Equilibrium Problem; Senior High School Level

Click For Summary
SUMMARY

The equilibrium problem discussed involves the reaction SO2Cl2(g) ⇌ SO2(g) + Cl2(g) in a 5.00 L container at 100°C, with an equilibrium constant (Kc) of 2.60. Initial concentrations were calculated as [Cl2]initial = 0.530 mol/L and [SO2Cl2]initial = 0.970 mol/L. The solution utilized an I.C.E. table to derive the equilibrium concentrations, leading to the quadratic equation 2.6 * (0.970 - y) = 0.530y + y². The final equilibrium concentration for y was found to be 0.619 mol/L, which was verified against the equilibrium constant.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with the I.C.E. table method for equilibrium calculations
  • Knowledge of quadratic equations and their solutions
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the application of the I.C.E. table in various equilibrium problems
  • Learn how to derive and solve quadratic equations in chemical contexts
  • Explore the concept of equilibrium constants and their significance in chemical reactions
  • Investigate the impact of temperature changes on equilibrium constants
USEFUL FOR

Chemistry students, educators, and anyone interested in mastering equilibrium calculations in chemical reactions.

General_Sax
Messages
445
Reaction score
0

Homework Statement


The reaction SO2CL2(g) == SO2(g)+ Cl2(g)
was studied in a 5.00 L container at 100C. Initially, 2.65 mol of Cl2(g) and 4.85 mol of SO2CL2(g) were present in the container. The equilibrium constant is 2.60. Calculate the equilibrium concentrations of the reactant and products.

[CL]initial = 2.65 mol / 5.00 L = 0.530 mol/L
[SOCL]initial = 0.970 mol/L



Homework Equations


aA + bB == cC + dD

kc = [A]a * b / [C]c * [D]d

Initial [ ] + Change_in [ ] = Equilibrium [ ]

[ ] = concentration (mol/L)



The Attempt at a Solution




I was taught a method to solve these problems that utilizes a "I.C.E. table". To me it resembles a matrix (at least I think it does).

[SOCL] ||| [SO] ||| [CL]

Initial: 0.970 ||| 0 ||| 0.530


Equilibrium: 0.970 - y ||| y ||| 0.530 + y



Kc = 2.60


Because the constant terms aren't 1000 times greater than the equilibrium constant, I can't simplify the equation by substituting 0.970 for the equilibrium concentration for SOCL.
So I have this to work with:

Kc = y * (0.530 +y) / ( 0.970 - y )

So I do this:

2.6 * ( 0.970 - y ) = 0.530y + y2

All I can see is a quadratic equation, so after a lil' bit of tinkering I set up a quadratic equation.

y = -2.6 +- (sqrt)[14.728] / 2

y = 0.619 mol/L

However, if that was true then I should be able to check that value against the equilibrium constant of 2.60. I don't want to rewrite another equation, but if you substitute in the new values for the equilibrium concentrations into the original formula you don't end up with 2.60, but instead you arrive at value of approximately 2.02.

Can someone please help me?
 
Physics news on Phys.org
Great after all that work typing my problem up, I've solved my problem!

If anyone is searching threads, for aid with one of their own problems, remember to be careful with your algebra.
 
General_Sax said:
Great after all that work typing my problem up, I've solved my problem!

That's one of the reasons why we want you to show you did. And that's no joke, you will be surprised how often people realize what they did wrong when they try to present their work to others.

Glad it worked for you.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
2K
Replies
1
Views
9K
Replies
1
Views
9K