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Chlorine gas [tex](Cl_2)[/tex] can be produced using this equilibrium reaction involving gaseous [tex]PCl_5[/tex] and [tex]PCl_3[/tex]. Which change would drive this process toward the product side?

[tex]PCl_5 + \text{energy} \rightleftarrows PCl_3 + Cl_2[/tex]

A. Increasing the pressure

B. Decreasing the temperature

C. Removing the [tex]PCl_5[/tex] as it forms

D. Decreasing the pressure

Sagot :

GinnyAnswer
Sure, let's analyze the given equilibrium reaction and explore how different changes would affect it.

The reaction given is:

[tex]\[ \text{PCl}_5 + \text{energy} \rightleftarrows \text{PCl}_3 + \text{Cl}_2 \][/tex]

According to Le Chatelier's principle, a system at equilibrium will adjust to counteract any changes applied to it. Let's examine each change proposed:

A. Increasing the pressure:

Increasing the pressure of a system will favor the side of the reaction with fewer moles of gas. In this reaction:
- On the reactant side, there is 1 mole of gas ([tex]$\text{PCl}_5$[/tex]).
- On the product side, there are 2 moles of gas ([tex]$\text{PCl}_3$[/tex] and [tex]$\text{Cl}_2$[/tex]).

Increasing the pressure would therefore shift the equilibrium towards the reactant side (where there are fewer moles of gas), which is not the direction towards more chlorine gas [tex]$\left(Cl_2\right)$[/tex] being produced.

B. Decreasing the temperature:

The reaction is endothermic (it requires energy to proceed towards the products). Decreasing the temperature would favor the exothermic direction, which is the reverse reaction in this case. Therefore, it would shift the equilibrium towards the reactant side and lead to the formation of more [tex]$\text{PCl}_5$[/tex] rather than [tex]$\text{Cl}_2$[/tex].

C. Removing the [tex]$\text{PCl}_5$[/tex] as it forms:

If you remove [tex]$\text{PCl}_5$[/tex] continuously, the reaction would try to produce more [tex]$\text{PCl}_5$[/tex] to replace what is being removed. This again does not drive the process towards producing more [tex]$\text{Cl}_2$[/tex].

D. Decreasing the pressure:

Decreasing the pressure of a system will favor the side with more moles of gas. For our reaction:
- The reactant side has 1 mole of gas.
- The product side has 2 moles of gas.

By decreasing the pressure, the equilibrium will shift towards the side with more moles of gas – the product side. Thus, this would drive the reaction towards producing more [tex]$\text{PCl}_3$[/tex] and [tex]$\text{Cl}_2$[/tex].

Therefore, the correct choice is:

D. Decreasing the pressure
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