Finding Kc for 2A(g) ⇌ B(g) + C(g) with initial amounts

[~angel~]

Could someone please help me with this problem?

Samples of A (2.5 mol) and B (1.0 mol) are placed in a 5.0L container and the following reaction takes place:


2A(g) <--> B(g) + C(g)

At equilibrium, the concentration of A is 0.20M. What is the value of Kc?

Thanks. It's probably really simple but I can't seem to get it. I'm not that strong on this stuff.

Also, apparently the relationship between Kc and Kp is Kc = Kp. Why is that the case?

Thanks again.

 

[Dr.Brain]

$K_p=K_c (RT)^{dn}$

where $dn$= No. of moles of products- No. of moles of reactants

Kc = Kp because the number of moles of the gaseous componentson both sides are the same.

For your first query , what is law of mass action?...How is Kc related to the products and reactants?

 

[GCT]

$$Kc= \frac{[C]<b>}{[A]^{2}} </b>$$

$$Kc= \frac{[C_b+.25mol/5.0L][.25mol/5.0L]}{[.200M]^{2}}$$

$$Kc= \frac{[1.0mol/5.0L+.25mol/5.0L][.25mol/5.0L]}{[.200M]^{2}}$$

solve for $$Kc$$

 

[siddharth]

Initially, the concentration of all the species are
[A] = 2.5(mol) / 5.0(L) = c1 -I
= 1.0(mol) / 5.0(L) = c2 -II
[C] = 0

Now, let 'a' be the extent of the reaction

Concentration of all species at equilibirium are

[A]=c1(1-2a) -III
=c2+c1(a)
[C]=c1(a)

Now it is given that the concentration of [A] at equilibirium is 0.2 mol/L.
From III, c1(1-2a)=0.2
Now, the value of c1 is known.Therefore the value of a can be found.
From this the value of Kc=([C])/([A]^2) {The concentration of species at equilibirium}

GCT:
Instead of 0.25, shouldn't it be 0.75? Initially, number of moles of A are 2.5. Finally, the number of moles of A 0.2*5=1 . Therefore the number of moles of A reacted = 2.5-1=1.5 . By molar ratio, the number of moles of B and C formed are 0.75.

 

[~angel~]

Ok, I get it. Thanks guys

 

[GCT]

GCT:
Instead of 0.25, shouldn't it be 0.75? Initially, number of moles of A are 2.5. Finally, the number of moles of A 0.2*5=1 . Therefore the number of moles of A reacted = 2.5-1=1.5 . By molar ratio, the number of moles of B and C formed are 0.75.

I'm not quite sure why you're trying to find the number of moles, when the equilibrium concentration of A is given. Anyways I neglected to use the appropriate initial concentration for A.

$$.25mol/5.0L=.5M,~mol~A~reacted~=~.5M-.20M=.3M$$

you'll need to divide by 2, to find the equilvalent moles of B and C

$$.3M~A~reacted(1~mol~B/2~mol~A)~=~.15M~B,C~formed$$

 

[siddharth]

I'm not quite sure why you're trying to find the number of moles, when the equilibrium concentration of A is given. Anyways I neglected to use the appropriate initial concentration for A.

$$.25mol/5.0L=.5M, mol A reacted~=~.5M-.20M=.3M$$

It should be 2.5mol/5L=.5M
Anyway, finding the number of moles is the same thing. If 0.75 moles of B and C are formed, then the concentration of each is 0.75/5= 0.15 M. That's why in your initial post, instead of 0.25/5, it should be 0.75/5=0.15M. (As you wrote in the last post)