find the change in entropy for this equation. And again, you're gonna have to do some work on your own. Either there is a table in your book, or you can find an online of the standard enter peas of each given element or compound. So let's begin by naming the standard entropy for and H three over here, which is going to be 1 92 point for five Jules over moles times, Kelvin and then for nitrogen. We have 1 91.61 Jules over moles times, Calvin. And then lastly for hydrogen, we have 1 30 time points 46 before shoes rules. Such time's coming. Okay, so what you're gonna do is you're going Teoh, multiply the entropy is the standard. Enter peas by its coefficient. So this one we're going to divide it by two for hydrogen. We're going, Teoh. Multiply it by three halves. And this were just going to leave alone. This is all good. Okay, so the standard enter Pete for the new nitrogen is now 95 point eight No. Five. The new standard entropy for hydrogen is going to be 1 96 0.86 And we're just going to do products minus reactant since get 1 92.45 minus to 91 point 83 one. This is B negative. 99.38 Jules over Moles. That's it. Thank you.

In this problem, we are given the values for the change in m pulpy, built th in the change in entropy Delta s at Sander conditions for a certain reaction. And we want to find what the temperature T is at which the bill to us of the reaction is equal to be built us of these surroundings. This is the equation to find deltas of the surroundings. And since we were told in the problem that we want to know the temperature at which l test of the surroundings is equal to delta us of the reaction and were given the value of Delta s reaction, we can just substitute that value in four Delta s of surroundings and for this equation and solve for the temperature. So when we do that and rearrange, we find that the temperature is equal to negative delta. Each of the reaction divided by don't s of the reaction. And now we plug in the values that were given to solve for the temperature. So there's a negative sign out in front of the numerator and the delta h were given is negative 112 times 10 to the third jewels. We can convert into Jules so that we can cancel off units with the denominator, which will be the value of Delta s reaction, which is also equal to Delta us of the surroundings 354 Jules per kelvin. So we see that when we cancel out these negatives, the temperature results in a positive value, which makes sense because we have units of Kelvin. After we cancel off jewels and bring that up to the numerator, Kelvin units can only be positive. So we just divide those two quantities and we find that the temperature comes out to be 316 Kelvin.


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