38 Practice Exercises

Solution

b) 9.5 kJ
Performing an energy balance on the system, we know that $\Delta H = Q + W$. Since this is an endothermic reaction and work is done on the system, both the enthalpy of the reaction and the work are positive:
$$\begin{align*} Q&= \Delta H -W \\ &= 15.8 kJ – 6.3 kJ\\ &=9.5kJ \end{align*}$$

Solution

a) 22.3 MW

Recall that $\Delta \dot{H} = \dot{n} \int_{T_{1}}^{T_{2}} CpdT$. We can expand this expression in order to include phase changes:

$\Delta \dot{H} = \dot{n} (\int_{T_{1}}^{T_{boiling}} Cp(liquid)dT + \Delta H_{vap} +\int_{T_{boiling}}^{T_{2}} Cp(vapour)dT )$

Plugging in our values, we get:

$\Delta \dot{H} = (500 mol/s) (\int_{25^{\circ}C}^{65^{\circ}C} 80 J/molK dT + 40,000 J/mol +\int_{65^{\circ}C}^{100^{\circ}C} 40 J/molK dT)$

$\Delta \dot{H} = (500 mol/s) (80 J/molK (65-25)K + 40,000 J/mol +40 J/molK (100-65)K )$

$\Delta \dot{H} = (500 mol/s)(44,600 J/mol)$

$\Delta \dot{H} = 22,300,000 J/s = 22,300,000 W = 22.3 MW$

Solution

c) Low-pressure steam

A minimum of $10^{\circ}C$ of difference in temperature is required for efficient heat exchange.

The cooling and refrigerated water are not hot enough to heat the toluene. Both the low and medium pressure steam are hot enough to heat the toluene to $60^{\circ}C$, but low-pressure steam is the cheaper option.

Solution

Using Hess’s Law we can multiply reaction 1 by -3/2, reaction 2 by 1/2, and reaction 3 by -1 to obtain our desired reaction:

$$\begin{align*} 3NO_{2}(g) → 3NO(g) + \frac{3}{2}O_{2}(g) \;\;\;\;\;&-3/2\Delta H_{1}\\ N_{2}(g) + \frac{5}{2}O_{2}(g) + H_{2}O(l) → 2HNO_{3}(aq)\;\;\;\;\;&1/2\Delta H_{2}\\ 2NO(g) → N_{2}(g) + O_{2}(g)\;\;\;\;\;&-\Delta H_{3} \end{align*}$$

Therefore, our change in enthalpy can be represented by:

$\Delta H = -3/2*\Delta H_{1} + 1/2 \Delta H_{2} - \Delta H_{3}$

Solution

Recall that

$\Delta H^{\circ}_{r} = \Sigma_{i}\nu_{i}\Delta\hat{H}^{\circ}_{f,i} = \Sigma_{products}|\nu_{i}|\Delta\hat{H}^{\circ}_{f,i}-\Sigma_{reactants}|\nu_{i}|\Delta\hat{H}^{\circ}_{f,i}$

Using the data provided and the reaction’s stoichiometry:

$\Delta H^{\circ}_{r} = 4 mol(-393.5 kJ/mol) + 2 mol(-241.8 kJ/mol) - 2 mol (226.7 kJ/mol) - 5 mol (0 kJ/mol)$

$\Delta H^{\circ}_{r} = -2511 kJ$

Solution

The following steps can be taken to calculate the overall change in enthalpy of going from liquid methanol at 1 atm and $25^{\circ}C$ to gas methanol at 1 atm and $100^{\circ}C$.

1. Enthalpy change from $25^{\circ}C$ to $65^{\circ}C$ using the liquid state heat capacity of methanol
2. Heat of vapourization of methanol at a constant temperature of $65^{\circ}C$
3. Enthalpy change from $65^{\circ}C$ to $100^{\circ}C$ using the gas state heat capacity of methanol

The enthalpy change of the process will be the sum of steps 1-3.

