The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...

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Chemistry

The equilibrium constant, K, for the following reaction is1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixtureof the three gases in a 1.00 L flask at 500 K contains 0.200 MPCl5, 4.90×10-2 M PCl3 and 4.90×10-2 M Cl2. What will be theconcentrations of the three gases once equilibrium has beenreestablished, if 3.12×10-2 mol of Cl2(g) is added to theflask?

[PCl5] =

[PCl3] =

[Cl2] =  

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4.1 Ratings (659 Votes)
Given K 12 x 102 0012equillibrium conc of PCl5 02 M since the volume is1 L number of mole at equilibrium 0200 molesimilarly equliibrium conc of PCl3 49 x102 M 0049 M number of mole at equillibrium 0049 moleequillibrium conc of Cl2 49 x 102 M 0049 M    See Answer
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