Question

Consider the following reaction where Kc = 1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) A reaction mixture was found to contain 0.121 moles of PCl5(g), 2.60×10-2 moles of PCl3(g), and 3.82×10-2 moles of Cl2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals . The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Answer 1

Answer:

The system is not in equilibrium. The reaction A. must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient (Qc or Qp) is a calculated quantity when a reaction has not yet reached equilibrium. In this way, the reaction quotient allows us to determine if a chemical system is in equilibrium, and if it is not, it allows us to determine in which sense the chemical system will evolve.

For the reaction:

aA + bB ⇔ cC + dD

Qc calculation is done by:  $Qc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

where the concentrations are not those of the equilibrium, but other concentrations given in a certain moment of the reaction.

When Qc <Kc the chemical system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. Then the system will evolve to the right, direct reaction prevailing, to increase the concentration of products (that is, reagents will be consumed and more products will be formed).

When Qc> Kc is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant, that is, it will reach equilibrium. Then the system will evolve to the left, the reverse reaction prevailing, to increase the concentration of reagents (that is, products will be consumed to form more reagents).

If Qc = Kc, then the reaction is in chemical equilibrium and will remain so if conditions do not change.

In the reaction:

PCl₅(g) ⇔ PCl₃(g) + Cl₂(g)

The constant Qc can be obtained by:

$Qc=\frac{[PCl_{3}]*[Cl_{2} ]}{[PCl_{5} ] }$

where:

• $[PCl_{3} ]=\frac{2.60*10^{-2} moles}{1 liter} =2.60*10^{-2} \frac{moles}{L}$

• $[Cl_{2} ]=\frac{3.82*10^{-2} moles}{1 liter} =3.82*10^{-2} \frac{moles}{L}$

• $[PCl_{5} ]=\frac{0.121 moles}{1 liter} =0.121 \frac{moles}{L}$

Replacing:

$Qc=\frac{(2.60*10^{-2} )*(3.82*10^{-2} )}{0.121}$

Qc= 8.21*10⁻³

So Qc <Kc and the system is not in equilibrium. The reaction A. must run in the forward direction to reach equilibrium.