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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.

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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

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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.

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