At 580 K, 6.42% of the..Calculate the equilibrium constant for the preceding rea

Question

At 580 K, 6.42% of the..Calculate the equilibrium constant for the preceding reaction as written. At 248 degree C and a total pressure of 1,000 atm, the fractional dissociation of SbCl_3 is 0.718 for the reaction SbCI_5(g) equivalent SbCI_3(g) + Cl_2(g) This means that 718 of every 1000 molecules of She, finally present have dissociated. Calculate the equilibrium constant. Sulfuric chloride (S0_2CI_2) is a colorless liquid that boils at 69 degree C. Above this temperature, the vapors dissociate into sulfur dioxide and chlorine: SO_2CI_2(g) equivalent S0_2(g) + Cl_2(g) This reaction is slow at 100 degree C. but it is accelerated by the presence of some FeCI_3 (which does not affect the position of the equilibrium). In an experiment, 3.174 g of SO_2Cl_2f) and a small amount of solid FeCl_3 are put evacuated 1.000-L flask, which is then scaled and heated to 100 degree C. The total pressure in the flask at that temperature s found to be 1.30 atm. Calculate the partial pressure of each of the three gases present. Calculate the equilibrium constant at this temperature.

Explanation / Answer

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24)
SbCl5 <---> SbCl3 + Cl2
                       0              0      (initial)
1-0.718          0.718       0.718   (at equilibrium)

Kp= p(Cl2)p(SbCl3) / p(SbCl5)
Kp = 0.718*0.718 / (1-0.718)
      = 1.83
Answer: 1.83

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