Question

Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation: PCl5(g) ⇌ PCl3(g) + Cl2(g) At 250° 13.0 g of PCl5 is added to the flask with a final solution volume of 0.500 L. If the value of Kc at this temperature is 1.80, what are the equilibrium concentrations of each gas?

Answer 1

`[PCl_3]_(eq)=0.117M`

`[Cl_2]_(eq)=0.117M`

`[PCl_5]_(eq)=8x10^(-3)M`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction at equilibrium, we can write the law of mass action as shown below:

`Kc=([PCl_3][Cl_2])/([PCl_5])`

That in terms of the ICE methodology is written by means of the change
`x` due to the reaction extent:

`Kc=(x*x)/([PCl_5]_0-x)`

Thus, we need to compute the initial concentration of phosphorous pentachloride:

`[PCl_5]_0=(13.0g*(1mol)/(208.25g) )/(0.500L) =0.125M`

So we write:

`1.80=(x^2)/(0.125-x)`

That we solve via either solver or quadratic equation to obtain the solution:

`x=0.117M`

Thereby, the equilibrium concentrations are:

`[PCl_3]_(eq)=x=0.117M`

`[Cl_2]_(eq)=x=0.117M`

`[PCl_5]_(eq)=0.125M-x=0.125M-0.117M=8x10^(-3)M`

Best regards.