The gas-phase dissociation of phosphorus pentachloride to the trichloride has Kp

Question

The gas-phase dissociation of phosphorus pentachloride to the trichloride has Kp = 3.60 at 540 degree Celsius: PCI_5(g) right arrow PCI_3(g) + CI_2(g) What will be the partial pressures of PCI_3 if 0.200 mole of PC1_5 and 3.00 moles of PCl_3 are combined and brought to equilibrium at this temperature and at a total pressure of 1.00 atm? 0.012 atm 0.94 atm 0.047 atm 0.094 atm 0.250 atm For Questions #7 -11, predict the effect of the following concentration changes on the reaction below. CH_4 (g) + 2S_2 (g) implies CS_2(g) + 2H_2 S(g)

Explanation / Answer

Answer.   (b) is right answer

given that Kp=3.6 atm

and the reaction is PCl5   -----> PCl3 + Cl2

           at t=0          0.2 moles      3.00moles               0 moles

         at t=equm     0.2-a moles      3+a moles             a moles

total number of moles at equm = 0.2-a+ 3+a + a =3.2+a moles

partial pressure of PCl5 = [0.2-a/3.2+a]xtotal pressure=[0.2-a/3.2+a]x1.00=0.2-a/3.2+a

partial pressure of PCl3 = [3+a/3.2+a]xtotal pressure=[3+a/3.2+a]x1.00 =3+a/3.2+a -----(1)

partial pressure of Cl2 = [a/3.2+a]xtotal pressure=[a/3.2+a]x1.00 =a/3.2+a

Equilibrium constant Kp =3.6 = ( pPCl3 x pCl2)/ pPCl5 --------(2)

substituting all the partial pressures in eq (2)

       3.6 = [ (a/3.2+a)(3+a/3.2+a)]/(0.2-a/3.2+a)

3.6 =[a(3+a)/(3.2+a)]/(0.2-a)/1

3.6(-a2 -3a+0.64) =3a+ a2

-4.6 a2 -13.8a +2.3=0    is a quadratic equation in" a"   (qa2 +ra +s=0)

solution for a ={ -r +,- sqrt(r2-4qs)}/2q

a = {-(-13.8)+,- sqrt[(13.82 -4x(-4.6)(2.3)]}/2x(-4.6)

a = {13.8+,- sqrt(232.76)}/(-9.2)

a ={13.8+,-(15.26)}/(-9.2)

a =29.06/(-9.2) ,   a=-1.46/-9.2

a =-3.16 ,     a=0.16  

take only positive value of a =0.16   substitute a value in eq (1) then

partial pressure of PCl3   ,pPCl3 = 3+a/3.2+a=3+0.16/3.2+0.16=0.94 atm

partial pressure of PCl3   ,pPCl3 = 0.94 atm

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