NotTim
  • NotTim
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.
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NotTim
  • NotTim
I tried this out three times, using 2 different methods.
NotTim
  • NotTim
Answer is that HBr=0.78 mol/L, H2=Br2=0.011 mol/L
NotTim
  • NotTim
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

NotTim
  • NotTim
ill just assume that you're doing the entire question rite now :}
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
NotTim
  • NotTim
imma be sleeping soon. At, E on the ICE table, He=Br2=0.5-x. Hbr IS 2x. You are right there.

Looking for something else?

Not the answer you are looking for? Search for more explanations.