Calculate the equilibrium concentrations of reactant and products when 0.303 moles of HI (g) are introduced into a 1.00 L vessel at 698 K.

Question

The equilibrium constant, K_c, for the following reaction is 0.0180 at 698 K.

2HI (g) ----> H₂ (g) + I₂ (g)

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Calculate the equilibrium concentrations of reactant and products when 0.303 moles of HI (g) are introduced into a 1.00 L vessel at 698 K.

 [HI].= _______M

 [H₂] = _______M

[I₂] = _________M

Answer 1

Set up an ICE table

2HI (g) <===> H₂ (g) + I₂ (g)

0.303..................0............0.........Initial

-2x....................+x...........+x........Change

0.303-2x.............x.............x.........Equilibrium

Kc = [H₂][I₂] / [HI]²

0.0180 = (x)(x) / (0.303 - 2x) and if we assume x and 2x are small relative to 0.303 we can ignore it

0.0180 = x2 / 0.303

x² = 5.94x10^-3

x = 3.53x10^-5 (which is small compared to 0.303 so above assumption was valid)

[HI] = 0.303 - 2x = 0.303 M

[H₂] = x = 5.94x10^-3 M

[I₂] = x = 5.94x10^-3 M