At a certain temperature, the equilibrium constant K for the following reaction

Question

At a certain temperature, the equilibrium constant K for the following reaction is 1.48 × 10 Suppose a 13. L reaction vessel is filled with 1.6 mof There will be very little Br2 and OCl2 Br2 and 1.6 mol of oCl2. What can you say about the composition of the mixture in the vessel at equilibrium? O There will be very little BroCl and BrcI. O Neither of the above is true. What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. BrOCI(g)+BrCl(g) Br2(g)+0C12(g) What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. K-

Explanation / Answer

For the reaction,

Br2(g) + OCl2(g) <==> BrOCl(g) + BrCl(g)

K = [BrOCl][BrCl]/[Br2][OCl2] = 1.48 x 10^-5

Using this information

--

A 13 L vessel has Br2 = 1.6 mol, OCl2 = 1.6 mol it it. At equilibrium : There will be very little BrOCl and BrCl

The K value is very small so the reaction goes only to a small extent in the forward direction to form products.

--

Equilibrium constant for the reverse reaction,

BrOCl(g) + BrCl(g) <==> Br2(g) + OCl2(g)

K = 1/1.48 x 10^-5 = 6.76 x 10^4

--

The equilibrium constant for the reaction,

2Br2(g) + 2OCl2(g) <==> 2BrOCl(g) + 2BrCl(g)

K = (1.48 x 10^-5)^2 = 2.19 x 10^-10

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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