At 500 °C, hydrogen iodide decomposes according to 2 HI ( g ) − ⇀ ↽ − H 2 ( g ) + I 2 ( g ) For HI ( g ) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [ H 2 ] = 0.325 M , [ I 2 ] = 0.325 M , and [ HI ] = 2.75 M . If an additional 1.00 mol of HI ( g ) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached?

[HI] = 3.55 M

[H₂] = [I₂] = 0.425 M

Explanation:

The strategy here is to calculate Kc given the concentrations at equilibrium.Then we will setup an equilibrium expression in terms of the new concentrations and the amount that will be consumed of HI to reach equilibrium again.

2 HI (g) ⇆ H₂(g) + I₂(g)

Kc = [H₂][I₂]/[HI]² = 0.325 x 0.325 / 2.75² = 0.014

To account for the new equilibrium after adding 1.oo mol HI lets use the following table to make it easy for us.

 mol                            HI               H₂                           I₂

Initially                 3.75                0.325                     0.325

Change                -2x                 + x                             + x

Equilibrium        3.75 - 2x          0.325 + x              0.325 + x

   These quantities at equilibrium has to obey the expression for Kc previously calculated. (Note we do not have to concern ourselves with calculating concentrations or switching to moles since the volume is 1 and M= mol/L)

We obtain the following equation:

(0.325 + x)² / (3.75 - 2x)² = 0.014

Taking square roots to both sides of this equation

( 0.325 + x )²  / (3.75 - 2x) = 0.118

Solving for x we get the value of  x =  0.10

Therefore the new equilibrium concentrations will be

[HI] = (3.75 - 0.20) M = 3.55 M

[H₂] = [I₂] = (0.325 + 0.10) M = 0.425 M

This answer can be checked by placing these values into the Kc. When we do it we get 0.014, so the answer is correct.

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