At 453K Al2Cl6 g reacts to form Al3Cl9 g according to the eq

At 453K, Al2Cl6 (g) reacts to form Al3Cl9 (g) according to the equation: 3Al2Cl6 <----> 2Al3Cl9 (g)

a. In an experiment at this temperature, the equillibrium partial pressure of Al2Cl6 (g) is 1.00atm, and the equilibrium partial pressure of Al3Cl9 (g) is 1.02E-02 atm. Compute the equilibrium constant

b. Some Al2Cl (g) (at a partial pressure of 0.473atm) is placed in a closed container at 454K with some Al3Cl9 (g) (at partial pressure 1.02E-02 atm). Enough Argon is added to raise the total pressure to 1 atm. Calculate the initial reaction quotient for the reaction 3Al2Cl6 (g) <---> 2Al3Cl9 (g)

c. As the gas mixture reaches equillbirium, will there be net production or consumption of Al3Cl9 (g)?

Solution

a)

3 Al2Cl6 <-------> 2 Al3Cl9 (g)

1.00 atm                     0

1.00 - 3x                     2x

At equilibrium : Al3Cl9 = 2x = 1.02 x 10^-2 atm

x = 0.0051

[Al2Cl6] = 0.9847 M

Kc = P²_Al3Cl9 / P³_Al2Cl6

       = (1.02 x 10^-2)^2 / (0.9847)^3

Kp = 1.09 x 10^-4

equilibrium constant = 1.09 x 10^-4

b)

3Al2Cl6 (g) <---> 2Al3Cl9 (g)

Q = (1.02 x 10^-2)^2 / (0.473)^3

    = 9.83 x 10^-4

reaction quotient = 9.83 x 10^-4