Question

The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and molecular chlorine (Cl2) is found to be 1.05 at 250oC. If the equilibrium partial pressure of PCl5 and PCl3 are 0.973 and 0.548 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250 oC?PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

Answer 1

Answer: The partial pressure of $Cl_2$ is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

The given balanced equilibrium reaction is,

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

Pressure at eqm.     0.973 atm                 0.548atm      x atm

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}$

Now put all the given values in this expression, we get :

$1.05=\frac{(0.548)\times (x)}{(0.973)}$

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of $Cl_2$ is 1.86 atm