IChO 1975 · Theory problem 6

Hydrogen iodide equilibrium at 600 °C

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The equilibrium constant of the reaction

Figure 1

is at 600 °C equal to 70.0.

Problems:

6.1 How much iodine (in %) is converted till the equilibrium is reached if the reactants are mixed in:

a) 1 : 1 molar ratio at 600 °C;

b) 2 : 1 molar ratio at 600 °C (the amount of hydrogen is twice as great as that of iodine).

6.2 How many moles of hydrogen should be mixed with one mole of iodine when 99 % of iodine is to be converted to hydrogen iodide till the equilibrium is reached at 600 °C?

Official English text of the problem.

Answers

  • 6.1 a) \(80.7\%\) of the iodine is converted.
  • 6.1 b) \(95.1\%\) of the iodine is converted.
  • 6.2 Initially mix \(6.59\ \mathrm{mol}\) of hydrogen with one mole of iodine.

Solution

The balanced equilibrium reaction is

\[ \ce{H2(g) + I2(g) <=> 2HI(g)}. \]

The species are hydrogen, \(\ce{H2}\), [H][H]; iodine, \(\ce{I2}\), II; and hydrogen iodide, \(\ce{HI}\), [H]I.

Take an initial amount of \(1.00\ \mathrm{mol}\) of iodine and \(r\ \mathrm{mol}\) of hydrogen, with no hydrogen iodide initially. Let \(x\) be the fraction of the initial iodine converted.

Species Initial amount / mol Equilibrium amount / mol
\(\ce{H2}\) \(r\) \(r-x\)
\(\ce{I2}\) \(1\) \(1-x\)
\(\ce{HI}\) \(0\) \(2x\)

At \(600\,^\circ\mathrm{C}\),

\[ K=\frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}=70.0. \]

The volume factors cancel because the total stoichiometric coefficients are equal on both sides:

\[ 70.0 =\frac{(2x/V)^2}{((r-x)/V)((1-x)/V)} =\frac{4x^2}{(r-x)(1-x)}. \]

6.1 a) Initial molar ratio \(1:1\)

Here \(r=1\), so

\[ 70.0=\frac{4x^2}{(1-x)^2}. \]

Since \(0<x<1\), taking the positive square root gives

\[ \sqrt{70.0}=\frac{2x}{1-x}, \qquad x=\frac{\sqrt{70.0}}{2+\sqrt{70.0}} =0.807073. \]

Therefore,

\[ \boxed{\text{Iodine conversion}=80.7\%.} \]

6.1 b) Initial molar ratio \(2:1\)

Here \(r=2\):

\[ 70.0=\frac{4x^2}{(2-x)(1-x)}. \]

Rearranging,

\[ 70(2-3x+x^2)=4x^2, \]

\[ 33x^2-105x+70=0. \]

Thus,

\[ x=\frac{105\pm\sqrt{1785}}{66}. \]

The larger root, \(2.23105\), is impossible because the iodine conversion cannot exceed \(1\). The physical root is

\[ x=\frac{105-\sqrt{1785}}{66}=0.950769. \]

Therefore,

\[ \boxed{\text{Iodine conversion}=95.1\%.} \]

6.2 Hydrogen required for \(99\%\) iodine conversion

Set \(x=0.99\). With one mole of iodine initially, the equilibrium amounts are

\[ n(\ce{I2})=0.01\ \mathrm{mol},\qquad n(\ce{HI})=1.98\ \mathrm{mol},\qquad n(\ce{H2})=(r-0.99)\ \mathrm{mol}. \]

Hence,

\[ 70.0=\frac{1.98^2}{(r-0.99)(0.01)}. \]

Solving for the initial hydrogen amount,

\[ r=0.99+\frac{1.98^2}{70.0(0.01)} =6.59057. \]

Therefore,

\[ \boxed{n_0(\ce{H2})=6.59\ \mathrm{mol} \quad\text{per }1\ \mathrm{mol}\text{ of iodine}.} \]

Of this hydrogen, \(0.99\ \mathrm{mol}\) reacts and approximately \(5.60\ \mathrm{mol}\) remains at equilibrium.

The solution was checked against the official solution.