Equilibrium Problem; Senior High School Level

AI Thread Summary
The discussion revolves around calculating the equilibrium concentrations for the reaction SO2Cl2(g) == SO2(g) + Cl2(g) in a 5.00 L container at 100°C, given initial amounts and an equilibrium constant of 2.60. The participant initially used an I.C.E. table to set up the problem but encountered difficulties when solving the quadratic equation derived from the equilibrium expression. After some algebraic manipulation, they found a value for y, which did not satisfy the equilibrium constant when checked. Ultimately, the participant realized the importance of careful algebra in solving equilibrium problems and successfully corrected their approach. The thread emphasizes the value of presenting work to identify errors in calculations.
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.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top