Asked by Chemistry
Method for getting correct answer for this? Thank you! A mixture containing an initial concentration of 0.1877 M for H2 and 0.1432 M for Cl2 is allowed to come to equilibrium (see reaction below). What must be the equilibrium concentration of HCl?
H2(g) + Cl2(g) ↔ 2HCl(g) Kc = 5.00e16
a. 0.1432 M
b. 0.0716 M
c. 0.2864 M
d. 0.0865 M
H2(g) + Cl2(g) ↔ 2HCl(g) Kc = 5.00e16
a. 0.1432 M
b. 0.0716 M
c. 0.2864 M
d. 0.0865 M
Answers
Answered by
DrBob222
........H2 + Cl2 ==> 2HCl
I...0.1877.0.1432......0
C.......-x....-x.......2x
E...0.1877-x.0.1432-x..2x
Substitute the E line into Kc expression and solve for x, then evaluate each individual term.
I...0.1877.0.1432......0
C.......-x....-x.......2x
E...0.1877-x.0.1432-x..2x
Substitute the E line into Kc expression and solve for x, then evaluate each individual term.
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