Question

An equilibrium mixture in a 16.2 L container at 712 K contains 0.218 mol NH4I(s), 0.189 M NH3 and 0.116 M HI. What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 6.79 L

Answer 1

The equilibrium constant of the mixture is not given hence I will provide a general answer

answer :

( new concentration at constant temp and volume of 6.79 L)

NH3 = 0.45 - X

HI = 0.28 - X

Step-by-step explanation:

Given data :

Equilibrium mixture volume = 16.2 L

temperature = 712 k

0.218 mol of NH4 I(s)

0.189 M NH3

0.116 M HI

Determine the concentrations of the two gases once equilibrium has been reestablished at a constant temperature and to a volume of 6.79 L

first step : calculate the number of moles for the gases

NH3 = 0.189 * 16.2 = 3.06 moles

HI = 0.116 * 16.2 = 1.88 moles

when the mixture is compressed to a volume of 6.79 L

the concentrations of the gases ( moles / volume )

NH3 = 3.06 / 6.79 = 0.45

HI = 1.88 / 6.79 = 0.28

Change in Concentrations of the two gases at compressed volume of 6.79 L at constant temperature can be expressed as

NH3 = 0.45 - X

HI = 0.28 - X

where X = K ( equilibrium constant ) =
`((0.45-X) (0.28-X))/(1)` ----- ( 1 )

Insert the value of K into equation 1 and solve for X

where : 1 = concentration of NH4 I(s)