Question

Can someone explain to me how the answer to these two problems are this..

Given the equilibrium below: PCl3(g) + Cl2(g) ⇔ PCl5(g) Kc = 0.18
(a) What is the equilibrium constant for PCl5(g) ⇔ PCl3(g) + Cl2(g)? answer 5.56
(b) What is the equilibrium constant for ⅓ PCl5(g) ⇔ ⅓ PCl3(g) + ⅓ Cl2(g)? answer=1.77

Answer 1

Remember, the equilibrium constant is just: \[K=\frac{[products]}{[reactants]}\] so if you look at the rate constant you're given, the product is PCl5 and the reactants are PCl3 and Cl2. Now if you look at a) you can see that it's almost the same thing, except now the products and reactants have reversed! That means the rate constant is also flipped upside down like a fraction. \[K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]}\] becomes \[\frac{1}{K_c} = \frac{[PCl_3][Cl_2]}{[PCl_5]} \] So just divide 1 by your rate constant to get the answer for part a.

If that makes sense, then we can move on to part b.

Answer 2

Oh I see! , yes i understand =)

Answer 3

I'm sorry I have to leave but perhaps @Somy can help you out; she's very knowledgeable!

Answer 4

okay think you

Answer 5

isn't it done?

Answer 6

b part is basically same as a
the only thing is that for b part all reactants and products have coefficient 1/3
thus this 1/3 in formula of Kc will be power of reactant or product it belongs to
meaning coefficient of a compound is a power of that compound in Kc formula

Answer 7

no she was gong to go into step two.. but she had to go

Answer 8

oh ok

Answer 9

Okay I will try, doing that, thank you =)

Answer 10

u r welcome :)

Answer 11

i hope u got it that it's just same as @Kainui said
in part a its 1/Kc
in part be however its \[(\frac{ 1 }{ Kc })^{1/3}\]

Answer 12

b*