Equilibrium problem, anyone?

In summary: A and B to get the most accurate answer.In summary, the equilibrium concentration of c, for a reaction containing equal volumes of A and B, is 0.65mol/L. The equilibrium concentration of c, for a reaction containing unequal volumes of A and B, is the greater of the two volumes of A and B. The attempt at a solution to solving this problem was a failure. The answer is 0.022 mmol/L.
  • #1
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1. Homework Statement

When equal volumes of A and B are combined in a 3.5L flask, their initial concentrations were each 1.75mol/L. Once equilibrium is reached, the equilibrium concentration of c, is [c] = 0.65mol/L . determine the Kc for this reaction




2. Homework Equations

Usually, there is an initial value given, then the equilibrium value given, then you subttract those and plug them into the rest of the table. However... I don't know what to do inorder to get the answer when jus the eq val is given of c , and no data for d is given (assuming you need to find x first)


3. The Attempt at a Solution

I created the ice table

2A : 1.75-2x
B : 1.75 - x
c: 0.65
d: x


Attempt to solution was a failure. The answer is 0.022

Help please!
 
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  • #2
What is the reaction equation?
 
  • #3
Sorry, the reaction equation is 2A + B ----> 3C + D

I still don't know how to do this problem.

Edit: "2 A (g) + B (g)-----> 3 C (g) + D (g) ΔH0 = -315.9kj
 
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  • #4
Use stoichiometry to calculate equilibrium concentrations of all substances. This will be not much different from creating ICE table.

Can you express C in terms of x?
 
  • #5
Okay... well, here's what I did. Since C is 0.65. And change in A and b is -x .

For A : i did I created the ice table

2A : 1.75-2(0.65) 0.45
B : 1.75 - (0.65) 1.1
c: 0.65
d: 0.0619

n = c/v , 0.65/3.5L 1/3 * that mol value = 0.0619 for d
answer still doesn't come out as 0.022.
 
  • #6
Ive tried it using several methods now... used stoic of 0.65 to calculate all values, then subtracted by x (ended up giving neg values). Used stoic of 0.65 to calculate D , giving 0.216, then used x to subtract the other two values (A and B) . Used 0.65 as X in d and used x to calculate A and B... nothing is working
 
  • #7
Solving the problem is not about plugging numbers randomly in hope something will work.

If you have produced x of D, can you tell - just by looking at stoichiometric coefficients - how much C was produced? Was it x? 2x? 3x? Something else?
 
  • #8
Yes... Like i said..

If X is 0.65 for c. Then d will be 0.65/3 Which will be 0.216.

Change for a : -2x. Change for b : -x. change for c : +3x . Change for d = x

I don't understand what youre trying to tell me

Apparently you neglect any coefficiant in change when you have x. So i did 1.7-0.62 , giving me 1.1 for both a and b. then plugged those values in and got to 0.04456 Which is still wrong. However. 0.04456 / 2 = 0.022. I don't know why you do divided by 2 though.
 
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  • #9
I understand it a bit more. Its saying that 3c= 0.65. for 3 moles of C. making D 0.216.

And A , since your neglecting the change value "2" coefficiant when subtraction. you need to use stoic : 2/3 * 0.65 to get the change avalue for that , which is 0.4333.

then doing 1.7-0.433 to get the equilib value for A. then for b you get 1.7- 0.216 and get that value.

I reinput the numbers and got 0.0245 . Which is a bit off...

figured maybe its a rounding error, made the eq. values whole numbers. got to 0.0233
 
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  • #10
does anyone know...?
 
  • #11
Exact value seems to be 0.022384... Perhaps you used rounded down numbers in your calculations. Unfortunately your explanation of what you did is chaotic to me, I see correct elements, but I am not sure everything is right.
 
  • #12
Okay. So this is what I did.

2A + b ---------> 3C + D.

Since 3C is 0.65 mol/l at equilibrium , per 3 moles, I used stoic calculations fo first figure out D at equilibrium.

