Considering H2(g) + I2(g) ↔ 2HI and temperature = 731K, 1.20 mol of H2 and 1.20 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of I2 in the gaseous mixture? The equilibrium constant is K = 49.0

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Considering H2(g) + I2(g) ↔ 2HI and temperature = 731K, 1.20 mol of H2 and 1.20 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of I2 in the gaseous mixture? The equilibrium constant is K = 49.0

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Answer 1 · 2014-10-08T00:59:47

The equilibrium concentration of I₂ is 0.933 mol/L.

We start with the balanced chemical equation for the equilibrium.

H₂ + I₂ ⇌ 2HI

Then we write the $K_{eq}$ $K_"eq"$ expression.

$K_{eq} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]}$ $K_"eq" = ["HI"]^2/(["H"_2]["I"_2])$

Next, we set up an ICE table:

$H_{2} + I_{2} ⇌ 2 HI$ $"H"_2 + "I"_2 ⇌ 2"HI"$

I/mol·L⁻¹: 1.20; 1.20; 0
C/ mol·L⁻¹: - $x$ $x$ ; - $x$ $x$ ; + $2 x$ $2x$
E/ mol·L⁻¹: 1.20- $x$ $x$ ; 1.20- $x$ $x$ ; $2 x$ $2x$

Insert these values into the $K_{eq}$ $K_"eq"$ expression:

$K_{eq} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]} = \frac{{(2 x)}^{2}}{(1.20 - x) (1.20 - x)} = 49.0$ $K_"eq" = ["HI"]^2/(["H"_2]["I"_2]) = (2x)^2/((1.20-x)(1.20-x)) = 49.0$

$\frac{4 x^{2}}{{(1.20 - x)}^{2}} = 49.0$ $(4x^2)/(1.20 - x)^2 = 49.0$

$\frac{2 x}{1.20 - x} = 7.00$ $(2x)/(1.20 – x) = 7.00$

$2 x = 8.40 - 7.00 x$ $2x= 8.40 – 7.00x$

$9.00 x = 8.40$ $9.00x = 8.40$

$x = 0.933$ $x= 0.933$

[I₂] = (1.20 – $x$ $x$ ) mol/L = (1.20 – 0.933) mol/L = 0.27 mol/L

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Considering H2(g) + I2(g) ↔ 2HI and temperature = 731K, 1.20 mol of H2 and 1.20 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of I2 in the gaseous mixture? The equilibrium constant is K = 49.0

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