Extra Credit 9

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Q16.3.4

“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is:

\[\ce{Fe2O3}(s)+\ce{2Al}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)\]

Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.

Solution:

Since the surrounding absorb 851.8kJ/mol means that system release 851.8kJ/mol of heat to surroundings. This means

\(\Delta H_{sys} = -\Delta H_{surr} = - 851.8kJ/mol.\)

\(\Delta S_{universe} \) is always greater than zero since entropy of universe is increasing. Therefore, for a spontaneous process, \(\Delta S_{total}\) is always greater than zero.

Also

\(\Delta S_{surr} = -\dfrac{q_{sys}}{T} = -\dfrac{\Delta H_{sys}}{T} \)

Therefore,

\(\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = \Delta S_{sys} - \dfrac{\Delta H_{sys}}{T} \)

From the above equation and Table T1, we can calculate the \(\Delta S_{sys} = \Delta S°_{rxn}\) at room temperature under standard condition:

\[\ce{Fe2O3}(s)+\ce{2Al}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)\]

\(\Delta S°_{f}\) : 87.4 28.3 50.9 27.3 in J/mol K

Therefore,

\(\Delta S°_{rxn} = \Delta S°_{product} - \Delta S°_{reactant} = [(2*27.3+50.9) - (2*28.3+87.4)] J/mol K = -48.5 J/mol K\)

and

\(\Delta S_{total} = \Delta S°_{rxn} - \dfrac{\Delta H_{sys}}{T} = -48.5 J/mol K -\dfrac{-851.8kJ/mol*\dfrac{1000J}{1kJ}}{273+25K} = 2810 J/mol K > 0\)

Since \(\Delta S_{total}\) is greater than 0, the reaction is spontaneous at room temperature under standard condition.

Q5.4.7

How much heat is produced by combustion of 125 g of methanol under standard state conditions?

Solution:

Step 1: Convert methanol into mole

Since molecular weight of methanol is 32.0419 g/mol from NIST

\(n_{methanol} = \dfrac{m_{methanol}}{MM Methanol} = \dfrac{125 g}{32.0419 g/mol} = 3.90 mol\)

Step 2: Writing the chemical equation for the combustion process of methanol

\[\ce{CH3OH}(l)+\dfrac{3}{2}\ce{O2}(g)⟶\ce{CO2}(g)+2\ce{H2O}(l)\]

Step 3: Calculation \(\Delta H°_{rxn}\) for this reaction

From Table T1 we can see \(\Delta H°_{f}\) for all the reactants and products of the chemical equation written in step 2. Therefore

\(\Delta H°_{rxn} = \Delta H°_{rea} - \Delta H°_{pro} \)

\(= 2*\Delta H°_{f}[\ce{H2O}(l)]+\Delta H°_{f}[\ce{CO2}(g)]-\dfrac{3}{2}*\Delta H°_{f}[\ce{O2}(g)]-\Delta H°_{f}[\ce{CH3OH}(l)]\)

\(= [2*(-285.8)+(-393.5)-\dfrac{3}{2}*0-(-239.2)] kJ/mol = -725.9 kJ/mol\)

Step 4: Calculate heat released in the specific combustion

Since the reaction quotient for methanol in the chemical equation is 1

heat released in this combustion = \(|1*\Delta H°_{rxn}|*n_{methanol} = 725.9 kJ/mol * 3.90 mol = 2832 kJ\)

Q10.3.13

A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?

Q13.2.25

What is the approximate value of the equilibrium constant K_P for the change \(\ce{C2H5OC2H5}(l) \rightleftharpoons \ce{C2H5OC2H5}(g)\) at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)

Solution:

referred to NIST (https://webbook.nist.gov/cgi/cbook.c...60297&Units=SI), for \(\ce{C2H5OC2H5}\):

\(S°_{gas} = 342.2 J/mol-K\), \(S°_{liquid} = 253.5 J/mol-K\)

\(\Delta H°_{f}gas = -244 kJ/mol\), \(\Delta H°_{f}liq = -271.2 kJ/mol\)

we can calculate out \(\Delta G°_{rxn}\) of the reaction \(\ce{C2H5OC2H5}(l) \rightleftharpoons \ce{C2H5OC2H5}(g)\) at 25 °C (298K):

\(\Delta G°_{rxn} = \Delta G°_{gas} - \Delta G°_{liquid} = (\Delta H°_{f}gas - T*\Delta S°_{gas}) - (\Delta H°_{f}liq - T*\Delta S°_{liquid})\)

