Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: webassignbuddy on October 23, 2013, 10:44:50 AM
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Given:
K1 = 1.00 x 10-14
T1 = 298.15 K
K2 = 3.12 x 10-15
T2 = 283.15 K
ΔH° = ?
ΔS° = ?
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) = ΔH - TΔS
What do I do though??? I have 2 K's! And the equation
ln(K2/K1) = (ΔH/R)(1/T2 - 1/T2) doesn't give me the right ΔH value!
The answer for ΔH° is supposed to +54.4.
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Anyone willing to help?
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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values
regards
Ingo
p.s.:"really complicated" ? you ain't seen nothin' yet...
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Your Van't Hoff equation is wrong. Using the right one, I get pretty close to your right answer. Also, don't forget to include the units.
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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values
regards
Ingo
p.s.:"really complicated" ? you ain't seen nothin' yet...
Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)? and what do you use it to solve?
It isn't on the equation sheet my teacher gave our class :(
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Van't Hoff is derived from the ΔG expression, not the other way around.
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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values
regards
Ingo
p.s.:"really complicated" ? you ain't seen nothin' yet...
Wait but...you mentioned 2 equations and 2 unknowns, right?
ΔG = ΔH - TΔS
-RT ln K = ΔH - TΔS
So,
-(0.08314)(298.15)ln(1.00 x 10-14) = ΔH - (298.15)ΔS
+[-(0.08314)(283.15)ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
––––––––––––––––––––––––––––––––––––––––––––––––
Nothing on the right side of the equation cancels so that I can solve for ΔH first ???
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Van't Hoff is derived from the ΔG expression, not the other way around.
Ok I'm more confused...can you please show me the equation I'm supposed to use? Because the main ΔG expression I think I'm supposed to use is ΔG = ΔH - TΔS
which is basically -RT ln K = ΔH - TΔS.
But I don't know where to go from there because that equation has the 2 unknowns that I'm looking for, and I can't get TΔS to cancel in the equation i posted above this post because the T's have 2 different values.
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Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)?
It isn't on the equation sheet my teacher gave our class :(
take a look at your own equations:
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) = ΔH - TΔS
now: try to combine those two equations
shouldn't be too hard to do...
and what do you use it to solve
with K(T) given, calculate [itex]\Delta[/itex] G (T) belonging to , for example:
K298 K = 10-14 :rarrow: [itex]\Delta G[/itex]298 K = - RT lnK = - R * 298 * ln (10-14)
so, you calculate [itex]\Delta G[/itex]298 K and [itex]\Delta G[/itex]283 K
... and put up your secondary equaitions of the general structure ΔG = ΔH - TΔS
for example: [itex]\Delta G[/itex]283 K = ΔH - 283 K* ΔS
:rarrow: two equations with two unknowns resulting thereof
regards
Ingo
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solve like this??
-(0.08314)(283.15)ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
-[-(0.08314)(298.15)ln(1.00 x 10-14) = ΔH - (298.15)ΔS]
Which becomes:
-23.54ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
-[-24.04ln(1.00 x 10-14) = ΔH - (298.15)ΔS]
Which becomes:
(786.25) = ΔH - (283.15)ΔS
-(774.95) = ΔH - (298.15)ΔS
________________________
(786.25)-(774.95) = 0 - (283.15)ΔS + (298.15)ΔS
11.3 = 15ΔS
11.3/15 = ΔS
0.753 = ΔS
^ That doesn't make any sense
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There are multiple, essentially equivalent ways to solve this problem. Here's how I did it.
First use the Van't Hoff equation to solve for ΔH°.
Then because ΔG° equals both -RT ln K AND ΔH° - TΔS°, therefore -RT ln K = ΔH° - TΔS°. You can insert your ΔH° value into this and use one of your K and T data pairs. Solve for ΔS°. Done. (Doesn't matter what K/T pair you use; ΔH° and ΔS° are approximated as roughly independent of temperature, which you'll see if you put in the numbers.)
Or you can solve directly from the -RT ln K = ΔH° - TΔS° equation, two equations with two unknowns. Both are essentially the same because the Van't Hoff equation is basically derived by integrating -RT ln K = ΔH° - TΔS°.
I got ΔH° = 54,400 J/mol and ΔS° = ~-85.6 J/mol K
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There are multiple, essentially equivalent ways to solve this problem. Here's how I did it.
First use the Van't Hoff equation to solve for ΔH°.
Then because ΔG° equals both -RT ln K AND ΔH° - TΔS°, therefore -RT ln K = ΔH° - TΔS°. You can insert your ΔH° value into this and use one of your K and T data pairs. Solve for ΔS°. Done. (Doesn't matter what K/T pair you use; ΔH° and ΔS° are approximated as roughly independent of temperature, which you'll see if you put in the numbers.)
