[webassignbuddy]

Ok so I'm given:

K_D = 0.520
T = 298.15 K
[RP]_initial = 0.0870 M

My initial guess was to just do K_D = [products]/[reactants] = [R][P] / [RP]

But I have to know the SPECIFIC concentrations for [RP] and [P] at 25° C (or 298.15 K)

[webassignbuddy]

I'm having trouble with determining exactly what equation contains the variables I'm given...

[Corribus]

Writing out the entire question with exact wording will help.

[webassignbuddy]

Writing out the entire question with exact wording will help.

Oh wait...did the picture not show up?

[Corribus]

Possibly it did. My ISP here blocks a lot of pictures, particularly if they're at file hosting sites like photobucket.

[webassignbuddy]

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  R + P       K_D = 0.520* at 25ºC

where the equilibrium constant K_D 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)

[Corribus]

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?

[webassignbuddy]

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          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]

[Borek]

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.

[webassignbuddy]

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

[webassignbuddy]

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!

[Corribus]

@webassign

Does your answer make sense?

[webassignbuddy]

@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?

[Corribus]

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.

[webassignbuddy]

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