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Chemistry LibreTexts

10: Thermodynamics

  • Page ID
    285680
  • Energy

    Exercise \(\PageIndex{1.1}\)

    1. Express the quantity of 422 J in calories.

    2. Express the quantity of 3.225 kJ in calories.

    Answer
    1. 101 cal

    \(422 J\left ( \frac{1 calorie}{4.184 } \right )=101 calorie\)

    1. 770.8 cal

    Exercise \(\PageIndex{1.2}\)

    1. Express the quantity 55.69 cal in joules.

    2. Express the quantity 965.3 kcal in joules.

    3. Express the quantity 965.3 Cal in joules.

    Answer
    1. 233.0 joules

    \(55.69 cal\left ( \frac{4.184 J}{1 cal} \right )=233.0 J\)

    1. 4.039 x 106  joules
    2. 4.039 x 106 joules  (Cal is kcal)

    Heat Capacity

    Exercise \(\PageIndex{2.1}\)

    The specific heat capacity of water(liquid) is 4.18 J/g⋅°C. What is the molar specific heat capacity of this substance? The molar mass of water is 18.01 g/mol.

    Answer

    75.3 J/mol * °C

    \(c_{m}=\frac{4.18 J}{\left ( \frac{1}{18.01} mol * °C\right )}=75.2818 J/mol * °C\)

    Exercise \(\PageIndex{2.2}\)

    Exactly 149.6 J will raise the temperature of 10.0 g of a metal from 25.0 °C to 60.0 °C. What is the specific heat capacity of the metal?

    Answer

    0.427 J/(g *°C)

    Exercise \(\PageIndex{2.3}\)

    A 100 g sample of each of the following metals is heated from 35°C to 45°C. Which metal absorbs the lowest amount of heat energy?

    Metal                        Specific Heat Capacity

    Copper                        0.385 J/(g*°C)

    Magnesium                  1.02 J/(g*°C)

    Mercury                       0.138 J/(g*°C)

    Silver                          0.237 J/(g*°C)

    Lead                           0.129 J/(g*°C)

    Answer

    Lead

     

    Exercise \(\PageIndex{2.4}\)

    35.0 kJ raises the temperature of an gold object from 22oC to its melting point of 1064.18oC.  What is the mass of the gold object?

    Answer

    \[q=mc\Delta T \] [\ m=\frac{q}{c\left ( T_f-T_i \right )}\] \[=\frac{35kJ\left ( \frac{1000J}{kJ} \right )}{\frac{0.129J}{g\cdot ^{o} C}\left (1064.18^{o}C-22^{o}C \right )}=260. g \nonumber\]

    Exercise \(\PageIndex{2.5}\)

    What is the specific heat capacity in cal/goC of a substance if it takes 5.09 kJ to raise a 30.1 g mass sample of the material from 25.0 to 400.0oC?

    Answer

    \[q=mc\Delta T\\c=\frac{q}{m\left ( T_f-T_i \right )}=\frac{5.09kJ\left ( \frac{1000J}{kJ} \right )\left ( \frac{cal}{4.184J} \right )}{30.1g\left (400.0^{o}C-25.0^{o}C \right )}=0.108\left ( \frac{cal}{g\cdot^{o}C} \right ) \nonumber\]

     

     

    Exercise \(\PageIndex{2.6}\)

    What is the final temperature if 20.0kJ is added to a 340 g of gold at 25.0oC?

    Answer

    \[q=mc\Delta T \\ T_F = T_i+\frac{q}{mc} \\ = 25.0^o+\frac{20.0kJ}{0.340kg(0.108)( \frac{kcal}{kg \cdot^oC)}}=545^oC\nonumber \]    

     

    Exercise \(\PageIndex{2.7}\)

    35.0 kJ raises the temperature of a 260.0g object from 22oC to its melting point of 1064.18oC.  Could the object be gold?

    Answer

    \[q=mc\Delta T\\c=\frac{q}{m\left ( T_f-T_i \right )}=\frac{35.0kJ\left ( \frac{1000J}{kJ} \right )\left ( \frac{cal}{4.184J} \right )}{260.0g\left (1064.18.0^{o}C-25.0^{o}C \right )}=0.108\left ( \frac{cal}{g\cdot^{o}C} \right ) \\=0.0309 \\ \text{wrong heat capacity, can not be gold}\nonumber\]    

    Exercise \(\PageIndex{2.8}\)

     How much heat is required to be removed from water for it to cool the temperature of 1.00 liter of water (density = 0.9982g/mL) from  25.0oC to 32.0oF?

    Answer

    \[q=mc\Delta T= 1.00L\left ( 0.9982\frac{kg}{L} \right )4.18J\frac{kJ}{kg \cdot^oC}\left ( 0.00^oC-25.0^oC \right )\\q=-104kJ\]

    Exercise \(\PageIndex{2.9}\)

    How much heat is required to be removed from water for it to cool the temperature of 1.00 liter of water (density = 0.9982g/mL) from  its freezing point to -5.00oC?