Ie) 1/3 * 0.65 = D . Which gave me the value for D.

Then, I used stoic calculations to figure out the equilibrium value of A.

Ie) 2/3 * 0.65 = Mol/L equilibrium of A.
. Since i found the equilibrium value of A, I then subtracted 1.7-0.43 which gave me 1.27.

For B I used 1/3 * 0.65 which gave me 0.216 mol/l at equilibrium of b.

So, I did 1.7 - 0.216 = 1.484 Mol/l of B .

Thus giving me a final keq quation of [0.65]^3[0.216] / [1.27]^2[1.484] = 0.245
 
  • #13
Nelo said:
Ie) 2/3 * 0.65 = Mol/L equilibrium of A.
. Since i found the equilibrium value of A, I then subtracted 1.7-0.43 which gave me 1.27.

Why 1.7 and not 1.75? Why 0.43 and not the exact value?

For B I used 1/3 * 0.65 which gave me 0.216 mol/l at equilibrium of b.

So, I did 1.7 - 0.216 = 1.484 Mol/l of B .

Again, why 1.7? And why 0.216? If anything, 0.21(6) rounds up to 0.217, but you shouldn't use neither of these values.

Use guard digits in your calculations, don't round down intermediate results.
 
  • #14
1. Homework Statement

A mixture of 9.22 moles of A, 1011 moles of B, and 27.83 moles of C is placed in a 1 litre container. constant temp. At equilibrium the number of moles of B is 18.32 Calculate the eq constant for this reaction


2. Homework Equations

A + 2B <-> 3C

3. The Attempt at a Solution

I need help.. on how to solve this.. here's what i did:

Ice table:

A + 2B <-> 3C

9.22 10.11 27.8
+x +2x -3x ( for some reason the change is backwords)
18.32

Since equilib moles of B is 18.32, I did 18.32 - 10.11 , giving me 8.21 moles of 2B at equilibrium. This means that A will be 8.21mol divided by 2 , giving me 4.105. 9.22 + 4.105= 13.32 moles of A at equlibrium.

For C i used stoic, 3/2 * 8.21 mol = 12.315. at equilibrium. So, 27.83 -3(12.315) Gives me a negetive value, So now I am stuck. Wth am i doing wrong?
Help please!
 
  • #15
"For C i used stoic, 3/2 * 8.21 mol = 12.315. at equilibrium. So, 27.83 -3(12.315)"
Huh why are you multiplying with 3 again?? you have already done that at the 3/2 part.

27.83 -12.315 = 15,51 Slow down man :D
 
  • #16
What? No i havent. When did i multiply with 3? Its said that value of equilibrium for 2 moles is 8.21. , ALl i did was 1/2 * 8.21 to give me the value of A. Then went back to the value it is for 2 moles (8.21) and did 3/2 * 8.21 , giving me 12.315. When did i already multiply by 3? oh, yes. So.. i see what youre saying. Why am i supposed to neglect the change on that value? Is it because the stoic ratio GIVES me that *3 value? so i don't need to do "change again" since i already calculated it?
 
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  • #17
Nelo said:
What? No i havent. When did i multiply with 3? Its said that value of equilibrium for 2 moles is 8.21. , ALl i did was 1/2 * 8.21 to give me the value of A. Then went back to the value it is for 2 moles (8.21) and did 3/2 * 8.21 , giving me 12.315. When did i already multiply by 3?

You know that you gain 8.21 moles of b. According to the reaction equation you gain 1/2 moles of A and lose 3/2 moles C for every mole of B you gain. Now you multiply 3/2 * 8.21 to know how many moles of C you lost. That numbers you arrived at was 12.315
Since you lost that amount, you subtract it from how many moles of C you had at the start, which was 27.83

27.83-12.315 = 15.51

When you did your final calculation you said "27.83 -3*(12.315)" and that's the error. 12.315 is the amount of moles of c lost, not 3*12.315

Edit: You ADD/SUB the change, but the change wasn't 3*12.315 but 12.315!
 
