Question 8.5.1: Production of Hydrochloric Acid Hydrochloric acid is produce......

Production of Hydrochloric Acid
Hydrochloric acid is produced by absorbing gaseous HCl (hydrogen chloride) in water. Calculate the heat that must be transferred to or from an absorption unit if HCl(g) at 100°C and H_{2}O(l) at 25°C are fed to produce 1000 kg/h of 20.0 wt% HCl(aq) at 40°C.

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It is advisable to determine the molar amounts or flow rates of the components of all feed and product solutions before drawing and labeling the flowchart. In this case

\begin{matrix} 1000  kg/h  of  20.0  wt\%  HCl(aq) \\ \left. \Large{\Downarrow} \right.\\\\ \dot{n}_{HCl}= \begin{array}{c|c|c}1000  kg& 0.200  kg  HCl &10^{3}  mol \\ \hline h &kg& 36.5  kg  HCl\end{array} = 5480  mol  HCl/h \\\\ \dot{n}_{H_{2}O}= \begin{array}{c|c|c}1000  kg& 0.800  kg  H_{2}O &10^{3}  mol \\ \hline h &kg& 18.1  kg  H_{2}O\end{array} = 44,400  mol  H_{2}O/h \end{matrix}

The enthalpy table for the process is shown below. As usual, physical property data valid at P = 1 atm are used and the effects on enthalpy of any pressure differences that may occur in the process are neglected.

Note that the value of \dot{n} for the product solution is the molar flow rate of the solute (HCl) rather than the solution, since the enthalpy will be determined in kJ/mol solute.

Calculate \hat{H}_{1} and \hat{H}_{2} :        HCl(g, 25°C) → HCl(g, 100°C)

\hat{H}_{1} = \Delta \hat{H} = \int_{25°C}^{100°C}{C_{p}  dT} \\ \left. \Large{\Downarrow} \right. C_{p}  for  HCl(g)  from  Table  B.2 \\ \hat{H}_{1} = 2.178  kJ/mol

For the product solution,

\begin{matrix}r= (44,400  mol  H_{2}O) /(5480  mol  HCl) = 8.10 \\ HCl(g, 25°C) + 8.10  H_{2}O(1, 25°C) \overset{\Delta \hat{H}_{a}}{\left. \Large{\longrightarrow } \right.} HCl(aq, 25°C) \overset{\Delta \hat{H}_{b}}{\left. \Large{\longrightarrow } \right.} HCl(aq, 40°C) \\ \Delta \hat{H}_{a} = \Delta \hat{H}_{s} (25°C,  r = 8.1) \overset{Table  B.11}{\left. \Large{\Longrightarrow } \right.} –  67.4  kJ/mol HCl\end{matrix}

The heat capacities of aqueous hydrochloric acid solutions are listed on p. 2-183 of Perry’s Chemical Engineers’ Handbook (see Footnote 5) as a function of the mole fraction of HCl in the solution, which in our problem is

\frac{5480  mol  HCl/h}{(5480 + 44,400)  mol/h} = 0.110  mol  HCl/mol \\ \left. \Large{\Downarrow} \right. \\ C_{p} = \begin{array}{c|c|c} 0.73  kcal &1000  kg  solution &4.184  kJ\\ \hline kg\cdot °C& 5480  mol  HCl &kcal\end{array} = 0.557  \frac{kJ}{mol  HCl \cdot°C} \\ \Delta \hat{H}_{b}= \int_{25°C}^{40°C}{C_{p}  dT} = 8.36  kJ/mol  HCl \\ \left. \Large{\Downarrow} \right. \\ \hat{H}_{2} = \Delta \hat{H}_{a} + \Delta \hat{H}_{b} +(-  67.4 + 8.36)  kJ/mol  HCl=-  59.0  kJ/mol  HCl 

Energy Balance:

\dot{Q}=\Delta \dot{H} = \sum\limits_{out}{\dot{n}_{i}\hat{H}_{i}} – \sum\limits_{in}{\dot{n}_{i}\hat{H}_{i}} \\ = (5480  mol  HCl/h) (-  59.0  kJ/mol  HCl)  –  (5480  mol  HCl/h)(2.178  kJ/mol  HCl) \\ =\boxed{ –  3.35 \times 10^{5}  kJ /h}

Heat must be transferred out of the absorber at a rate of 335,000 kJ/h to keep the product temperature from rising above 40°C.

References: HCl(g), H_{2}O(l) at 25°C and 1 atm

Substance \dot{n}_{in} \hat{H}_{in} \dot{n}_{out} \hat{H}_{out}
HCl(g) 5480 mol HCl \hat{H}_{1} (kJ/mol HCl)
H_{2}O (l) 44,400 mol H_{2}O 0
HCl(aq) 5480 mol HCl \hat{H}_{2} (kJ/mol HCl)
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