Question

Question #377c6

Answer 1

`["Cl"_2] = "0.0239 M"`

Explanation:

You are dealing with the following equilibrium reaction

`"PCl"_ (5(g)) rightleftharpoons "PCl" _ (3(g)) + "Cl"_ (2(g))`

You also know that the equilibrium constant for this reaction at `327^@"C"` is equal to

`K_(eq) = 2.24 * 10^(-2)`

Notice that you have `K_(eq) < 1`; this should tell you that at equilibrium, the reaction vessel should contain more reactant than products.

In other words, you can expect the equilibrium concentration of phosphorus pentachloride, `"PCl"_5`, to decrease only slightly compared to its initial value.

You will have less reactant at equilibrium than what you started with because the vessel doesn't contain chlorine gas, `"Cl"_2`, so right from the start, you can predict that the reaction will produce some chlorine gas.

Given the value of `K_(eq)`, we can expect a very low concentration of chlorine gas at equilibrium.

Since the vessel has a volume of `"1.00 L"`, you can treat moles and molarity interchangeably. Set up an ICE table to help you find the equilibrium concentration of the three species

`" ""PCl"_ (5(g)) " "rightleftharpoons" " "PCl" _ (3(g))" " + " ""Cl"_ (2(g))`

`color(purple)("I")color(white)(aaaacolor(black)(0.235)aaaaaaaaaacolor(black)(0.174)aaaaaaaaacolor(black)(0)`
`color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaacolor(black)((+x))aaaaaacolor(black)((+x))`
`color(purple)("E")color(white)(aacolor(black)(0.235-x)aaaaaaacolor(black)(0.174+x)aaaaaaacolor(black)(x)`

By definition, the equilibrium constant is equal to

`K_(eq) = (["PCl"_3] * ["Cl"_2])/(["PCl"_5])`

In your case, you will have

`2.24 * 10^(-2) = ((0.174 + x) * x)/(0.235 - x)`

This is equivalent to

`x^2 + 0.1964x - 0.005264 = 0`

Tis quadratic equation ahs two solutiuons, one positive and one negative. Since `x` represents concentration, discard the negative value and go with

`x = 0.0239`

Therefore, you can say that at equilibrium, the reaction vessel will contain

`color(darkgreen)(ul(color(black)(["Cl"_2] = "0.0239 M")))`

The answer is rounded to three sig figs.

As predicted, the equilibrium concentration of chlorine gas is significantly lower than that of phosphorus pentachloride and that of phosphorus trichloride.