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For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 2.89...

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Chemistry

For the following reaction, Keq is 0.00183 at 390. K: PCl5(g)PCl3(g) + Cl2(g) If 2.89 g of PCl5 is placed in a 2.86 L bulb at390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013bar

answer in bar

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3.8 Ratings (337 Votes)

PCl5(g)------------> PCl3(g) + Cl2(g)

no of moles of PCl5 = W/G.M.Wt

                                 = 2.89/208 = 0.0139moles

PV = nRT

P   = nRT/V

     = 0.0139*0.0821*390/2.86   = 0.156atm

             PCl5(g) ---------->PCl3(g) + Cl2(g)

I            0.156                    0                0

C           -x                        +x               +x

E        0.156-x                 +x                +x

           Kp   = PPCl3 PCl2/PPCl5

          0.00183   = x*x/0.156-x

          0.00183*(0.156-x) = x2

            x    = 0.016

         PXl2     = x    = 0.016atm = 0.016*1.013   = 0.016208bar

     

    
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