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sample of 0.43 mol of HI is placed in a 1.00 L vessel heated to 500 °C. When equilibrium is reached what is the molar concentration of H2

2HI(g)  H2(g) + I2(g)

The equilibrium constant for HI decomposition at 500 °C is 5.8×10-3.

Oct 17th, 2017

Please note that the concentration og HI is [HI]= 0.43/1 = 0.43 mol/liter, so the concentration of H2, is [H2]=0.215 mol/liter, since we have 0.43/2 = 0.215 mol of H2. Now at the Equilibrium we have the subsequent concentrations:

[H2] = y

[I2]= y

[HI]= 0.43- 2y

Next I substitute all those quantities into the expression for the Equilibrium constant, and I get this:

58 * 10^(-4) =[ (0.43 -2y)^2]/ (y^2), and taking the square roots of both sides I can write:

7.62 * 10^(-2) = (0.43 - 2y)/y, or:

2.0762 y = 0.43, and finally I get:

[H2] = y = 0.207 mol/liter at the equilibrium

Feb 2nd, 2015

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