Question

In: Chemistry

For the reaction H2(g) + I2(g) ↔ 2HI the value of the equilibrium constant is 25....

For the reaction

H2(g) + I2(g) ↔ 2HI

the value of the equilibrium constant is 25.

Starting with 1.00mol of each reactant in a 10.L vessel, how many moles of HI will be present at equilibrium?

Expert Solution

H₂ (g) + I₂ (g) 2 HI (g)

Kc = 25

Initial concentrations,

[H₂] = [I₂] = [HI] = 1.00 / 10.0 = 0.100 M

at equilibrium, [H₂] = [I₂] = 0.100 - x and [HI] = 0.100 + 2x

Equilibrium constant expression can be written as,

Kc = [HI]² / [H₂][I₂]

Kc = ( 0.100 + 2x)² / (0.100 - x)(0.100 - x)

25 ( 0.0100 + x² - 0.200x) = 0.0100 + 4x² + 0.400x

0.250 + 25x² - 5.00x = 0.0100 + 4x² + 0.400x

21x² -5.4x + 0.24x = 0

it is similar to ax² + bx +c = 0

So, on applying quadratic equation,

x = [-(-5.4)((-5.4)² - (4*21*0.24))] / (2 *21)

x = 0.0571 M

So, at equilibrium, [HI] = 0.100 + 2(0.0572) = 0.214 M

But, number of moles = Molarity * volume = 0.214 * 10 = 2.14 mol

queen_honey_blossom answered 2 years ago

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