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Sulfuryl chloride (SO2Cl2) is a colorless liquid that boils at 69°C. Above this temperature, the vapors dissociate into sulfur dioxide and chlorine: SO2Cl2(g) uv SO2(g) 1 Cl2(g) This reaction is slow at 100°C, but it is accelerated by the presence of some FeCl3 (which does not affect the final position of the equilibrium). In an experiment, 3.174 g of SO2Cl2(,) and a small amount of solid FeCl3 are put into an evacuated 1.000-L flask, which is then sealed and heated to 100°C. The total pressure in the flask at that temperature is found to be 1.30 atm. (a) Calculate the partial pressure of each of the three gases present. (b) Calculate the equilibrium constant at this temperature.

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Answer:

(a) pSOâ‚‚Clâ‚‚ = 0.14 atm

    pSO₂ = pCl₂ = x = 0.58 atm

(b) Kp = 2.4

Explanation:

Let's consider the following reaction.

SO₂Cl₂(g) ⇄ SO₂(g) + Cl₂(g)

We can calculate the initial pressure of SOâ‚‚Clâ‚‚ using the ideal gas equation.

[tex]P.V=n.R.T=\frac{m}{M} .R.T\\P=\frac{m.R.T}{M.V} =\frac{3.174g\times (0.08206atm.L/mol.K) \times 373.15K}{(134.97g/mol) \times 1.000L} =0.7201atm[/tex]

We can find the partial pressures at equilibrium using an ICE chart. In this chart, we complete each row with the pressure or change in pressure in each stage.

           SO₂Cl₂(g)      ⇄       SO₂(g)      +         Cl₂(g)

I             0.7201                      0                        0

C               -x                          +x                       +x

E          0.7201 - x                    x                         x

At equilibrium, the sum of partial pressures is equal to the total pressure.

pSOâ‚‚Clâ‚‚ + pSOâ‚‚ + pClâ‚‚ = 1.30 atm

(0.7201 - x) + x + x = 1.30 atm

x = 0.58 atm

pSOâ‚‚Clâ‚‚ = 0.7201 - x = 0.14 atm

pSOâ‚‚ = pClâ‚‚ = x = 0.58 atm

The equilibrium constant (Kp) is:

[tex]Kp=\frac{pSO_{2}.PCl_{2}}{pSO_{2}Cl_{2}} =\frac{0.58^{2} }{0.14} =2.4[/tex]

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