Question

At 250°C the equilibrium constant (Kp) for the reaction below is 1.81. PCl5(g) PCl3(g) + Cl2(g) Sufficient PCl5 is put into a vessel to give an initial pressure of 2.75 atm at 250°C. What will be the final pressure at this temperature after the system has reached equilibrium?

Answer 1

Answer:

the partial pressure of the $PCl_5$ at equilibrium is 1.25 atm

Explanation:

Initial pressure of the $PCl_5 =2.75 atm$

The value of the equilibrium constant =$K_p=1.81$

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

Initially

2.75 atm      0       0

At equilibrium

(2.74- x)      x         x

The expression for $K_p$ for the given reaction

$K_p=\frac{p_{PCl_3}\times p_{Cl_2}}{p_{PCl_5}}$

we get:

$1.81=\frac{x\times x}{2.75-x}x=1.50$

Pressure of the $PCl_5$ at equilibrium :

$= (2.75 - p) atm \\= 2.75 - 1.50 atm \\= 1.25 atm$

Hence, the partial pressure of the $PCl_5$ at equilibrium is 1.25 atm