Question

Consider the system: 2HBr <>H2+Br2. Intially, 0.24 mol of hydrogen and 0.25 mol of bromine are placed in a 500 mL reaction vessel that is heated electrically. K for the reaction at the temperature used is 0.020/ calculate the concentrations at equilibrium.

Answer 1

I tried this out three times, using 2 different methods.

Answer 2

Answer is that HBr=0.78 mol/L, H2=Br2=0.011 mol/L

Answer 3

Results for:
Method 1: x= 0.32985
Method 2:x=4.16
Method 3: x= 0.0329
Not including x values that are negated due to being negative.
Currently editing: Method 3. Maybe I can figure it out.
Concentration values (Given;solved already): C(H2)=0.5 mol/L
C(Br2)=0.5 mol/L

Answer 4

ill just assume that you're doing the entire question rite now :}

Answer 5

let x mole of h2 and br2 react
0.02=(0.24-x)*(0.25-x)/(2x)^2
x=0.19
h2 conc=(0.24-0.19)/0.5
=0.1
br2 conc=0.12
hbr conc=2*0.19/0.5
=0.76
i dont remember using this method so dont trust this one

Answer 6

Let x mol of H2 reacts with x mol of Br2 to form 2x mol of HBr.
So the equilibrium concentrations are-
HBr- 2x
H2 - 0.24-x
Br2- 0.25-x

K = (0.24-x)(0.25-x)/(2x)^2

Substituting the value of K and solving for x, we get
x= 0.342 mol and x= 0.191 mol.

Since x cannot be greater than 0.24 mol, hence x= 0.191 mol.
Concentrations at equilibrium:
HBr- 0.382 mol
H2 - 0.049 mol
Br2- 0.059 mol
(Use more decimals for higher accuracy)

Answer 7

imma be sleeping soon.
At, E on the ICE table, He=Br2=0.5-x. Hbr IS 2x. You are right there.