Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

I tried this out three times, using 2 different methods.
Answer is that HBr=0.78 mol/L, H2=Br2=0.011 mol/L
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question