For the exothermic reaction PCl3(g) Cl2(g)PCl5(g) K p 0 160 at a cer...

Given that PCl3g Cl2g PCl5g Kp 016 Initial 05 atm 05 atm 03 a ....

For the exothermic reaction PCl3(g)+Cl2(g)â‡ŒPCl5(g) Kp = 0.160 at a certain temperature. A flask is charged with 0.500...

3.6 Ratings (599 Votes)

Given thatÂ Â

Â Â Â Â Â Â Â Â Â Â Â Â Â PCl3(g)Â Â Â Â +Â Â Â Â Â Â Cl2(g) â‡Œ PCl5(g)Â Â Â Â Â Â Â Â Kp = 0.16

InitialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5 atmÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.5 atmÂ Â Â Â Â Â Â Â Â Â 0.3 atm

At equilibriumÂ Â Â Â Â Â 0.5-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5-xÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.3+x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Kp = (0.3+x)/ [ (0.5-x) (0.5-x)]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.16 = (0.3+x)/ (0.5-x)²

Â Â Â Â Â Â Â Â Â Â Â On solving,

Â Â Â Â Â Â Â Â Â Â Â Â Â x = -0.217 atm

Therefore,

Equilibrium partial pressures are

PCl3 = 0.5-x = 0.5 - (-0.217) = 0.717 atm

Cl2 = 0.5-x = 0.5 - (-0.217) = 0.717 atm

PCl5 = 0.3+x = 0.3-0.217 = 0.083 atm

--------------------------------------------------------------

Given reaction reaction exothermic.

All exothermic reactions are favoured by decrease in temperature i.e. forward reaction reaction is favourable ( Equilibrium shifts to right)

Hence,

X Cl2 increases when Â Â Â increase the temperature

X Cl2 decreases whenÂ Â Â decrease the temperature

X Cl2 stays the same Â Â increase the volume of the container Â Â Â Â Â Â Â Â Â Â Â Â Â

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