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...0.36-x..0.44-x...2x
At equil (H2) = 0.157M; therefore, we know
0.36-x = 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.
A mixture of 1.441 g of H2 and 70.24 g of Br2 is heated in a 2.00-L 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
2 answers
Be sure and check (confirm) those numbers. I just ran them one time on my calculator.