Question

Calculate equilibrium pressure of A2, C, A2C2 and A2B

Answer 1

Assume that Initially 12 m moles of A2 are present. Then 4m moles of C and 15m moles of B are mixed in to the container.  Volume of the container is fixed.

At equilibrium we will have x moles of A8, y moles of A2C2 and z moles of A2B from the three reactions.  That means 12m-4x-y-z moles of A , 15m-z moles of B , 4m-2y moles of C are present.

The total number of moles in the mixture = N
 N  = 12m-4x-y-z + 15m-z + 4m-2y + x + y + z = 
 N  =  31 m - 3x - 2y - z    ---(1)

Partial pressure of A8 = 0.3 atm.   Total Pressure = 1.8 atm.

Mole fraction of A8 = x/N = 0.3/1.8 = 1/6.
    => N = 6 x
    => 31 m = 9x + 2y + z    --- (2)

Kp of reaction    4 A2 <==> A8  is 3.6/atm³

Partial pressure of A2 = p_A2.
    p_A2 = (12m-4x- y -z)/(6x) * 1.8 
     => Kp = p_A8 / (p_A2)⁴ = 3.6
     p_A2 = (0.3/3.6)^0.25 = 0.537 atm. --(3)

p_A2 = 1.8 * (12m - 4x - y - z)/6x = 0.537
        => 10.4 x + 1.35 y + 2.35 z = 9.2 m  ---(4)

Continue further for the partial pressures of other gases.
p_B = 15m/6x   ,   p_C = (4m-2y)/6x      p_A2C2 = y/6x

Answer 2

compounds are dissolved in water, the following equilibrium is established. V2+ + H2O V3+ + H2 2 1 + OH– What would alter the composition of the equilibrium mixture in favour of the V2+ ions?