Consider the reaction: 2 hi(g) ⇄ h2(g) + i2(g) it is found that, when equilibrium is reached at a certain temperature, hi is 35.4 percent dissociated. calculate the equilibrium constant kc for the reaction at this temperature.

Kc = 0,075

Explanation:

For a reaction:

A + B ⇄ xC + yD

Kc =

For the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

Kc = (1)

In equilibrium:

[HI] = 1- 2x

[I₂] = x [H₂] = x

If 35,4% is dissociated:

[HI] = 1-2x = 0,646 ⇒ x = 0,177

[I₂] = 0,177 [H₂] = 0,177

Replacing in (1):

Kc =

kc = 0,075

I hope it helps!

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.