A00104 mol sample eso.-placed i.100 1 comme-and hered to tempersture to so, and

Question

A00104 mol sample eso.-placed i.100 1 comme-and hered to tempersture to so, and o 21. 100 t equilibrium, the partial pressures of so, so, and o, re 0 3176 am, o7126 tm, and 0 3608 atm to return to equilibriom and the volume of the doutled to 2.00 b. Construct an ICE table to show how the system reaches equiibrium at 1100 c Calculate a, 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

Explanation / Answer

21. Decomposition of SO3 gas,

2SO3 (g) <==> 2SO2 (g) + O2 (g)

a. When the volume of the system was increased from 1 L to 2 L, pressure would decrease to half. So according to the LeChatellier's principle, the reaction would shift to the direction which has least number of moles of gas, that is towards the reactant side (left handside), and we would get formation of more SO3 from SO2 and O2, until it reaches equilibrium state again.

b. ICE chart for the reaction would be,

                           2SO3 (g) <==> 2SO2 (g) + O2 (g)

initial                    0.0104                -               -

change                  -2x                  +2x            +x

equilibrium        0.0104-2x             2x             2x

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

          = 3.87 at 1100 K

d. Kc at 1100 K

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

e. Total pressure at 1100 K

Initial pressure of SO3 = nRT/V

                                     = 0.0104 x 0.0821 x 1100/1 = 0.94 atm

Total pressure in terms of change "x"

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

22. pH of vinegar solution

initial [CH3COOH] = 0.875 M

ICE chart,

                  CH3COOH <===> CH3COO- + H+

I                    0.875                        -               -

C                    -x                            +x            +x

E                0.875-x                         x              x

So,

Ka = x^2/(0.875-x)

let x be a small amount, can be neglected from denominator

So,

1.8 x 10^-5 = x^2/0.875

x = [H+] = 3.97 x 10^-3 M

pH = -log[H+] = 2.40

b. When vinegar was diluted with water

Volume of CH3COOH = 1 ounce = 29.6 ml

Total volume = 8 ounces = 236.6 ml

new [CH3COOH] = 0.875 M x 29.6 ml/236.6 ml = 0.11 M

ICE chart,

                  CH3COOH <===> CH3COO- + H+

I                    0.11                         -               -

C                    -x                            +x            +x

E                0.11-x                          x              x

Ka = x^2/(0.11-x)

let x be a small amount, can be neglected from denominator

So,

1.8 x 10^-5 = x^2/0.11

x = [H+] = 1.41 x 10^-3 M

pH = -log[H+] = 2.85

c. percent ionization = [H+] x 100/[HA]

                                 = 1.41 x 10^-3 x 100/0.11 = 1.3%

Percent ionization is less than 5% and thus approximation is valid.

d. excess alka-seltzer tablet would neutralize acetic acid completely and would make the solution basic in nature. pH expected in greater than 7.

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