IChO 1991 · Theory problem 6

Production of sulphuric acid and oleum

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Sulphuric acid is produced by catalytic oxidation of \(\ce{SO2}\) to \(\ce{SO3}\), absorption of \(\ce{SO3}\) in concentrated sulphuric acid forming oleum (containing 20 % \(\ce{SO3}\) by mass) and appropriate dilution hereafter. The gas leaving the catalyst chamber contains nitrogen, oxygen, a trace of \(\ce{SO2}\) and 10 % (by volume) of \(\ce{SO3}\). Sulphur trioxide, \(\ce{SO3}\), is converted into sulphuric acid (98 % by mass) and/or oleum.

6.1 Assuming that oleum is the only product formed, calculate the mass of water which is required for \(1000\ \mathrm{m^3}\) of gas leaving the catalyst chamber (273 K, 101.3 kPa).

6.2 Assuming that only 98 % sulphuric acid is produced, calculate the necessary mass of water and the mass of product produced thereby.

6.3 In the industry both oleum and 98 % sulphuric acid are produced in a mass ratio of \(x=m_1/m_2\), where \(m_1\) denotes the mass of oleum, \(m_2\) the mass of 98 % sulphuric acid. Find an expression \(y=f(x)\) to describe the relation between the mass of water consumed for \(1000\ \mathrm{m^3}\) gas (denoted \(y\)) and the value of \(x\). Show that the results of 6.1) and 6.2) are in good agreement with your mathematical expression.

Official English text of the problem.

Answers

  • 6.1: Water required: \(\boxed{61.5\ \mathrm{kg}}\).
  • 6.2: Water required: \(\boxed{89.3\ \mathrm{kg}}\); mass of 98 % sulphuric acid produced: \(\boxed{446\ \mathrm{kg}}\).
  • 6.3: For \(x=m_1/m_2\geq0\), \[ \boxed{y\approx(357\ \mathrm{kg})\frac{36x+49}{209x+196}}. \] Only 98 % acid corresponds to \(x=0\), giving \(y=89.3\ \mathrm{kg}\). Only oleum corresponds to \(x\to\infty\), giving \(y\to61.5\ \mathrm{kg}\).

Solution

The reaction species are sulphur dioxide (O=S=O), oxygen (O=O), sulphur trioxide (O=S(=O)=O), sulphuric acid (OS(=O)(=O)O), and water (O).

Catalytic oxidation produces sulphur trioxide: \[ \ce{2SO2 + O2 -> 2SO3}. \] The net reaction consuming water during acid production is \[ \ce{SO3 + H2O -> H2SO4}. \]

Use the molar masses \[ M(\ce{SO3})=80.0,\qquad M(\ce{H2O})=18.0,\qquad M(\ce{H2SO4})=98.0 \quad\mathrm{kg\,kmol^{-1}}. \]

We balance the overall process, so circulating sulphuric acid cancels from the input and output streams. The stated oleum composition means 20 % excess \(\ce{SO3}\) and 80 % \(\ce{H2SO4}\) on an analytical mass basis. Oleum and aqueous sulphuric acid are mixtures and therefore have no single molecular SMILES.

6.1

Treating the gas as ideal, the volume fraction of \(\ce{SO3}\) equals its mole fraction. Using \[ R=8.314\ \mathrm{kPa\,m^3\,kmol^{-1}\,K^{-1}}, \] the amount of sulphur trioxide available is \[ n(\ce{SO3}) =\frac{0.10\,pV}{RT} =\frac{0.10(101.3)(1000)}{8.314(273)} =4.4631\ \mathrm{kmol}. \] Thus, its mass, denoted \(S\), is \[ S=4.4631(80.0)=357.048\ \mathrm{kg}. \]

Let \(y_1\) be the water required and \(m_1\) the oleum produced. All added water reacts, producing \((98/18)y_1\) kilograms of sulphuric acid. Since the oleum contains 80 % sulphuric acid, \[ \frac{98}{18}y_1=0.80m_1. \] The total mass balance is \[ m_1=S+y_1. \] Combining these equations, \[ 98y_1=14.4(S+y_1), \] so \[ y_1=\frac{14.4}{83.6}S =\frac{36}{209}(357.048) =61.501\ \mathrm{kg}. \] Therefore, \[ \boxed{m_{\text{water}}=61.5\ \mathrm{kg}}. \]

6.2

All sulphur trioxide now reacts to form sulphuric acid. The mass of pure acid formed is \[ m(\ce{H2SO4})=4.4631(98.0)=437.384\ \mathrm{kg}. \] Since this constitutes 98 % of the product, \[ m_2=\frac{437.384}{0.98} =446.310\ \mathrm{kg}. \]

The required water has two contributions: water incorporated into sulphuric acid and water remaining in the 98 % solution: \[ \begin{aligned} y_2 &=n(\ce{SO3})M(\ce{H2O})+0.02m_2\\ &=4.4631(18.0)+0.02(446.310)\\ &=80.336+8.926\\ &=89.262\ \mathrm{kg}. \end{aligned} \]

Hence, \[ \boxed{m_{\text{water}}=89.3\ \mathrm{kg}}, \qquad \boxed{m_{\text{98 \% acid}}=446\ \mathrm{kg}}. \]

6.3

For an oleum mass \(m_1\), the incoming sulphur trioxide supplies both the excess \(\ce{SO3}\), of mass \(0.20m_1\), and the \(\ce{SO3}\) incorporated into \(0.80m_1\) of sulphuric acid. Thus its total contribution is \[ \left(0.20+0.80\frac{80}{98}\right)m_1 =\frac{209}{245}m_1. \]

For a mass \(m_2\) of 98 % sulphuric acid, the required sulphur trioxide mass is \[ 0.98\frac{80}{98}m_2=\frac45m_2. \]

The sulphur trioxide balance is therefore \[ S=\frac{209}{245}m_1+\frac45m_2. \] The overall mass balance gives the water input: \[ y=m_1+m_2-S =\frac{36}{245}m_1+\frac15m_2. \]

Substituting \(m_1=xm_2\), \[ S=\left(\frac{209}{245}x+\frac45\right)m_2, \qquad y=\left(\frac{36}{245}x+\frac15\right)m_2. \] Eliminating \(m_2\) yields \[ \boxed{ y=S\frac{36x+49}{209x+196} \approx(357\ \mathrm{kg})\frac{36x+49}{209x+196} }. \]

The limiting cases reproduce the earlier answers:

  • Only 98 % sulphuric acid: \(m_1=0\), so \(x=0\): \[ y(0)=S\frac{49}{196}=\frac{S}{4} =89.3\ \mathrm{kg}. \] The product mass is \(m_2=S+y=446\ \mathrm{kg}\), agreeing with 6.2.

  • Only oleum: \(m_2\to0\), so \(x\to\infty\): \[ \lim_{x\to\infty}y =S\frac{36}{209} =61.5\ \mathrm{kg}, \] agreeing with 6.1.

The solution was checked against the official solution.