Or you can solve directly from the -RT ln K = ΔH° - TΔS° equation, two equations with two unknowns. Both are essentially the same because the Van't Hoff equation is basically derived by integrating -RT ln K = ΔH° - TΔS°.
I got ΔH° = 54,400 J/mol and ΔS° = ~-85.6 J/mol K
Yeah but THAT'S the problem though....i have absolutely no idea how to go about solving for ΔH FIRST because I'm having trouble manipulating a specific formula that I'm supposed to be using... :(
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Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)?
It isn't on the equation sheet my teacher gave our class :(
take a look at your own equations:
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) = ΔH - TΔS
now: try to combine those two equations
shouldn't be too hard to do...
and what do you use it to solve
with K(T) given, calculate [itex]\Delta[/itex] G (T) belonging to , for example:
K298 K = 10-14 :rarrow: [itex]\Delta G[/itex]298 K = - RT lnK = - R * 298 * ln (10-14)
so, you calculate [itex]\Delta G[/itex]298 K and [itex]\Delta G[/itex]283 K
... and put up your secondary equaitions of the general structure ΔG = ΔH - TΔS
for example: [itex]\Delta G[/itex]283 K = ΔH - 283 K* ΔS
:rarrow: two equations with two unknowns resulting thereof
regards
Ingo
I know that much but do I set TΔS = to 0 for both??
I did this:
ΔG298.15 K = 774.95
ΔG283.15 K = 786.25
I know how to do that no problem.
But what I DON'T know is how to set up both equations in such a way that I can find ONE of the variables I'm looking for, especially when none of them cancel out...
(786.25) = ΔH - (283.15)ΔS
-(774.95) = ΔH - (298.15)ΔS
^ when I do this, I get an extremely small number for ΔS which is completely 100% wrong.
(786.25) = ΔH - (283.15)ΔS
+(774.95) = ΔH - (298.15)ΔS
When I do this, i go nowhere...because I end up with 1561.12 = 2 ΔH - 581.3 ΔS. I STILL have two unknown variables.
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This is just plug and chug, so there's really not much I can do to help you.
You know
[tex]\ln \frac{k_1}{k_2}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
Put in your values for k1, k2, T2, T1, solve for ΔH°. If you're not getting the right answer, you're making a basic arithmetic or algebra mistake.
EDIT: In the post above, try subtracting the equations rather than adding them, if you want to do the direct method.
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This is just plug and chug, so there's really not much I can do to help you.
You know
[tex]\ln \frac{k_1}{k_2}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
Put in your values for k1, k2, T2, T1, solve for H°. If you're not getting the right answer, you're making a basic arithmetic or algebra mistake.
I tried that originally but for some reason didn't get the right answer on paper and then I erased it. I'm gonna try it again.
Also,
R = 0.08314 kJ/mol Kthough, right? not 8.314 J/mol K?
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8.31446 J/mol K. You can convert to kJ/mol K, but it's not 0.0831446; try 0.00831446. Looks like you need to start being a lot more careful with the basics. This is why you are not getting correct answers.
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8.31446 J/mol K. You can convert to kJ/mol K, but it's not 0.0831446; try 0.00831446. Looks like you need to start being a lot more careful with the basics. This is why you are not getting correct answers.
Ok thank you!
I'm gonna try it out
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LOL, and then I go and I type it into LA Tex wrong. Try this one. :D
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
(I put the k2 and k1 in the wrong order earlier...)
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(...)
^ That doesn't make any sense
your right here: numbers without dimensions are meaningless (and you better should have put in dimensions in your calculatons right from the start), esp. if you miscalculate additionally
... and by doing better, my results next to perfectly match with @Corribus' results
regards
Ingo
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LOL, and then I go and I type it into LA Tex wrong. Try this one. :D
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
(I put the k2 and k1 in the wrong order earlier...)
Wait but, I know T2 and T1, But what does T equal in ΔH's denominator?
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/facepalm
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
That's what I get for scolding you about being careful...
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/facepalm
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
That's what I get for scolding you about being careful...
Ohh...haha thank you!
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/facepalm
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
That's what I get for scolding you about being careful...
Ok so I'm stumped again.
I got ΔH fine (54.7 kJ/mol) but for some reason I'm getting ΔS = 2.41.
I did.
