Question 9.5.6: Energy Balance on a Neutralization Process A 10.0 wt% aqueou......

Basis: 1 kg H_{2}SO_{4} Solution

H_{2}SO_{4}(aq, 10\%) + 2NaOH(aq, 20\%) \rightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)

This is a straightforward problem, but the number of intermediate calculations required to solve it might make it appear more difficult than it is. Let us summarize what must be done:
1. Solve for m_{1}, m_{2}, and m_{3} by material balances.
2. Calculate the solvent-to-solute mole ratios of all solutions. (The quantities are needed to determine the solution enthalpies from tabulated heats of solution.)
3. Calculate the enthalpies of the solutions. (This will require additional composition calculations to allow the use of tabulated solution heat capacities.)
4. Write the energy balance equation and solve it for the heat removal rate.
Observe that nothing here is really new, and, as we work our way through to the final result, recognize that most of the calculations are simply conversions of solution compositions from mass fractions to mole ratios back to mass ratios—conversions required by the nature of the available data for properties of solutions.
1. Solve for m_{1}, m_{2}, and m_{3} by material balances, and calculate the amount of water formed.

S  balance: \begin{array}{c|c}100  g  H_{2}SO_{4} &32.0  g  S \\ \hline & 98.1  g  H_{2}SO_{4}\end{array} = \begin{array}{c|c} m_{2}(g  Na_{2}SO_{4}) & 32.0  g  S\\ \hline & 142  g  Na_{2}SO_{4}\end{array} \\ \left. \Large{\Downarrow} \right.\\ m_{2}= 145  g  Na_{2}SO_{4} \\ Na  balance: \begin{array}{c|c}0.200  m_{1}(g  NaOH)&23.0  g  Na \\ \hline & 40.0  g  NaOH\end{array}=\begin{array}{c|c}145  g  Na_{2}SO_{4} &46.0  g  Na\\ \hline& 142  g  Na_{2}SO_{4}\end{array} \\ \left. \Large{\Downarrow} \right.\\ m_{1}= 408  NaOH  (aq)

Total mass balance:  1000  g + 408  g = 145  g+m_{3} \Longrightarrow m_{3}=  1263  g  H_{2}O(l)

Mass of product solution:   m = m_{2}+ m_{3}= 1408  g

Water formed by reaction

= 2.04  mol  H_{2}O

2. Calculate solvent/solute mole ratios (needed to determine heats of solution).

H_{2}SO_{4}(aq):  (900  g  H_{2}O)/(18.0  g/mol)= 50.0  mol  H_{2}O \\ (100  g  H_{2}SO_{4})/(98.1  g/mol = 1.02  mol  H_{2}SO_{4} \\ \left. \Large{\Downarrow} \right.\\ r= 50.0  mol  H_{2}O/1.02  mol  H_{2}SO_{4} = 49.0  mol  H_{2}O/mol  H_{2}SO_{4} \\ NaOH(aq):   [(0.800 \times 408)g  H_{2}O]/ (18.0  g/mol) = 18.1  mol  H_{2}O \\ [(0.200 \times 408) g  NaOH]/(40.0  g/mol)= 2.04  mol  NaOH \\ \left. \Large{\Downarrow} \right.\\ r= 18.1  mol  H_{2}O/2.04  mol  NaOH = 8.90  mol  H_{2}O/mol  NaOH \\ Na_{2}SO_{4}(aq):  (1263  g  H_{2}O)/(18.0  g/mol) = 70.2  mol  H_{2}O \\ (145  g  Na_{2}SO_{4})/(142  g/mol) = 1.02  mol  Na_{2}SO_{4} \\ \left. \Large{\Downarrow} \right.\\ r=70.2  mol  H_{2}O/1.02  mol  Na_{2}SO_{4} = 68.8  mol  H_{2}O/mol  Na_{2}SO_{4}

3. Calculate extent of reaction. To calculate ξ, we note that 1.02 mol H_{2}SO_{4} reacted. From Equation 9.1-3,

\xi = \left. \Large{\{} \right. \begin{matrix} Batch: \frac{(n_{i,out} – n_{i,in})(mol  i)}{\nu _{i}(mol  i)} = \frac{n_{ir}}{\left|\nu _{i}\right|} \\Continuous: \frac{(\dot{n}_{i,out} – \dot{n}_{i,in})(mol  i/ time)}{\dot{\nu} _{i}(mol   i/ time)} = \frac{\dot{n}_{ir}}{\left|\dot{\nu} _{i}\right|} \end{matrix}                 (9.1-3)

\xi =\frac{(n_{H_{2}SO_{4}})_{reacted}}{\left|\nu _{H_{2}SO_{4}}\right| } = \frac{1.02  mol}{1  mol}=1.02

4. Calculate ΔH. This problem is made tricky by the fact that water is not just the solvent in the solutions involved but is also formed as a reaction product. We will take as references the reactant and product solutions at 25°C and evaluate ΔH using Equation 9.5-1a:

It is convenient in solution chemistry calculations to tabulate the products n\hat{H} rather than n and \hat{H} separately. The completed enthalpy table is shown below, followed by the calculations that led to the entries.

H_{2}SO_{4}(aq, r = 49, 40°C): From Table 2.174, p. 2-184 of Perry’s Chemical Engineers’ Handbook (see Footnote 2), the heat capacity of a sulfuric acid solution with the given composition is by interpolation, approximately 0.916 cal/(g·°C), or 3.83 J/(g·°C).

NaOH(aq, r = 8.9, 25°C):      n\hat{H}=0

Na_{2}SO_{4}(aq, r = 69, 35°C): In the absence of better information, we will assume that the heat capacity of the solution is that of pure water, 4.184 J/(g·°C).

The heats of formation of H_{2}SO_{4}(l) and NaOH(c) are given in Table B.1, and the heats of solution of these species are given in Table B.11. Perry’s Chemical Engineers’ Handbook (see Footnote 2) on p. 2-192 gives the standard heat of formation of Na_{2}SO_{4}(aq, r = 1100) as – 330.82 kcal/mol Na_{2}SO_{4} =- 1384 kJ/mol Na_{2}SO_{4}. In the absence of heat of solution data, we will assume that this value also applies to the solution for which r = 69 moles of water per mole of solute. The standard heats of formation of the species involved in the reaction

H_{2}SO_{4}(aq, r= 49) + 2NaOH(aq, r=8.9) \rightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)

are obtained from Equation 9.5-3 (heat of formation of the solution equals heat of formation of the solute plus heat of solution) as

\boxed{(\Delta \hat{H}^{°}_{f})_{solution}= (\Delta \hat{H}^{°}_{f})_{solute} +\Delta \hat{H}^{°}_{s}(n) }              (9.5-3)

and the standard heat of reaction is therefore

5. Energy balance.

Q= \Delta H= \xi \Delta H^{°}_{r} + \sum{n_{out}\hat{H}_{out}} – \sum {n_{in}\hat{H}_{in}}\\ = (1.02)(- 134.9  kJ) + (58.9  –  57.5)  kJ = \boxed{- 136  kJ}