Solution

Because there are no other forms of energy transfer other than the heat transfer between the streams, the energy balance simplifies to:

$\Delta\dot{H}_{water}+\Delta\dot{H}_{acetic \;acid}=0$

The enthalpy change for steam  $\Delta\dot{H}$ in the energy balance can be expanded to:

$\Delta\dot{H}_{water}=\dot{H}_f-\dot{H}_i$

$$\begin{align*} \Delta\dot{H}_{benzene}&=\int_{T_i}^{T_f}C_pdT\\ &=\int_{T_i}^{T_f}A+BT+CT^2+DT^3 dT\\ &=AT+\frac{1}{2}BT^2+\frac{1}{3}CT^3+\frac{1}{4}DT^4\;\Bigg|_{T_i}^{T_f}\\ &= A(T_f-T_i)+\frac{1}{2}B(T_f-T_i)^2+\frac{1}{3}C(T_f-T_i)^3+\frac{1}{4}D(T_f-T_i)^4 \end{align*}$$

Step 1: Input given information that will be used in the calculation. This includes the given data to calculate enthalpy change, the molar flow rate of both substances, molecular weights, and the initial temperature of the benzene stream.

Step 2: Calculate the molar enthalpy change. For water, this is calculated by subtracting the molar enthalpy values calculated by subtracting the initial enthalpy from the final. We also convert the units by multiplying the water’s mass enthalpies by the molecular weights at the final and initial temperatures to keep it consistant with the benzene stream.
For benzene, we need to provide a guess of the final temperature to substitute into the equation. For example, we can guess 400 K.

Step 3: Calculate the total enthalpy change for both substances by multiplying the molar flow rate by the molar enthalpy change.

Step 4: Use a cell to represent the sum of enthalpy change. If we have guessed the correct final temperature, the value of this cell should be 0. We will use Solver to let excel perform the guess-and-check process for us.

Step 5: Select “Solver” from the “Data” menu in the header, and indicate that we want to solve for the final temperature of benzene that makes our $\text{sum of}\; \Delta H=0$:

• Set objective to be the cell containing $\text{sum of}\; \Delta H$
• To: value of 0
• By changing variable cells containing the final temperature of benzene
• Click “Solve”

Step 6: Click “OK” if Solver finds a solution. If the solver cannot find a solution, it is maybe because our initial guess is too far away from the actual final temperature, if so, we can change the temperature and try to solve it again.

Step 7: After the window closes, the value of the final temperature of benzene and all cells affected by that temperature are changed to the corresponding values calculated with the actual final temperature.

Therefore, the final temperature of the benzene stream is $342.1K$, which is equal to $69.1°C$. This temperature is lower than the boiling temperature of benzene at 1 atm, so the stream will not start to vapourize.

Solution

Step 1: Determine the combination of reactions that will give us the desired combustion reaction.

We know that we need C₂H₄ in the reactants, so we multiply reaction 1 by -1. We notice that by adding the other 2 reactions to reaction 1 (multiplied by -1), we get our desired reaction:

Step 2: Calculate the heat of combustion from the reaction enthalpies.

$\Delta H^{\circ} =-1*\Delta H^{\circ}_{r1} + \Delta H^{\circ}_{r2} + \Delta H^{\circ}_{r3}$

$\Delta H^{\circ} = -1* (-174.19 kJ) + (-2511 kJ) + 1013.7 kJ$

$\Delta H^{\circ} = -1323.11 kJ$

Solution

Step 1: Calculate the enthalpy of reaction for each of the reactions using the enthalpy of formation of the reactants and products.

$\Delta H^{\circ}_{r} = \Sigma_{i}\nu_{i}\Delta\hat{H}^{\circ}_{f,i} = \Sigma_{products}|\nu_{i}|\Delta\hat{H}^{\circ}_{f,i}-\Sigma_{reactants}|\nu_{i}|\Delta\hat{H}^{\circ}_{f,i}$

$\Delta H^{\circ}_{r} = 2 mol (-393.5 \frac{kJ}{mol}) + 2 mol (-241.8 \frac{kJ}{mol}) - 1 mol (52.51 \frac{kJ}{mol} - 3 mol (0 \frac{kJ}{mol}) = -1323.11 kJ$

Step 2: Compare the calculated enthalpy of combustion using heats of formation to the enthalpy of combusion using Hess’s Law from the previous exercise.

They are both equal to -1323.11 kJ!