Part 2 21. A 00104 mol sample of SO, is placed in a 1.00L container and heated t

Question

Part 2 21. A 00104 mol sample of SO, is placed in a 1.00L container and heated to 1100 k temperature to SO, and O to 1100 K. The SO, decomposes at that 250)(g)#2502(g) + O2(g) At equilibrium, the partial pressures respectively. Now, imagine that the of so, sO, and O, are 0.2176 atm, 0.7126 atm, and 0.3608 atm, volume of the container suddenly doubled to 2.00L a. In what direction will the system go to return to equilibrium and explain your reasoning? b. Construct an ICE table to show how the system reaches equilibrium at 1100 K. c. Calculate Q, for this reaction after the volume expansion. d. Write K, in terms of"x for this reaction after the volume expansion at 1100 K e. Determine the total pressure of the system once it reaches equilibrium in terms of "x

Explanation / Answer

21. For the given reaction,

2SO3 <==> 2SO2 + O2

after the volume change to 2 L from initial 1 L

a. The reaction would shift towards the direction where it has lowest moles of gas. Increase in volume reduces the pressure of the system. So according to the LeChatellier's principle, the reaction would shift towards left handside and we would get more SO3 formation until it reaches the equilibrium state again.

b. at 1100 K

ICE chart

                         2SO3 <==> 2SO2 + O2

initial                0.0104             -          -

change              -2x                +2x      +x

equilibrium   0.0104-2x            2x        x

So,

Kc = (2x)^2.(x)/(0.0104-2x)^2

c. Qp at 1100 K

Qp = [SO3]^2/[SO2]^2/[O2]

feeding the partial pressure value of each gas at equilibrium 1100 K

Qp = (0.7126)^2.(0.3608)/(0.2176)^2 = 3.87

d. Kc in terms of "x" would be,

Kc = (2x)^2.(x)/(0.0104-2x)^2

e. Total pressure of system at 1100 K in terms of "x" would be,

initial pressure of SO3 = 0.0104 x 0.0821 x 1100/1 = 0.94 atm

x be the change in pressure of SO3 at 1100 K

So,

Total pressure = (0.94 - 2x) + 2x + x

as pressure of SO2 at equilibrium = 2x and pressure of O2 at equilibrium = x

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