When 0.152 mol of solid PH3BCl3 is introduced into a 3.0 Lcontainer at a certain

Question

When 0.152 mol of solid PH3BCl3 is introduced into a 3.0 Lcontainer at a certain temperture, 8.44 x 106-3 mol of PH3 ispresent at equilibrium:

PH3BCl3 (s) <---> PH3 (g) + BCl3 (g)

Construct a reaction table for the process, and use it to calculateKc at this temperature.

I thought maybe 'd have to use and ice table and i don't get whatto do to find the kc value from this. and i don't understand what areaction table is, is it exothermic drawing of the reactionprocess...help me thank will rate life saver

Explanation / Answer

Molarity of PH3 at equilibrium = 8.44 * 10-3 / 3.0                                                =2.81 * 10-3 M                                               PH3BCl3(s) <---> PH3 (g) + BCl3 (g)
Initial                                             0            0 Change                                          +x         +x Equilibrium                                    x            x We know molarity of PH3, x = 2.81 * 10-3M Since PH3BCl3 is a solid, its concentration does notchange, So Kc = [PH3] [ BCl3]            =2.81 * 10-3 M * 2.81 * 10-3M            =7.89 * 10-6

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