Question

In: Chemistry

The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g)

The dissociation of molecular iodine into iodine atoms is represented as

I₂(g) ⇌ 2I(g)

At 1000 K, the equilibrium constant K_c for the reaction is 3.80

×

10⁻⁵. Suppose you start with 0.0461 mol of I₂ in a 2.27−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I₂?

M

What is the equilibrium concentration of I?

M

Expert Solution

I₂(g) ⇌ 2I(g)

I 0.0461 0

C -x 2x

E 0.0461-x 2x

[I-] = 2x/2.27

[I2] = 0.0461-x/2.27

KC = [I-]^2/[I2]

3.8*10^-5 = (2x/2.27)^2/ 0.0461-x/2.27

3.8*10^-5 = 4x^2/2.27(0.0461-x)

3.8*10^-5 *(0.0461-x)*2.27 = 4x^2

x = 0.000986

[I-] = 2x/2.27 = 2*0.000986/2.27 = 0.000868M

[I2] = 0.0461-x/2.27 = 0.0461-0.000986/2.27 = 0.0451/2.27 = 0.0198M

queen_honey_blossom answered 2 years ago

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