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irina [24]
3 years ago
7

For the reaction PCl5(g) 4 PCl3(g) 1 Cl2(g) Kp 5 23.6 at 500 K a. Calculate the equilibrium partial pressures of the reactants a

nd products at 500 K if the initial pressures are PPCl5 5 0.560 atm and PPCl3 5 0.500 atm.
Chemistry
1 answer:
ollegr [7]3 years ago
6 0

Answer: Cl_2 = 42.9 atm, PCl_3 = 93.4 atm and PCl_5 = 7.66 atm

Explanation: The given balanced equation is:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

K_p=523.6

Initial pressure of PCl_5 = 50.560 atm  

initial pressure of PCl_3 = 50.500 atm

Let's say the change in pressure is p. Then:

equilibrium partial pressure of PCl_5 = (50.560 - p) atm

equilibrium partial pressure of PCl_3 = (50.500 + p) atm

equilibrium partial pressure of Cl_2 = p atm

K_p=\frac{(PCl_3)(Cl_2)}{PCl_5}

Let's plug in the values in it:

523.6=\frac{(50.500+p)(p)}{50.560-p}

on cross multiply:

26473.216-523.6p=50.500p+p^2

on rearranging the above equation:

p^2+574.1p-26473.216=0

It's a quadratic equation. On solving this equation:

p = 42.9

So, the equalibrium partial pressure of Cl_2 = 42.9 atm

equilibrium partial pressure of PCl_3 = 50.500 + 42.9 = 93.4 atm

equilibrium partial pressure of PCl_5 = 50.560 - 42.9 = 7.66 atm

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