# Considering H2(g) + I2(g) â†” 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

Dec 13, 2014

The answer is $0.53 M$.

Starting from the balanced chemical equation

${H}_{2 \left(g\right)} + {I}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 H {I}_{\left(g\right)}$

Since the reaction's equlibrium constant, ${K}_{e q}$, greater than 1, the reaction will favor the formation of the product, $H I$, so we would expct the equilibrium concentrations of ${H}_{2}$ and ${I}_{2}$ to be smaller than the concentration of $H I$.

From the data given we can determine the starting concentrations of both ${H}_{2}$ and ${I}_{2}$ to be

${C}_{{H}_{2}} = {n}_{{H}_{2}} / V = \frac{2.40 m o l e s}{1.00 L} = 2.40 M$
${C}_{{I}_{2}} = {n}_{{I}_{2}} / V = \frac{2.40 m o l e s}{1.00 L} = 2.40 M$

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

...${H}_{2 \left(g\right)} + {I}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 H {I}_{\left(g\right)}$
I: 2.40........2.40...........0
C: (-x).............(-x)..........(+2x)
E: 2.40-x.....2.40-x........2x

We know that ${K}_{e q} = \frac{{\left[H I\right]}^{2}}{\left[{H}_{2}\right] \cdot \left[{I}_{2}\right]}$, so we get

${K}_{e q} = {\left(2 x\right)}^{2} / \left(\left(2.40 - x\right) \left(2.40 - x\right)\right) = \frac{4 {x}^{2}}{2.40 - x} ^ 2 = 49.0$

Rearranging this equation will give us

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

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

$\left[{H}_{2}\right] = 2.40 - 1.87 = 0.53 M$
$\left[{I}_{2}\right] = 2.40 - 1.87 = 0.53 M$
$\left\{H I\right] = 2 \cdot 1.87 = 3.74 M$

Notice how the final concentrations match the estimate derived from ${K}_{e q}$'s value - the reaction indeed favors the product ,$H I$.

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