Answer to Question #222180 in General Chemistry for Meagan

Question #222180

Kp of PH3BCl3


1
Expert's answer
2021-08-02T12:27:20-0400

Solution


Kp = Kc(RT)Dn


Dn = (moles of gaseous products -moles of gaseous reactants)


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


Kc = 6.96 "\\times10^5" @ 333 K


Dn = (2 moles of gaseous products - 0 moles of gaseous reactants) = 2


"Kp = (6.96 \\times10^{-5})[(0.0821)(333)]^2 = 0.052"


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