ΔG298.15 K = ΔH - TΔS
774.95 = 54.7 - (298.15)ΔS
774.95 - 54.7 = -(298.15)ΔS
720.25 = -(298.15)ΔS
720.25/(-298.15) = ΔS
-2.4157 = ΔS
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man, use DIMENSIONS!
and RECALCULATE ΔG298,15K : you miscalculated there!
regards
Ingo
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man, use DIMENSIONS!
and RECALCULATE ΔG298,15K : you miscalculated there!
regards
Ingo
ok...
ΔG298.15 K = -RTln K
ΔG298.15 K = -(0.008314 kJ/mol K)(298.15 K)ln (1.00 x 10-14)
ΔG298.15 K = -(0.008314 kJ/mol K)(298.15 K)(-32.236)
ΔG298.15 K = (-2.478 kJ/mol)(-32.236)
ΔG298.15 K = 79.9?
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ΔG298.15 K = 79.9?
79.9 of WHAT ? EGGS? years of unconciousness?
USE DIMENSIONS ... else you NEVER get the difference ΔG(T) - ΔH right, and hence NEVER will find a meaningfull solution for ΔS
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ΔG298.15 K = 79.9?
79.9 of WHAT ? EGGS? years of unconciousness?
USE DIMENSIONS ... else you NEVER get the difference ΔG(T) - ΔH right, and hence NEVER will find a meaningfull solution for ΔS
79.9 kJ/mol :)
And I got the right answer for ΔS and I ALSO included dimensions!
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congratulations!
regards
Ingo
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LOL, well done. How many more problems in your problem set now? :)
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LOL, well done. How many more problems in your problem set now? :)
2 more! :D
And thanks again to both of you for ALL of the help though and the clarification!
This next one is kind of throwing me off also...
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi11.photobucket.com%2Falbums%2Fa179%2Fslourg%2FNumber5_zps13446b72.png&hash=9846691b8b59cbbe801c9885c0e13916f8951885)
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Ok so I'm given:
KD = 0.520
T = 298.15 K
[RP]initial = 0.0870 M
My initial guess was to just do KD = [products]/[reactants] = [R][P] / [RP]
But I have to know the SPECIFIC concentrations for [RP] and [P] at 25° C (or 298.15 K)
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I'm having trouble with determining exactly what equation contains the variables I'm given...
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Writing out the entire question with exact wording will help.
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Writing out the entire question with exact wording will help.
Oh wait...did the picture not show up?
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Possibly it did. My ISP here blocks a lot of pictures, particularly if they're at file hosting sites like photobucket.
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Possibly it did. My ISP here blocks a lot of pictures, particularly if they're at file hosting sites like photobucket.
Oh ok I'll see if I can copy and paste.
V. The antibiotic paromomycin (P) forms a complex with a particular 27-nucleotide RNA construct (R), (see Anal. Biochem. 2000, 280(2), 264-71). Dissociation of this complex can be described schematically (in aqueous solution) as:
RP :resonance: R + P KD = 0.520* at 25ºC
where the equilibrium constant KD is commonly referred to as the dissociation constant. Assuming one begins with 0.0870M of RP complex (only):
(a) Determine the equilibrium amounts of [RP] and [P] at 25ºC.
(b) Suppose that the value of KD changes to 1.47* when the temperature is raised to 37 ºC. Does the equilibrium of part (a) shift to the right (R); shift to the left (L); or remained unchanged (U).
(c) Is this process (reaction) exothermic or endothermic or is H = zero at equilibrium (put correct answer/word in the blank)?
(d) Starting from the final equilibrium amounts determined in part (a) at 25ºC, determine the equilibrium amounts of [RP] and [P] if the temperature is raised to 37ºC.
Hints: quadratics should use the positive roots and keep three significant digits. If pressed for time, set up rxn table/quadratics for partial credit. (25 points)
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Alright. Part a.
You know the initial amounts of RP, R and P. Can you set up an expression for the final amounts at equilibrium?
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Alright. Part a.
You know the initial amounts of RP, R and P. Can you set up an expression for the final amounts at equilibrium?
I can try.
Step 1) Make a table (?)
RP :resonance: R + P
initial: 0.0870 M 0 M 0 M
Δ: -x +x +x
final: 0.0870 - x x x
Step 2) Kd = [R][P]/[RP]
0.520 = (x)(x)/(0.0870 - x)
0.520 = x2/(0.0870 - x)
0.520(0.0870 - x) = x2
0.04524 - 0.520x = x2
0 = x2 + 0.520x - 0.04524
Step 3) Quadratic formula to solve for "x"
x = [-b ± √(b2 - 4ac)] / 2a
a = 1
b = 0.520
c = 0.04524
x = [-0.520 ± √(0.5202 - 4(1)(0.04524)] / 2(1)
x = [-0.520 ± √(0.2704 - 0.18096)]/2
x = [-0.520 ± √(0.48827)]/2
x = [-0.520 ± 0.6987]/2
x = [-0.520 + 0.6987]/2 = 0.178/2 = 0.089
x = [-0.520 - 0.6987]/2 = -1.2187/2 = -0.609[/s]
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Oh wait...did the picture not show up?
