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

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`.

Step-by-step explanation:

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Step-by-step explanation:

Equilibrium constant of the reaction =
`K_c=83.3`

Moles of
`PCl_3 = 0.280 mol`

Concentration of
`[PCl_3]=(0.280 mol)/(1.00 L)=0.280 M`

Moles of
`Cl_ = 0.280 mol`

Concentration of
`[Cl_2]=(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=([PCl_5])/([PCl_3][Cl_2])`

Putting values in above equation, we get:

`83.3=(x)/((0.280-x)* (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`.