A mixture containing an initial concentration of 0.1726 M for H2 and 0.1361 M for I2...

A mixture containing an initial concentration of 0.1726 M for H2 and 0.1361 M for I2 is allowed to come to equilibrium (see reaction below). What must be the equilibrium concentration of HI? H2(g) + I2(g) ↔ 2HI(g) Kc = 48.7000

0.0726 M
	0.0582 M
	0.1164 M
	0.2327 M

Expert Solution

H₂(g) + I₂(g) ↔ 2HI(g)

initial conc. 0.1726 0.1361 0

change -a -a +2a

Equb conc 0.1726-a 0.1361-a 2a

Equilibrium constant , K_c = [HI(g)]² / ([H₂(g)][I₂(g)] )

K_c = 48.7000 = (2a)² / [(0.1726-a)(0.1361-a)]

Solving we get a = 0.1164M

Therefore the equilibrium concentration of HI = 2a

= 2 x 0.1164 M

= 0.2327 M

queen_honey_blossom answered 2 years ago

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