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21 January, 15:46

The compound pcl5 decomposes into cl2 and pcl3. the equilibrium of pcl5 (g) cl2 (g) pcl3 (g) has a keq of 2.24 x 10-2 at 327°c. what is the equilibrium concentration of cl2 in a 1.00 liter vessel containing 0.235 mole of pcl5 and 0.174 mole of pcl3? remember to use the correct number of significant digits

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  1. 21 January, 15:56
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    The equilibrium constant for this problem is Kp = [cl2] [pcl3]/[Pcl5]. Kp equal to 2.24 x 10-2 is given, while the initial concentration of pcl5 is 0.235 M and that of pcl3 is 0.174 M. The equation now becomes 2.24x10-2 = (0.174+x) x / (0.235-x) where x is the concentration produced from rxn and that of cl2. The equilib concentration of cl2 is then 0.22 M.
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