Answer to Question #222180 in General Chemistry for Meagan

Question

Kp of PH3BCl3

Answer 1

Solution

Kp = Kc(RT)^Dn

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

PH₃BCl_3(s) --> PH_3(g) + BCl_3(g)

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

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

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