Attach pictures to your posts on the forum - click on "Additional options" below the edit field in which you type your message. This is the best way, as your posts are not depending on other systems, if the forum works - images show.
Not to mention the fact we prefer question to be typed, not posted as images. When they are typed they become searcheable.
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Oh wait...did the picture not show up?
Attach pictures to your posts on the forum - click on "Additional options" below the edit field in which you type your message. This is the best way, as your posts are not depending on other systems, if the forum works - images show.
Oh the picture is showing for me, it just wasn't showing for Corribus for some reason
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Oh wait...did the picture not show up?
Attach pictures to your posts on the forum - click on "Additional options" below the edit field in which you type your message. This is the best way, as your posts are not depending on other systems, if the forum works - images show.
Not to mention the fact we prefer question to be typed, not posted as images.
Oh ok I'll keep that in mind!
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@webassign
Does your answer make sense?
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@webassign
Does your answer make sense?
No it doesnt...
My answer is wrong...it's kind of close to the actual answer though. Here are the answers:
a)
[RP]25º = 0.0111
[P]25º = 0.0759
b) At 37ºC, equilibrium shifts to the Right
c) Reaction is Endothermic
d)
[RP]37º = 0.00459
[P]37º = 0.0824
Hmm....maybe I did my table wrong in part a?
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Well I didn't ask if it was wrong; I asked if it makes sense. Not exactly the same thing. It's nice to have the answer ahead of time for checking if you did the problem right, but this will not always be the case. It's good to develop an understanding of whether an answer makes sense. This is a quick and easy way to determine if your answer is likely to be incorrect. Your answer (positive root, 0.089) can't possibly be right, because if it was, your equilibrium concentration of RP would be negative, which is impossible.
Now, how do we troubleshoot where you went wrong? Here's an idea. Take your root, 0.089, and put it back in your quadratic equation. Does the right hand side equal zero? If yes, you know you solved the quadratic equation correctly, and it is likely your table is wrong. If no, you know you've made some algebra mistake when solving your equation. This doesn't mean your table is necessarily right (you could make mistakes in more than one place), but I'm trying to teach you how to troubleshoot your own wrong answers and find out where you went astray.
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Well I didn't ask if it was wrong; I asked if it makes sense. Not exactly the same thing. It's nice to have the answer ahead of time for checking if you did the problem right, but this will not always be the case. It's good to develop an understanding of whether an answer makes sense. This is a quick and easy way to determine if your answer is likely to be incorrect. Your answer (positive root, 0.089) can't possibly be right, because if it was, your equilibrium concentration of RP would be negative, which is impossible.
Now, how do we troubleshoot where you went wrong? Here's an idea. Take your root, 0.089, and put it back in your quadratic equation. Does the right hand side equal zero? If yes, you know you solved the quadratic equation correctly, and it is likely your table is wrong. If no, you know you've made some algebra mistake when solving your equation. This doesn't mean your table is necessarily right (you could make mistakes in more than one place), but I'm trying to teach you how to troubleshoot your own wrong answers and find out where you went astray.
Well I plugged it back in...
0 = (0.089)2 + 0.520(0.089) - 0.04524
0 = 0.007921 + 0.04628 - 0.04525
0 ≠ 0.008961
ehh...
I'm gonna check my algebra
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Right. So let's assume for the moment you set the problem up right and only made a math mistake, because you clearly made a math mistake. At this point all you can do is scrutinize it until you figure out what you did wrong. I'll help you out. Your "a, b and c" values you used for the quadratic equation: one of them is not right.
Beyond that, here's a tip: if you know for a fact your answer isn't right, always look first to see if you made a math mistake before you question whether you approached the problem incorrectly. Only once you've ruled out a math error should you start to question your conceptual approach to the problem.
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Right. So let's assume for the moment you set the problem up right and only made a math mistake, because you clearly made a math mistake. At this point all you can do is scrutinize it until you figure out what you did wrong. I'll help you out. Your "a, b and c" values you used for the quadratic equation: one of them is not right.
Beyond that, here's a tip: if you know for a fact your answer isn't wrong, always look first to see if you made a math mistake before you question whether you approached the problem incorrectly. Only once you've ruled out a math error should you start to question your conceptual approach to the problem.
