(HBr) = 3.20mols/21.3L = about 0.150 but that's an estimate.
It's just like it sounds. I = initial; C = change; E = equilibrium.
I assume you made a typo and meant Br2.
..........H2 + Br2 ==> 2HBr
I.........0.....0......0.150
C.........x.....x.......-x
E.........x.....x......0.150-x
Substitute E line into Kc expression and solve for x and 0.150-x.
equilibrium constant = 2.180 x 10^6 @ 730 C.
H2 (g) + Br (g) <=> 2HBr (g) (reversible)
startin with 3.20 moles of HBr in a 21.3-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium
idk how to start my ice table
2 answers
shouldn't it be -2x?
4 answers