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  • #18
I some what get it... not really though, does that mean I never really bother subt/adding the change? only lookign at if its positive or negetive (from what you determine) .

For example: if i say C is instead +3x, the only difference is i add 27.83 + 12.315?

If A was -x I would just subtract by that single value? I don't need to do +2(4.12) etc, since i already have that +2 right?
 
  • #19
What does it mean if the reaction moves to the right or left? I don't understand. Heres a sample question.

C7H14 ----> C7H8 +3H2

Using only the info given. which of the following changes would increase the molar concentration of C7H8

a) increase pressure at constant temp, b) inc temperature , c ) decrease concentration. d) add a catalyst

Apparently the answer is a, but i chose C. I know this relationship. : decreasing the reactant conc will make the reaction move to the left, , does that mean the conc of nh3 will decrease?
 
  • #20
Nelo said:
c ) decrease concentration

Decrease concentration of what? If you decrease concentration, concentration decreases - that's quite logical.

Please try to be less chaotic, it is impossible to help you not knowing what is the question, what is the information given and what you mean.

does that mean the conc of nh3 will decrease?

First: you were already told twice to properly capitalize formulas. Second: assuming you mean NH3, there is no ammonia in the reaction. That probably means you meant something else, but I am not going to guess what.
 
  • #21
What does it mean if the reaction moves to the right or left? I don't understand. Heres a sample question.

C7H14 ----> C7H8 +3H2

Using only the info given. which of the following changes would increase the molar concentration of C7H8

a) increase pressure at constant temp,
b) inc temperature

c ) decrease concentration on C7H14
d) add a catalyst

Apparently the answer is A, but i chose C. I know this relationship. : decreasing the reactant conc will make the reaction move to the left, , does that mean the conc of C7H8 will increase or decrease?
 
  • #22
In general you are right that removing reactant shifts the equilibrium to the left, but it will never go further than the starting point - so decreasing the concentration would not increase it above the initial value.
 
  • #23
How about this one?

23) If 3 moles of HI are placed in a 5 litre vessel and allowed to reach equilibrium, what is the equilibrium concentration of H2?

What i did::

Created the ice table, since values are not at equilibrium change must be represented

Keq is : 0.0183

expression: 0.0183[sqroot] = x^2 / (0.6-x)^2 [sqroot]

I squarerooted the equation to get rid of the power.

I then cross multiplied the 0.1352(which is the sqroot of 0.0183) with 0.6 and -x.

0.08112 -0.1352x = x

Moved x to the other side.

0.08112 = 1.1352x

0.08112 / 1.1352 = 0.0714. However, the answer is 0.064. What have i done wrong?
 
  • #24
You were told to start new threads for new problems. Twice. Topic locked.
 

What is the Equilibrium Problem?

The Equilibrium Problem refers to a situation in which opposing forces or factors are balanced and there is no net change in the system. It is often used in physics and chemistry to describe the state of a system at rest.

Why is the Equilibrium Problem important?

Understanding the Equilibrium Problem is essential for predicting how a system will behave under different conditions. It allows scientists to determine the stability of a system and make predictions about how it will respond to external changes.

What factors can affect the Equilibrium Problem?

Factors such as temperature, pressure, and concentration can all affect the Equilibrium Problem. Changing these variables can shift the balance of the system and result in a new equilibrium state.

How do scientists study the Equilibrium Problem?

Scientists use experimental and theoretical methods to study the Equilibrium Problem. They can manipulate the variables of a system and observe how it responds, or use mathematical models to predict the behavior of a system at equilibrium.

What are some real-life examples of the Equilibrium Problem?

The Equilibrium Problem is seen in many natural and man-made systems. Examples include the balance between predator and prey populations in an ecosystem, the equilibrium between supply and demand in economics, and the equilibrium of chemical reactions in a closed system.

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