1. For the reaction in the previous problem, that is, 2HI(g) ↔ H2(g) + I2(g) Keq...

1. For the reaction in the previous problem, that is,

2HI(g) ↔ H2(g) + I2(g) Keq = 0.016

Initially a container contains 0.69 M HI and no product. What is the equilibrium concentration of H2?

2. For the reaction:

2HBr(g) ↔ H2(g) + Br2(g)

Initially a container contains 0.87 M HBr and no product. What is the equilibrium concentration of HBr if the equilibrium concentration of H2 is 0.16 M? (Hint provided in feedback.)

Expert Solution

1)
ICE Table:

[HI] [H2] [I2]

initial 0.69 0 0

change -2x +1x +1x

equilibrium 0.69-2x +1x +1x

Equilibrium constant expression is
Kc = [H2]*[I2]/[HI]^2
0.016 = (1*x)^2/(0.69-2*x)^2
sqrt(0.016) = (1*x)/(0.69-2*x)
0.1265 = (1*x)/(0.69-2*x)
0.08728-0.25298*x = 1*x
0.08728-1.25298*x = 0
x = 0.06966

At equilibrium:
[H2] = x = 0.06966 M

Answer: 0.070 M

Only 1 question at a time please

queen_honey_blossom answered 1 year ago

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