    Answer

    \[q=mc\Delta T= 1.00L\left ( 0.9982\frac{kg}{L} \right )2.06\frac{kJ}{kg \cdot^oC}\left ( -5^oC-0^oC \right )\\q=-10.3kJ\]

     

     

     

    Energy and Phase Changes

    Exercise \(\PageIndex{1}\)

    How much energy is needed to convert 5.88 g of ice at 0.00 °C to liquid water at 85.0 °C?

    Specific heat capacity (ice) = 2.10 J/g*°C
    Specific heat capacity (liquid water) = 4.18 J/g*°C
    Heat of fusion = 333 J/g

    Answer

    4.05 * 103 J

    \(q_{1} = m\Delta H_{f} = \left ( 5.88 g \right )\left ( 333 J/g \right ) = 1958.04 J\)

    \(q_{2} = mc\Delta T = \left ( 5.88 g \right )\left ( 4.18 J/g*°C \right )\left ( 85 °C - 0 °C \right ) = 2089.164 J\)
    qTotal = q1 + q2 = 1958.04 J + 2089.164 J = 4047.204 J

    Exercise \(\PageIndex{2}\)

    Calculate the energy in the form of heat (in kJ) required to convert 125 grams of liquid water at 20.0 °C to steam at 115 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: liquid water = 4.184 J/g*°C, steam = 1.92 J/g*°C)

    Answer

    3.28 * 102 kJ

    Exercise \(\PageIndex{3}\)

    Calculate the energy in the form of heat (in kJ) required to change 81.5 g of liquid water at 30.6 °C to ice at –13.5 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g*°C, liquid water = 4.184 J/g*°C)

    Answer

    -39.8 kJ

    Exercise \(\PageIndex{4}\)

    What is the minimum mass of ice at 0.0 °C that must be added to 2.50 kg of water to cool the water from 28.0 °C to 9.0 °C? (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/g*°C, liquid water = 4.184 J/g*°C)

    Answer

    m = 536 g

    \(mc\Delta T = mc\Delta T + m\Delta H_{f}\)

    \((2500g)(4.184 J/g*°C)(28 °C - 9 °C) = x(4.184 J/g*°C)(9 °C - 0 °C)+ x(333 J/g)\)

    \(198740 = x370.656\)

    \(x = \frac{198740}{370.656}\)

    x = 536.184 g

    Exercise \(\PageIndex{5}\)

    How much heat is required to be removed from 1.00L water to freeze it at it's freezing point

    Answer

    \[\Delta H=-n\Delta H_f=1.00L\left ( 998.2\frac{g}{L} \right )\frac{mol_{H_2O}}{18.015g}\left ( \frac{6.01kJ}{mol_{H_2O}} \right )= 332kJ\]

     

    First Law of Thermodynamics

    Exercise \(\PageIndex{1}\)

    Calculate ΔU of a gas for a process in which the gas absorbs 60 J of heat and does 15 J of work by expanding.

    Answer

    Δ U = 45 J

    \(q = + 60 J\) (positive, system absorbs heat)

    \(w = - 15 J\) (negative, system does work)

    \(\Delta U = q + w = 60 J - 15 J = 45 J\)

    Exercise \(\PageIndex{2}\)

    What is the change in internal energy of the system (ΔU) if 110 kJ of heat energy is evolved by the system and 62 kJ of work is done on the system for a certain process?

    Answer

    Δ U = -48 kJ

    \(q = - 110 J\) (negative, system releases heat)

    \(w = + 62 J\) (positive, system has work done on it)

    \(\Delta U = q + w = -110 J + 62 J = - 48 J\)

    Exercise \(\PageIndex{3}\)

    A sample of an ideal gas in the cylinder of an engine is compressed from 750.0 mL to 30.5 mL during the compression stroke against a constant pressure of 9.05 x 105 Pa. At the same time, 250 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy Δof the gas in joules?

    Answer

    Δ U = 401 J

    \(\Delta U = \Delta H -P\Delta V\)

    \(\Delta H = -250 J\) 

    \(W = P\Delta V\)

    \(\Delta V = \left ( 0.0305 L - 0.750 L \right ) = -0.7195 L * \left ( \frac{1 m^{3}}{1000 L} \right ) = -7.195*10^{-4} m^{3}\)

    Note: A Pascal is also defined as kg/m*s2. A joule is defined as kg*m2/s2

    \(W = \left ( 9.05 * 10^{5} Kg/m*s^{-2} \right )\left ( -7.195*10^{-4} m^{3} \right ) = -651.1475 J\)

    \(\Delta U = \Delta H + P\Delta V\)
    \[\Delta U = -250 J - (-651 J) = 401 J\]

     

    Enthalpy of Reaction

    Exercise \(\PageIndex{1}\)

    Decide whether each of these reactions is exothermic or endothermic:

    1. When two chemicals mix their temperature rises
    2. A solid burns brightly and releases heat, light and sound
    3. When two chemicals are mixed their temperatures drops
    4. Two chemicals will only react if you heat them continually
    5. Plants take in light energy for photosynthesis
    Answer
    1. exothermic
    2. exothermic
    3. endothermic
    4. endothermic
    5. endothermic

    Exercise \(\PageIndex{2}\)

    What quantity, in moles, of hydrogen is consumed when 185.6 kJ of energy is evolved from the combustion of a mixture of H2(g) and O2(g)?

