Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
NotTim
Group Title
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.
 2 years ago
 2 years ago
NotTim Group Title
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.
 2 years ago
 2 years ago

This Question is Closed

NotTim Group TitleBest ResponseYou've already chosen the best response.0
I tried this out three times, using 2 different methods.
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
Answer is that HBr=0.78 mol/L, H2=Br2=0.011 mol/L
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
ill just assume that you're doing the entire question rite now :}
 2 years ago

samjordon Group TitleBest ResponseYou've already chosen the best response.1
let x mole of h2 and br2 react 0.02=(0.24x)*(0.25x)/(2x)^2 x=0.19 h2 conc=(0.240.19)/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
 2 years ago

samjordon Group TitleBest ResponseYou've already chosen the best response.1
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  0.24x Br2 0.25x K = (0.24x)(0.25x)/(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)
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
imma be sleeping soon. At, E on the ICE table, He=Br2=0.5x. Hbr IS 2x. You are right there.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.