Question

Consider the reaction: H2(g) + I2(g)->2HI(g)
A reaction mixture in a 3.67-L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2. At equilibrium, the flask contains 90.4 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answer 1

Answer:

K= 616.34

Explanation:

Considering the above reaction, we should consider the initial conditions (when we first have H₂ and I₂ only), and the final conditions (when we have the remaining H₂ and I₂, and the product: HI, all species in equilibrieum)

We also should consider molecular masses:

M H₂ = 2 g/mol

M I₂ = 254 g/mol

M HI = 128 g/mol

In order to calculate the amount of each species at equilibrium:

Initial:

0.763 g H₂

96.9 g I₂

Final (equilibrium):

90.4 g HI

To calculate the final concentration of I₂ and H₂, we use the chemical reaction stoichiometry:

256 g HI -----> 2 g H₂

90.4 g HI -----> x = 0.706 g H₂, so at equilibrium: 0.763-0.706= 0.057 g H₂= 0.0285 moles H₂

256 g HI -----> 254 g I₂

90.4 g HI -------> x = 89.69 g I₂ , so at equilibirum: 96.9-89.69= 7.21 g I₂ = 0.0284 moles I₂

and then 90.4 g HI = 0.7062 moles HI

Using the volume of the flask, we should calculate the concentration of all species, in molar units (Molar= moles/Lt)

[H₂] = 0.0285/3.67 = 0.00777M

[I₂] = 0.0284/3.67 = 0.00773 M

[HI] = 0.7062/3.67 = 0.1924 M

So: Kc = [HI]² /[I₂][H₂] = (0.1924)² / (0.00777)(0.00773)

Kc = 616.34

Remember that:

- If Kc > 1, this reaction is more run to products

- If Kc<1 , this reaction is more run to reagents