At a particular temperature, Kc = 54 for the reaction H2(g) + I2 (g) <=> 2 HI(g).

One mole of HI is placed in a 3.0-L container. What would be the equilibrium concentration of HI?
Choose one answer.

a. 0.035

b. 0.071

c. 0.29

d. 0.30

e. 0.33

Post your work for this problem. I don't get ANY of the answers listed.

.26

To find the equilibrium concentration of HI, we can use the equation:

Kc = [HI]^2 / ([H2][I2])

First, let's establish the initial concentration of HI. We are told that one mole of HI is placed in a 3.0-L container, so the initial concentration of HI ([HI]_initial) is:

[HI]_initial = 1 mol / 3.0 L

Now, we can substitute the initial concentrations into the equation and solve for [HI]:

54 = [HI]^2 / ([H2][I2])

Since the reaction starts with only HI and no H2 or I2, their initial concentrations ([H2]_initial and [I2]_initial) are both zero. Therefore, we can simplify the equation:

54 = [HI]^2 / (0 * 0)

Since anything divided by zero is undefined, we can conclude that the equilibrium concentration of HI is also zero.

Therefore, the correct answer is not provided in the options.