Question

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 and products at 500 K if the initial pressures are PPCl5 5 0.560 atm and PPCl3 5 0.500 atm.

Answer 1

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