At temperature t kelvin, PCl5 is 50% disassociated at an equilibrium pressure of 4 atm. At what pressure it would disassociate to the extent of 80% at the same temperature?
Answers
Given data:
At Equilibrium pressure P = 4 atm, dissociation x = 0.5
Chemical reaction behind the given scenario is
PCl5 ---> PCl3 + Cl2
No. of moles initially will be 1 --> 0 , 0
No. of moles at equilibrium will be 1-x ----> x , x
Therefore, total no. of moles will be 1 -x + x + x = 1 + x
Kp = Pp(PCl3) * Pp(PCl2) / Pp(PCl5)
= (x/(1+x))P * (x/(1+x))P / (1-x)/(1+x) P
= x^2 * P / 1 - x^2
= 0.5^2 * 4 / (1 - 0.5^2)
Kp = 1.33 ( For 5.% dissociation)
Therefore for 80% dissociation,
Kp = x^2 * P / (1 - x^2)
1.33 = 0.8^2 * P / ( 1 - 0.8^2)
P = 1.33 * 0.36 / 0.64 = 0.75 atm
Hence, for dissociation of 80% of PCl5 at same temperature, pressure of 0.75 atmosphere is essential.
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the pressure would be
0.75 atm.
