At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.
2SO3(g)<--> 2SO2(g)+O2(g)
At equilibrium, 0.130 mol of O2 is present. Calculate Kc.
Kc= ?
Please post the answer if possible or the steps in detail; I have attempted this problem many times and have been unsuccessful.
0.900mol/2L = 0.45M
.........2SO3(g)<--> 2SO2(g)+O2(g)
I........0.45.........0.......0
C........-2x..........2x......x
E.......0.45-2x.......2x......x
The problem tells you that x = 0.130mol O2 is present which makes (O2) = 0.130/2L = 0.065M and (SO2) is twice that (i.e., 0.260mol/2L) = 0.130M. That makes (SO3) = 0.45-0.260 = 0.190 M
Substitute and solve for Kc. I suspect that you forgot it was in 2L and you didn't correct 0.130 mols O2 to M O2 by dividing by 2 L. Of course that messed up SO2 and SO3 too. Hope this helps.
What is the correct answer?
Most of the tutors here don't work the problem We just help solve it without going through the actual calculation.
I have Kc= [(0.065)(0.130)^2]/(0.190^2)
Kc= 0.030429
This answer was incorrect. Can you please help correct me?
To calculate Kc, the equilibrium constant, we need to set up the expression using the stoichiometric coefficients of the balanced equation.
The balanced equation is:
2SO3(g) <--> 2SO2(g) + O2(g)
Initially, the number of moles of SO3 is 0.900 mol, and there are 0 moles of SO2 and O2 present.
At equilibrium, 0.130 mol of O2 is present. We need to find the equilibrium concentrations of SO3 and SO2.
Let's define the change in the number of moles for each substance:
- Change in SO3 = -2x (since 2 moles of SO3 react to give 2 moles of SO2 and 1 mole of O2)
- Change in SO2 = +2x (since 2 moles of SO2 are formed for every 2 moles of SO3 reacted)
- Change in O2 = +x (since 1 mole of O2 is formed for every 2 moles of SO3 reacted)
Now, let's write the expressions for the equilibrium concentrations:
- [SO3] = 0.900 - 2x
- [SO2] = 0 + 2x
- [O2] = 0 + x
We are given that at equilibrium, [O2] = 0.130 mol. Substituting this into the expression for [O2]:
0.130 = x
Now we can substitute this value in the expressions for [SO3] and [SO2]:
[SO3] = 0.900 - 2(0.130) = 0.900 - 0.260 = 0.640 mol
[SO2] = 0 + 2(0.130) = 0.260 mol
Finally, we substitute the equilibrium concentrations into the equilibrium constant expression:
Kc = [SO2]^2 * [O2] / [SO3]^2
= (0.260)^2 * (0.130) / (0.640)^2
Calculating this value gives:
Kc ≈ 0.006108
Therefore, the value of Kc for this equilibrium is approximately 0.006108.
Thanks for showing your work.
I screwed up.
In my calculation I said x = 0.130 mols according to the problem. The problem DID state that O2 was 0.130 mols; however, in my calculation I let x = MOLARITY (not mols). I think the easiest way to explain this is to start over and use mols and not molarity.
.........2SO3 ==> 2SO2 + O2
I........0.9........0......0
C........-2x........x.......x
E.......0.9-2x.......x......x
where 0.9 = mols we started with and x = mols O2 with 2x = mols SO2.
(O2) = mols/L = 0.130/2 = 0.065M (the same number as before).
(SO2) = 2x/2L = 2*0.130/2 = 0.130M (the same number as before).
(SO3) = (0.9-2*0.130)/2 = 0.32 (not the same number as before).
Kc = (0.065)(0.130)^2/(0.32)^2 = ?
Check my work.
Sorry to have given you a bummer. The lesson to learn from this is to check what we do. Sometimes it's late, sometimes we just get in a hurry, sometimes we read the problem too fast (or read it wrong), and sometimes we just screw up as I did in this problem..