[jubba]

Q11 A sample of phosphorus pentachloride is placed in a sealed container, where it decomposes into phosphorus trichloride and chlorine gas according to: PCl5 (g) <-> PCl3 (g) + Cl2 (g)

Once the mixture is at equilibrium, a small quantity of helium gas is added, at constant pressure and temperature, and the mixture is allowed to return to equilibrium. Which of the following best describes the second equilibrium position with respect to the first?


A The second equilibrium position is the same as the first, because helium is much lighter than any of the other molecules in the container and will have a negligible effect on the reaction.

B The second equilibrium position is the same as the first, because helium does not react with any of the other molecules in the container.

C The second equilibrium position has more PCl3 than the first.

D The second equilibrium position has more PCl5 than the first.

E It is impossible to tell what will happen without knowing the equilibrium constant for this reaction.


The real answer is c but i don't know why

[sdekivit]

what whill happen when we add He in the sytem with the pressure ? And what is the effect of this on the equilibrium ?

[Donaldson Tan]

now your question is easier to read.

PCl5 (g) <-> PCl3 (g) + Cl2 (g)

K_p = P_PCl3.P_Cl2/P_PCl5

Adding helium reduces the mole fraction of PCl5, PCl3 and Cl2. However, the effect of reducing the mole fraction of the products is greater than that of of the reactants. Therefore, the favoured reaction is one that increases the molar fraction of the products, ie. the forward the reaction.

This means there will be more PCl3 than the first equilibrium.

[jubba]

Would adding helium an inert gas effect the concentration of the other chemicals?
For the equilibrium constant concentration is in moles per litre.
And the number of moles per litre is unaffected by the addition of an inert gas.
The pressure is also uneffected by the addition of a inert gas

[oldddog]

I think the answer lies with Le Chateliers Principle.

[jubba]

I think the answer lies with Le Chateliers Principle.

of course
But the question is how do we apply it to the equilibrium
 

[Donaldson Tan]

did anyone read my previous post in this topic? it is based on the Le Chatelier's Principle.

[jubba]

But the reduced molefraction does not effect the reaction quotient. The partial pressure of each component remains the same as far as i know. So shouldn't the equilibrium return to its original position.

[Donaldson Tan]

The partial pressure of each component remains the same as far as i know.

No! The pressure of the system at the second equilibrium is the same as the first equilibrium, so the partial pressure of each component reduces with its corresponding mole fraction.

[jubba]

i overlooked the bit on the question about constant pressure.