\(= (-244 kJ/mol - 298 K * 342.2 J/mol-K * \dfrac{1kJ}{1000J}) - (-271.2 kJ/mol - 298 K * 253.5 J/mol-K * \dfrac{1kJ}{1000J}) = 0.7674 kJ/mol = 767.4 J/mol\)

For a phase change reaction, \(\Delta G_{rxn} = 0\). Therefore:

\(\Delta G_{rxn} = \Delta G°_{rxn} + RTlnQ = 0\)

where \(Q = \dfrac{P_{gas}}{P_{liquid}} = P_{gas}\) = # vapor pressure at 298 K for \(C_{2}H_{5}OC_{2}H_{5}\)

Therefore:

\(Q = e^{\dfrac{-\Delta G°_{rxn}}{RT}} = e^{\dfrac{-767.4 J/mol}{8.314 J/mol-K * 298 K}} = 0.734\)

Therefore, vapor pressure equals 0.734 atm.

Q14.2.4

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

0.200 M HCl
0.0143 M NaOH
3.0 M HNO₃
0.0031 M Ca(OH)₂

Solution:

All of them ionized completely

Since at 25 °C

\(K_{w} = [H_{3}O^+]*[OH^-] = 1*10^-14\)

\(pK_{w} = pH + pOH = 14\)

\([H_{3}O^+] = [HCl] = 0.200 M\)

Since \(\dfrac{[H_{3}O^+]_{water}}{[H_{3}O^+]_{HCl}} = \dfrac{1*10^-7}{0.200 M} = 5*10^-7 <<0.05 \)

\(H_{3}O^+\) in water is negligible. Therefore,

\(pH = -log_{10}[H_{3}O^+] = -log_{10}[0.200] = 0.699\)

and \(pOH = pK_{w} - pH = 14 - 0.699 = 13.301\)

\([OH^-] = [NaOH] = 0.0143 M\)

Since \(\dfrac{[OH^-]_{water}}{[OH^-]_{NaOH}} = \dfrac{1*10^-7}{0.0143 M} = 7*10^-6 <<0.05\)

\(OH^-\) in water is negligible. Therefore,

\(pOH = -log_{10}[OH^-] = -log_{10}[0.0143] = 1.84\)

and \(pOH = pK_{w} - pH = 14 - 1.84 = 12.16\)

\([H_{3}O^+] = [HNO_{3}] = 3.0 M\)

Since \(\dfrac{[H_{3}O^+]_{water}}{[H_{3}O^+]_{HCl}} = \dfrac{1*10^-7}{3.0 M} = 3.3*10^-8 <<0.05 \)

\(H_{3}O^+\) in water is negligible. Therefore,

\(pH = -log_{10}[H_{3}O^+] = -log_{10}[3.0] = -0.48\)

and \(pOH = pK_{w} - pH = 14 - (-0.48) = 14.48\)

\([OH^-] = 2*[Ca(OH)_{2}] = 2*0.0031 M = 0.0062 M\)

Since \(\dfrac{[OH^-]_{water}}{[OH^-]_{Ca(OH)_{2}}} = \dfrac{1*10^-7}{0.0143 M} = 7*10^-6 <<0.05\)

\(OH^-\) in water is negligible. Therefore,

\(pOH = -log_{10}[OH^-] = -log_{10}[0.0062] = 2.2\)

and \(pOH = pK_{w} - pH = 14 - 2.2 = 11.8\)

Short Answers

Q5.4.16

90.8 kJ

Q10.3.21

Q13.3.9

a. T increase, K will increase, reaction goes forward; V decrease, K will decrease, reaction goes reverse.

b. T increase, K will decrease, reaction goes reverse; V decrease, K will increase, reaction goes forward.

c. T increase, K will decrease, reaction goes reverse; V decrease, K will increase, reaction goes forward.

d. T increase, K will increase, reaction goes reverse; V decrease, K will decrease, reaction goes forward.

Q14.3.3

CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3

a. Strong B-L bases: \(Ca(OH)_{2}\), \(NaOH\); Strong B-L acids: \(HCl\), \(HNO_{3}\), \(H_{2}SO_{4}\)

b. \(CH_{3}CO_{2}H\), \(CO_{2}\), \(H_{2}CO_{3}\), \(HF\), \(HNO_{2}\), \(H_{3}PO_{4}\)

c. \(CaO\), \(NH_{3}\), \(Na_{2}CO_{3}\)