Question

4.00 moles of HI are placed in an evacuated 5.00 L flask and then heated to 800 K. The system is allowed to reach equilibrium. What will be the equilibrium concentration of each species

Answer 1

Initial concentration of HI = number of moles/volume = 4 mol / 5 L = 0.8 M

the reaction taking place is:
2 HI(g) <----------> H2(g) + I2(g)    Kc = 0.016 @ 800 K
   0.8                                 0           0     (initial concentration)
   -2x                                +x         +x     (change)
0.8-2x                              x           x        (equilibrium concentration)

Kc = [H2]*[I2] / [HI]^2
0.016 = x*x / (0.8-2x)^2
0.016* (0.8-2x)^2 = x^2
0.016* (0.64 - 3.2*x + 4*x^2) = x^2
0.016* (0.64 - 3.2*x + 4*x^2) = x^2
0.01024 - 0.0512*x + 0.064*x^2 = x^2
0.936*x^2 + 0.0512*x - 0.01024 = 0
solving above expression, we get positive value of x as x = 0.08 M

At equilibrium:
[H2] = x = 0.08 M
[I2] = x = 0.08 M
[HI] = 0.8-2x = 0.8 - 2*0.08 = 0.64 M