Yeah it was my algebra :(:(:(:(:(:(:(:(:( :'(
Ughhhhh. I can't afford to make these careless mistakes on the test.
I used an online quadratic calculator and input all of my A B and C values and it got x = 0.0759, which is the answer...
And c is supposed to be -0.04524
At least the table was right :D!
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Good. I hope you see how, lacking the correct answer as a check, you could easily go through and find that you got an answer that didn't make sense, and then troubleshooted to see where your mistake was. BTW, if you have a graphing calculator, most have a root function for a polynomial which can be helpful to avoid that kind of error.
Now, onto part B: What are your thoughts here? I suppose we'll verify your answer in part D, so you can go ahead and take a stab at that while you're at it. (You have the answer to B already, but maybe you can think of WHY the Eq shifts to the right when the temperature is raised.
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Good. I hope you see how, lacking the correct answer as a check, you could easily go through and find that you got an answer that didn't make sense, and then troubleshooted to see where your mistake was. BTW, if you have a graphing calculator, most have a root function for a polynomial which can be helpful to avoid that kind of error.
Now, onto part B: What are your thoughts here? I suppose we'll verify your answer in part D, so you can go ahead and take a stab at that while you're at it. (You have the answer to B already, but maybe you can think of WHY the Eq shifts to the right when the temperature is raised.
For part B, I was actually gonna ask you :D But I'll take a guess...
Well if the Kd (which is referred to as the DISSOCIATION CONSTANT) of the RP is HIGHER, than MORE products are formed, which would mean that the product side is favored and thus the equilibrium shifts to the right.
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Seems like a reasonable answer to me.
What about C?
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Seems like a reasonable answer to me.
What about C?
Reaction is endothermic because the INPUT of heat is required to BREAK bonds and for RP to dissociate into R and P (?)
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Input of heat is almost always required to drive a reaction initially, be it endothermic or exothermic. It's not a bad thought, but there's a way to show it explicitly. Here's a hint: you're given two equilibrium constants at two temperatures. Note: you don't actually have to calculate anything, just determine whether the enthalpy change is positive or negative.
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Input of heat is almost always required to drive a reaction initially, be it endothermic or exothermic. It's not a bad thought, but there's a way to show it explicitly. Here's a hint: you're given two equilibrium constants at two temperatures. Note: you don't actually have to calculate anything, just determine whether the entropy change is positive or negative.
The entropy change should be negative, which would mean that Gibbs free energy would be positive (based on the equation ΔG = ΔH - TΔS ), and the reaction would be endothermic.
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Try again:
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
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Try again:
[tex]\ln \frac{k_2}{k_1}=\frac{-\Delta H^o}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]
Ohhh, if you plug everything in and solve for ΔH, it would be a positive value, which means that it's endothermic.
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Exactly. But you don't actually have to plug anything in. Let K2 be the higher temperature and K1 be the lower temperature. THis means K2/K1 is greater than one, and the log of any number greater than 1 is positive. Thus the left hand side is positive. Likewise, the temperature term has to be negative because if T2>T1, then the reciprocal of T2 has to be smaller than the reciprocal of T1. Just by looking at the signs, then, it's easy to see how the enthalpy change must be positive, or the reaction is endothermic. In general, if the equilibrium constant increases when the temperature increases, the process is endothermic. There's a way to rationalize this with a reaction coordinate diagram, but I wouldn't worry about it now. Just remember the Van't Hoff eqn and you can't go wrong.
Now have a shot at D. Should be pretty easy, since you should solve it exactly like you did A.
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Exactly. But you don't actually have to plug anything in. Let K2 be the higher temperature and K1 be the lower temperature. THis means K2/K1 is greater than one, and the log of any number greater than 1 is positive. Thus the left hand side is positive. Likewise, the temperature term has to be negative because if T2>T1, then the reciprocal of T2 has to be smaller than the reciprocal of T1. Just by looking at the signs, then, it's easy to see how the enthalpy change must be positive, or the reaction is endothermic. In general, if the equilibrium constant increases when the temperature increases, the process is endothermic. There's a way to rationalize this with a reaction coordinate diagram, but I wouldn't worry about it now. Just remember the Van't Hoff eqn and you can't go wrong.
Now have a shot at D. Should be pretty easy, since you should solve it exactly like you did A.
I wrote down that Van't Hoff equation in bold letters and with a pen on a flashcard :D Definitely will not forget that one.
And my teacher told us to skip d since it's pretty much the same as a except with a different K value (1.47).
Thanks a BUNCH for helping me out!!
Imma try to give the last 2 problems a shot and If I need more help I'll most likely come back here... but they seem pretty straightforward
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Sure no problem.