Considering H2(g) + I2(g) ↔ 2HI and a temperature of 731K; 2.40 mol of H2 and 2.40 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of H2 in the gaseous mixture? The equilibrium constant is K = 49.0

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Considering H2(g) + I2(g) ↔ 2HI and a temperature of 731K; 2.40 mol of H2 and 2.40 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of H2 in the gaseous mixture? The equilibrium constant is K = 49.0

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Answer 1 · 2014-12-13T18:45:07

The answer is $0.53 M$ $0.53M$ .

Starting from the balanced chemical equation

$H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}$ $H_(2(g)) + I_(2(g)) rightleftharpoons2HI_((g))$

Since the reaction's equlibrium constant, $K_{e q}$ $K_(eq)$ , greater than 1, the reaction will favor the formation of the product, $H I$ $HI$ , so we would expct the equilibrium concentrations of $H_{2}$ $H_2$ and $I_{2}$ $I_2$ to be smaller than the concentration of $H I$ $HI$ .

From the data given we can determine the starting concentrations of both $H_{2}$ $H_2$ and $I_{2}$ $I_2$ to be

$C_{H_{2}} = \frac{n_{H_{2}}}{V} = \frac{2.40 m o l e s}{1.00 L} = 2.40 M$ $C_(H_2) = n_(H_2)/V = (2.40 mol es)/(1.00L) = 2.40M$
$C_{I_{2}} = \frac{n_{I_{2}}}{V} = \frac{2.40 m o l e s}{1.00 L} = 2.40 M$ $C_(I_2) = n_(I_2)/V = (2.40 mol es)/(1.00L) = 2.40M$

We can now determine the equilibrium concentrations for this reaction by using the ICE method (more here: http://en.wikipedia.org/wiki/RICE_chart)

... $H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}$ $H_(2(g)) + I_(2(g))rightleftharpoons2HI_((g))$
I: 2.40........2.40...........0
C: (-x).............(-x)..........(+2x)
E: 2.40-x.....2.40-x........2x

We know that $K_{e q} = \frac{{[H I]}^{2}}{[H_{2}] \cdot [I_{2}]}$ $K_(eq) = ([HI]^2)/([H_2]*[I_2])$ , so we get

$K_{e q} = \frac{{(2 x)}^{2}}{(2.40 - x) (2.40 - x)} = \frac{4 x^{2}}{{(2.40 - x)}^{2}} = 49.0$ $K_(eq) = (2x)^2/((2.40-x)(2.40-x)) = (4x^2)/(2.40-x)^2 = 49.0$

Rearranging this equation will give us

$45 x^{2} - 235.2 x + 282.24 = 0$ $45x^2 - 235.2x + 282.24 = 0$ , which produces two values for $x$ $x$ , $x_{1} = 3.36$ $x_1 = 3.36$ and $x_{2} = 1.87$ $x_2 = 1.87$ ; we cannot choose $x_{1}$ $x_1$ , since that would imply negative concentration values at equilibrium for both $H_{2}$ $H_2$ and $I_{2}$ $I_2$ (2.40 -3.36 = -0.96);

Therefore, $x = 1.87$ $x = 1.87$ , which means that, at equilibrium,

$[H_{2}] = 2.40 - 1.87 = 0.53 M$ $[H_2] = 2.40 - 1.87 = 0.53M$
$[I_{2}] = 2.40 - 1.87 = 0.53 M$ $[I_2] = 2.40 - 1.87 = 0.53M$
${H I] = 2 \cdot 1.87 = 3.74 M$ ${HI] = 2 * 1.87 = 3.74M$

Notice how the final concentrations match the estimate derived from $K_{e q}$ $K_(eq)$ 's value - the reaction indeed favors the product , $H I$ $HI$ .

A quick word on the ICE table...the starting concentration of $H I$ $HI$ is 0 M because only $H_{2}$ $H_2$ and $I_{2}$ $I_2$ are present in the vessel; x simply represents the change in concentrations in accordance to the balanced chemical equation -> 1 mole of $H_{2}$ $H_2$ and 1 mole of $I_{2}$ $I_2$ combine to form 2 moles of $H I$ $HI$ - that is where the +2x comes from...

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Considering H2(g) + I2(g) ↔ 2HI and a temperature of 731K; 2.40 mol of H2 and 2.40 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of H2 in the gaseous mixture? The equilibrium constant is K = 49.0

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