g chem problem

[mrdorky]

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When 14.25 moles of PCl5 gas is placed in a 3 L container and comes to equilibrium at constant temperature, 40% of the PCl5 decomposes according to equation:
PCl5(g) <--> PCl3 (g) + Cl2(g)
What is the Kc for this rxn?



how would you set this up? I'm kind of confused with this problem. 😴

 

[Baylor2011]

When 14.25 moles of PCl5 gas is placed in a 3 L container and comes to equilibrium at constant temperature, 40% of the PCl5 decomposes according to equation:
PCl5(g) <--> PCl3 (g) + Cl2(g)
What is the Kc for this rxn?

I got this wrong today as well...just went back and figured it out.

First - write the equilibrium expression: Kc = [PCl3][Cl2]/[PCl5]

Then, you have to convert the moles to molarity (moles/L)

14.25 moles PCl5/3L = 4.75M

Now, since it says only 40% decomposes @ equilibrium, that means only 40% of the starting concentration goes to the product side of the equation

4.75M x .4 (just solve longhand) - 1.9M

Since PCl5 breaks down into equal concentrations of products...can just look at it as 1.9^2

Now, PCl5's concentration at equilibrium is 100% - 40% that decomposed = 60%

Take 4.75M x .6 and you get 2.85

Kc = 1.9^2 / 2.85

 

[mrdorky]

Thanks Baylor! Really appreciate it.
For the longest time I was wondering how they calculated the 2.85, but now it makes sense! 👍

 

[Baylor2011]

Haha me too! When I was taking the test I was thinking to myself "This question doesn't make any sense at all WTF?!?!" Then I guessed and marked it.

 

[rpatel8]

Another good way to think about how to get 2.85 more easily is that you have 4.75 M and only 1.9 M decomposes so 4.75-1.9= 2.85M actually decomposes to products.

 

[mrdorky]

Simplified enough! thanks man!