[aficc]

The equilibrium reaction H_2 (g) + Br_2 (g) <=> 2HBr (g) has an equilibrium constant of 4.1 x 10^18 at 298 K.

1. If you begin with 10 mol of Br_2 and 10 mol of H_2 in a 5.0-L container, what is the partial pressure of HBr at equilibrium?

2. What is the partial pressure of H_2 at equilibrium?

3. If H_2 is removed from the system, what is the effect on the partial pressure of Br_2?

for 1. I did :

AT equilibrium H_2 (g) + Br_2 (g) <=> 2HBr (g)
                            (2-x)        (2-x)            (2x)

2x^2 / (2-x)^2 = 4.1 x 10^18
                   x = 1.9999

However this Phbr = 2x1.9999 = 3.9999.... however this is not right...

Any help would be appreciated. thanks.

[Sev]

AT equilibrium H_2 (g) + Br_2 (g) <=> 2HBr (g)
                            (2-x)        (2-x)            (2x)

2x^2 / (2-x)^2 = 4.1 x 10^18
                   x = 1.9999

However this Phbr = 2x1.9999 = 3.9999.... however this is not right...

Where did you get (2-x)?  There would be 10-x mol at equilibrium.  For P_HBr, use gas law.

[aficc]

       H_2 (g) + Br_2 (g) <=> 2HBr (g)
I |    2 M          2M               0
C|     -x            -x              +2x
E|    (2-x)       (2-x)            2x

Thats how I got (2-x). 2 M is the concentration of H2 and Br2 (10 mol/5 L)

[Sev]

       H_2 (g) + Br_2 (g) <=> 2HBr (g)
I |    2 M          2M               0
C|     -x            -x              +2x
E|    (2-x)       (2-x)            2x

Thats how I got (2-x). 2 M is the concentration of H2 and Br2 (10 mol/5 L)

Sorry, my mistake (It's early here   )

Looking at it again, you seem to be right.  Although you have confused partial pressure with concentration.

[aficc]

      H_2 (g) + Br_2 (g) <=> 2HBr (g)
I |    10         10               0
C|     -x            -x              +2x
E|    (10-x)       (10-x)            2x

(2x)^2 / (10-x)^2 = 4.1 x 10^18
                       x = 10

Partial pressure of Hbr = nRT/v
                               = 20 x 0.0821 x 298 / 5
                               = 97.62 atm

Well, I've figured out #1, but for #2, (10-10) = 0 mol... how would I solve it?

[Sev]

      H_2 (g) + Br_2 (g) <=> 2HBr (g)
I |    10         10               0
C|     -x            -x              +2x
E|    (10-x)       (10-x)            2x

(2x)^2 / (10-x)^2 = 4.1 x 10^18
                       x = 10

Partial pressure of Hbr = nRT/v
                               = 20 x 0.0821 x 298 / 5
                               = 97.62 atm

Well, I've figured out #1, but for #2, (10-10) = 0 mol... how would I solve it?

1.  Your initial ICE table was fine.  Solve: 4x²/(2-x)² = 4.1*10¹⁸

P_HBr = nRT/V = cRT (c=2x)

2. Use same method as above.