Question #377c6
1 Answer
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
#K_(eq) = 2.24 * 10^(-2)#
Notice that you have
In other words, you can expect the equilibrium concentration of phosphorus pentachloride,
You will have less reactant at equilibrium than what you started with because the vessel doesn't contain chlorine gas,
Given the value of
Since the vessel has a volume of
#" ""PCl"_ (5(g)) " "rightleftharpoons" " "PCl" _ (3(g))" " + " ""Cl"_ (2(g))#
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 = 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.