For the system, I2(g) <==> 2I(g) (a) moles I2 = 20 g/253.8 g/mol = 0.08 mol mole

Question

For the system,

I2(g) <==> 2I(g)

(a) moles I2 = 20 g/253.8 g/mol = 0.08 mol

moles I = 1 g x 2/253.8 = 0.008 mol

mole fraction I2 = 0.08/0.088 = 0.91

mole fraction I = 0.008/0.088 = 0.09

(b) at equilibrium,

[I2] = 0.08 mol/10 L = 0.008 M

[I] = 0.008 mol/10 ml = 0.0008 M

ICE chart,

                I2(g) <==> 2I(g)

I              0.008         0.0008

C              -x                +2x

E          0.008-x       0.0008+2x

So,

Kc = [I]^2/[I2]

0.0258 = (0.0008 + 2x)^2/(0.008 - x)

6.4 x 10^-7 + 0.0032x + 4x^2 = 2.064 x 10^-4 - 0.0258x

4x^2 + 0.029x - 2.06 x 10^-4 = 0

x = 0.0044 M

equilibrium concentration of,

[I2] = 0.008 - 0.0044 = 0.0036 M

[I] = 0.0008 + 2 x 0.0044 = 0.0096 M

Total moles of system = 0.0036 M x 10 L + 0.0096 M x 10 L = 0.132 mol

Total pressure of system (P) = nRT/V

                                              = 0.132 x 0.08206 x 1473/10 L

                                              = 1.60 atm

Explanation / Answer

31 33.317 10 points) Consider the dissociation of iodine. 12(g)--21(g) t 1473 2mixture exerts a pressure 1.30 atm in a volume of 10.OL (a) At e mole fracti ressure of 1.30 atm, what are the e system reaches its equilibrium state, what is the pressure of the system at equibrium Hint: The molar mass of 12- 253.8 g/mol) ons of the two gases in the 1 M mihureé? b) Aher a cemain pchio

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

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