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The equilibrium constant K=0.36 for the reaction PCl5 (g)  PCl3 (g) +Cl2 (g). (a) Given...

The equilibrium constant K=0.36 for the reaction PCl5 (g)  PCl3 (g) +Cl2 (g). (a) Given that 2.0 g of PCl5 was initially placed in the reaction chamber of volume 250 cm3, determine the molar concentration in the mixture at equilibrium. (b) What is the percentage of PCl5 decomposed.

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Expert Solution

PCl5 moles = mass / Molar mass of PCl5 = ( 2g) / ( 208.24g/mol) = 0.0096        

volume = 250 cm3 = 250 ml = 0.25L

[PCl5] = ( 0.0096/0.25) = 0.0384 M                 

                              PCl5 (g)    <------>    PCl3 (g)   +   Cl2 (g)

Initial                   0.0384                            0              0

equilibrium         0.0384 -X                          X                   X

Kc = [PCl3] [Cl2] /[PCl5]

0.36 = ( X^2) / ( 0.0384-X)

X^2 + 0.36X - 0.01383 = 0

X = 0.035

at equilibrium [PCl3] =[Cl2] = 0.035M

[PCl5] = 0.0384-0.035 = 0.0034 M

b) % of PCl5 decomposed = 100 x ( PCl5 decomposed (i,e X) / PCl5 initial amount)

( 100 x 0.035/0.0384) = 91.15 %

queen_honey_blossom answered 2 years ago
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