Question

Kc = 54 at some temperature for the reaction:

H2(g) + I2(g) ⇌ 2HI(g)
If 17.8 mol of HI are introduced into a 10.0 L vessel, how many moles of I2 are present at equilibrium?
a. 0.74 mol
b. 1.48 mol
c. 8.5 mol
d. 17.8 mol

Answer 1

Using an ICE table and the given equilibrium constant, we find that 0.74 moles of I2 are present at equilibrium in a 10.0 L vessel when starting with 17.8 moles of HI.

Step-by-step explanation:

To find the number of moles of I2 present at equilibrium, we can use an ICE (Initial, Change, Equilibrium) table and the given equilibrium constant (Kc).

The reaction is:
H2(g) + I2(g) ⇌ 2HI(g)

Let's assume that 'x' moles of H2 and I2 react to form HI at equilibrium. Therefore, at equilibrium, the number of moles of HI will have decreased by 2x (since 2 moles of HI are produced from 1 mole of H2 and 1 mole of I2).

The equilibrium concentrations in the 10.0 L vessel can be represented as follows:

• [H2] = x / 10 L
• [I2] = x / 10 L
• [HI] = (17.8 - 2x) / 10 L

Given that Kc = 54, we can set up the equilibrium expression:

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

Plugging in the expressions from our ICE table gives us:

54 = ((17.8 - 2x) / 10)^2 / (x / 10 * x / 10)

By solving this quadratic equation for x, which represents the moles of I2 at equilibrium, we find that x = 0.74 moles, which corresponds to option a.