Question

The compound PCl5 decomposes into Cl2 and PCl3. The equilibrium of PCl5(g) Cl2(g) + PCl3(g) has a Keq of 2.24 x 10-2 at 327°C. What is the equilibrium concentration of Cl2 in a 1.00 liter vessel containing 0.235 mole of PCl5 and 0.174 mole of PCl3? Remember to use the correct number of significant digits. Cl2 = Are the products or reactants favored?

Answer 1

Take note of the reaction formula which is PCl5=Cl2+PCl3.
The Keq = [Cl2] * [PCl3] / [PCl5]=2.24*10^-2.
For the reason that the volume is 1 liter, the concentration of Cl2 will be computed through: (2.24 * 10^-2) * 0.235 / 0.174  = 0.0303 mol/L is the answer.

Answer 2

The equation you’re dealing with is this:

PCl5(g)⇌PCl3(g)+Cl2(g)

The equation is already balanced so you can now proceed to the next step.

The given equilibrium constant at 327°C is Keq = 2.24 x 10^-2.

Since Keq<1, this means that at equilibrium, the reaction vessel theoretically contains more reactant than the products. Given the value of Keq, we can expect a very low concentration of chlorine gas at equilibrium.

Set up an ICE table to help you find the equilibrium concentration of the three species.

    PCl5(g) ⇌ PCl3(g) + Cl2(g)

I   0.235         0.174       0

C (−x)            (+x)          (+x)

E 0.235 −      0.174 +     x

 

Based on the law,

Keq= (PCl3) x (Cl2)/PCl5

Substitute:

2.24x10^−2 =[(0.174+x)(x)]/( 0.235-x)

Compute for x.

X=0.0239

Therefore, the reaction vessel will contain Cl2=0.0239 M at equilibrium.

Regarding the second question, reactants are favored the equilibrium concentration of chlorine gas is significantly lower than that of phosphorus pentachloride.