2) The value of the equilibrium constant for the reaction H2 (g)+ Br2 (8)2 HBr (

Question

2) The value of the equilibrium constant for the reaction H2 (g)+ Br2 (8)2 HBr (8) at 25°C is K-1.9 x10 a) Write the expression for the equilibrium constant for this reaction. b) A certain mixture containing these three gases is at equilibrium. The mixture contains 2.0x 103 mol hydrogen and 2.0x 10 mol bromine in a 4.0L. What is the concentration of hydrogen bromide in this equilibrium mixture? c) How does the amount of the product compare to the amount of reactants? Would you have predicted that by ust considering the value of ke?

Explanation / Answer

2)

H2 (g) + Br2 (g)   <------------> 2 HBr (g)

a)

Equilibrium constant Expression :

Kc = [HBr]^2 / [H2] [Br2]

b)

H2 (g) + Br2 (g)   <------------> 2 HBr (g)

[H2] = 2.0 x 10^-3 / 4 = 5 x 10^-4 M

[Br2] = 2.0 x 10^-3 / 4 = 5 x 10^-4 M

Kc = [HBr]^2 / [H2] [Br2]

1.9 x 10^19 = [HBr]^2 / (5 x 10^-4 )^2

[HBr] = 2.2 x 10^6 M

c)

here Kc value is very high. so

amount of products > amount of reactants

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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