Question

At 500°C, hydrogen iodide decomposes according to 2 HI(g) = H2(g) +12(9) For HI(9) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [H, 1 = 0.383 M, L1=0.383 M, and [HI] = 3.24 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?

Answer 1

At 500 ⁰C Keq = [H₂][I₂] / [HI]² = [0.383 x 0.383] / [3.24]² = 0.01397357491

Now we will setup ICE

R.......................2HI....<==>......H2...........+...........I2

I.........................4.24................0...........................0

C........................-2x..................+x........................+x

E.......................4.24-2x.............+x........................+x

Keq = [H₂][I₂] / [HI]² = [x]² / [4.24 - 2x]² = 0.01397357491

[x]² = [4.24 - 2x]² x 0.01397357491

x² = [4x² - 16.96 x + 17.9776 ] x 0.01397357491 = 0.05589429964 x² - 0.23699183047 x + 0.2512113403

0.94410570036 x² + 0.23699183047 x - 0.2512113403 = 0 , solving this quadratic equation gives

x = 0.405 M

[H₂] = [I₂] = x = 0.405 M

[HI] = 4.24 - (2 x 0.405) = 3.43

[HI] = 3.43 M

At 500°C, hydrogen iodide decomposes according to 2 HI(g) = H2(g) +12(9) For HI(9) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [H, 1 = 0.383 M, L1=0.383 M, and [HI] = 3.24 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? [HIJ = 3.43 [H] = 0.405 1 - 0.405

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