Question

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide. H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430 ° C. Calculate the equilibrium concentrations of H2, I2, and HI at 430 ° C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.445 M.

Answer #1

**H2 (g) + I2 (g)
<—> 2 HI (g)**

**0
0
0.445 (initial)**

**x
x
0.445-2x (at equilibrium)**

**Kc = [HI]^2 / [H2][I2]**

**54.3 = (0.445-2x)^2 / x^2**

**sqrt(54.3) = (0.445-2x) / x**

**7.37 = (0.445-2x) / x**

**7.37*x = 0.445 - 2x**

**9.37*x = 0.445**

**x = 0.0475 M**

**At equilibrium:**

**[H2] = x = 0.0475 M**

**[I2] = x = 0.0475 M**

**[HI] = 0.445 - 2x = 0.445 - 2*0.0475 = 0.350
M**

Kc for the reaction of hydrogen and
iodine to produce hydrogen iodide.
H2(g) + I2(g) ⇌
2HI(g)
is 54.3 at 430°C. Calculate the
equilibrium concentrations of H2,
I2, and HI at 430°C if
the initial concentrations are [H2] =
[I2] = 0 M,
and [HI] = 0.483 M.

Be sure to answer all parts.
Kc for the reaction of hydrogen and iodine to produce hydrogen
iodide.
H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430°C.
Calculate the equilibrium concentrations of H2, I2, and HI at
430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI]
= 0.567 M.
[H2] = M
[I2] = M
[HI] = M

The equation for the formation of hydrogen iodide from
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H2(g) + I2(g) <--> 2HI(g)
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and I2 are both 0.1600 atm and initially there is no HI
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H2(g) + I2(g) double arrows 2HI (g)
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