Question

The equilibrium constant, K., for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC13(E) Calculate the equilibrium concentrations of reactant and products when 0.280 moles of PCl, and 0.280 moles of Cl, are introduced into a 1.00 L vessel at 500 K. [PC13] = [Cl] = [PC13] =

Answer 1

Answer:

The equilibrium concentration of $PCl_5=0.228 M$.

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$.

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$.

Explanation:

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = $K_c=83.3$

Moles of $PCl_3 = 0.280 mol$

Concentration of  $[PCl_3]=\frac{0.280 mol}{1.00 L}=0.280 M$

Moles of $Cl_ = 0.280 mol$

Concentration of $[Cl_2]=\frac{0.280 mol}{1.00 L}=0.280M$

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

Initial:            0.280      0.280                            0

At eq'm:         (0.280-x)   (0.280-x)                     x

We are given:

$[PCl_3]_{eq}=(0.280-x)$

$[Cl_2]_{eq}=(0.280-x)$

$[PCl_5]_{eq}=x$

Calculating for 'x'. we get:

The expression of $K_{c}$ for above reaction follows:

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

Putting values in above equation, we get:

$83.3=\frac{x}{(0.280-x)\times (0.280-x)}$

On solving this quadratic equation we get:

x = 0.228, 0.344

0.228 M < 0.280 M< 0.344 M

x = 0.228 M

The equilibrium concentration of $PCl_5=0.228 M$.

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$.

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$.