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Answers all supposed to be in M For the reaction I2(g)+Br2(g)←−→2IBr(g), Kc=280 at 150 ∘C. Suppose...

Answers all supposed to be in M

For the reaction
I2(g)+Br2(g)←−→2IBr(g),
Kc=280 at 150 ∘C. Suppose that 0.550 molIBr in a 2.00-L flask is allowed to reach equilibrium at 150 ∘C.

What is the equilibrium concentration of IBr?

What is the equilibrium concentration of I2?

What is the equilibrium concentration of Br2?

Expert Solution

initial concentration of IBr,

[IBr] = number of mol of IBr / volume in L

= 0.550 mol/ 2.00 L

= 0.275 M

I2(g) + Br2(g) <—> 2 IBr(g)

0 0 0.275 (initial)

x x 0.275-2x (at equilibrium)

Kc = [IBr]^2 / [I2][Br2]

280 = (0.275-2x)^2 / (x)^2

(0.275-2x)/x = sqrt(280)

(0.275-2x)/x = 16.73

0.275-2x = 16.73*x

18.73*x = 0.275

x = 0.0147 M

At equilibrium,

[IBr] = 0.275-2x = 0.275-2*0.0147 = 0.0246 M

[I2] = x = 0.0147 M

[Br2] = x = 0.0147 M

queen_honey_blossom answered 2 years ago

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