Posted by Mariah on .
A mixture of 1.441 g of H2 and 70.24 g of Br2 is heated in a 2.00L vessel at 700 K. These substances react as follows.
H2(g) + Br2(g) arrow 2 HBr(g)
At equilibrium the vessel is found to contain 0.627 g of H2.
calculate their equilibrium concentrations and Kc

chemistry 
DrBob222,
mols H2 @ equil = 0.627g/2 = 0.3135
M H2 @ equil = 0.3135/2L = 0.1567M
M H2 begin = (1.441/4) = 0.360M
M Br2 begin = (70.24/159.8) = 0.440M
........H2 + Br2 ==> 2HBr
I.....0.36..0.44.....0
C.....x.....x.......2x
E...0.36x..0.44x...2x
At equil (H2) = 0.157M; therefore, we know
0.36x = 0.157 and we solve for x. Use that to calculate each of the equilibrium values, substitute those into the Kc expression and calculate Kc. 
chemistry 
DrBob222,
Be sure and check (confirm) those numbers. I just ran them one time on my calculator.