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 ] = 0.335 M , [ I 2 ] = 0.335 M , and [ HI ] = 2.83 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] = 3.61 mol·L⁻¹; [H₂] = [I₂] = 0.446 mol·L⁻¹  

Explanation:

1. Calculate Keq

                     H₂   +      I₂    ⇌ 2HI

E/mol·L⁻¹:    0.335    0.335    2.83

2. Set up an ICE table.

Step 3. Calculate the equilibrium concentrations

[HI] = (3.83 - 2x) mol·L⁻¹ = (3.83 - 0.2223) mol·L⁻¹ = 3.61 mol·L⁻¹

[H₂] = [I₂] = (0.335 + x) mol·L⁻¹ = (0.335 + 0.1111) =  0.446 mol·L⁻¹  

Check:

Close enough.

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