Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, h2(g) i2(g) ⇌ 2hi(g) is 54.3 at 430°c. determine the initial and equilibrium concentration of hi if initial concentrations of h2 and i2 are both 0.11 m and their equilibrium concentrations are both 0.052 m at 430°c.

Answer : The initial concentration of HI and concentration of at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of and = 0.11 M

Concentration of and at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of and at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of will be,

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

i'm not sure

explanation:

i'll try whatever you got

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun