Consider the following reaction,which is involved in the catalytic destruction o

Question

Consider the following reaction,which is involved in the catalytic destruction of ozone by chlorineatoms: ClO(g) + O3(g) --> Cl(g) +2O2(g)
Kp=2.5 x 106 atm at 25oC. In an experiment at25oC, ClO(g) at 1.0 x 10-3 atm was mixed withO3(g) at 2.0 x 10-5 atm in a rigid vessel.Calculate the equilibrium pressure of O3(g) andO2(g).

The problem is, I used an ice table and found that my equation tosolve for x ended up being:

2.5 x 106 = 4x3/(2 x 10-8-0.00102x + x2)

so: 4x3 - 2500000x2 + 2550x - 0.05 = 0

How do I solve for an equation where x is cubed? We have NOTlearned this in class and I do not think that I am expected to knowit. Is there any other way of doing this question to avoid this?Thanks!

I'm beginning to think I've just done the entire thing wrong..Please help!
Consider the following reaction,which is involved in the catalytic destruction of ozone by chlorineatoms: ClO(g) + O3(g) --> Cl(g) +2O2(g)
Kp=2.5 x 106 atm at 25oC. In an experiment at25oC, ClO(g) at 1.0 x 10-3 atm was mixed withO3(g) at 2.0 x 10-5 atm in a rigid vessel.Calculate the equilibrium pressure of O3(g) andO2(g).

The problem is, I used an ice table and found that my equation tosolve for x ended up being:

2.5 x 106 = 4x3/(2 x 10-8-0.00102x + x2)

so: 4x3 - 2500000x2 + 2550x - 0.05 = 0

How do I solve for an equation where x is cubed? We have NOTlearned this in class and I do not think that I am expected to knowit. Is there any other way of doing this question to avoid this?Thanks!

I'm beginning to think I've just done the entire thing wrong..Please help!

Explanation / Answer

ClO(g) + O3(g) --> Cl(g) + 2O2(g) k = 2.5e6 = [Cl][O2]2 / [ClO][O3] 2.5e6 = [(x)(2x)2] / [(1.0e-3 - x)(2.0e-5 - x)] x = .001 or .00002 ; .001 is too big, so use x = .00002 O3: 2.0e-5 - 2.0e-5 = 0 O2 = 2(2.0e-5) = .00004

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.