    H2(g) + O2(g)  →  H2O(l); ΔrH° = –285.8 kJ/mol-rxn

    Answer

    0.6494 moles

    \(\frac{185.6 kJ}{285.8 kJ/mol} = 0.649405 moles\)

    Exercise \(\PageIndex{3}\)

    The following reaction of iron oxide with aluminum is an exothermic reaction.

    Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s)

    The reaction of 7.50 g of Fe2O3 with excess Al(s) evolves 31.5 kJ of energy in the form of heat. Calculate the enthalpy change per mole of Fe2O3.

    Answer

    -6.71*102 kJ*mol-1 (heat evolved, negative sign)

    Exercise \(\PageIndex{4}\)

    How much heat is liberated at constant pressure if 0.675 g of calcium carbonate reacts with 45.1 mL of 0.498 M hydrochloric acid?

    CaCO3(s) + 2HCl(aq)    CaCl2(aq) + H2O(l) + CO2(g); ΔrH° = –15.2 kJ/mol-rxn

    Answer

    ΔH = -0.103 kJ

    Enthalpy Calculations

    Hess's Law And Enthalpy of Reaction

    Exercise \(\PageIndex{1}\)

    What is the overall chemical equation that results from the sum of the given steps?

    2 C(s) + 2 H2O(g) → 2 CO(g) + 2 H2(g)

    CO(g) + H2O(g) → CO2(g) + H2(g)

    CO(g) + 3 H2(g) → CH4(g) + H2O(g)

    Answer

    2 C(s) + 2 H2O(g) → CO2(g) + CH4(g)

    Exercise \(\PageIndex{2}\)

    Determine the heat of evaporation of carbon disulfide,
    CS2(l) → CS2(g)
    given the enthalpies of reaction below.

    C(s) + 2 S(s) → CS2(l)                         ΔrH° = +89.4 kJ/mol-rxn

    C(s) + 2 S(s) → CS2(g)                        ΔrH° = +116.7 kJ/mol-rxn

    Answer

    +27.3 kJ

    Reverse (1) and add to (2)                    ΔHrxn = -ΔH1 + ΔH2

    (1) CS2(l) → C(s) + 2 S(s)                               = -89.4 kJ/mol + 116.7 kJ/mol = 27.3 kJ/mol

    (2) C(s) + 2 S(s) → CS2(g) 


    CS2 (l) → CS2 (g)

     

    Enthalpy of Formation

    Exercise \(\PageIndex{3}\)

    Determine the standard enthalpy of formation of Fe2O3(s) given the thermochemical equations below.

    Fe(s) + 3 H2O(l) → Fe(OH)3(s) + 3/2 H2(g)                    ΔrH° = +160.9 kJ/mol-rxn

    H2(g) + 1/2 O2(g) → H2O(l)                                          ΔrH° = –285.8 kJ/mol-rxn

    Fe2O3(s) + 3 H2O(l) → 2 Fe(OH)3(s)                              ΔrH° = +288.6 kJ/mol-rxn

    Answer

    –824.2 kJ/mol-rxn

    1. (Fe(s) + 3 H2O(l) → Fe(OH)3(s) + 3/2 H2(g)                    ΔrH° = +160.9 kJ/mol-rxn) * 2

        2 Fe(s) + 6 H2O(l) → 2 Fe(OH)3(s) + 3 H2(g)                    ΔrH° = +321.8 kJ/mol-rxn

    1. (H2(g) + 1/2 O2(g) → H2O(l)                                          ΔrH° = –285.8 kJ/mol-rxn) * 3

         3 H2(g) + 3/2 O2(g) → 3 H2O(l)                                     ΔrH° = –857.4kJ/mol-rxn

    1. (Fe2O3(s) + 3 H2O(l) → 2 Fe(OH)3(s)                              ΔrH° = +288.6 kJ/mol-rxn) Reverse Equation or ( * -1)

        2 Fe(OH)3(s) → Fe2O3(s) + 3 H2O(l)                               ΔrH° = -288.6 kJ/mol-rxn

    1. Add the equations and ΔrH°.

    2 Fe(s) + 6 H2O(l)2 Fe(OH)3(s) + 3 H2(g)                    ΔrH° = +321.8 kJ/mol-rxn

    3 H2(g) + 3/2 O2(g) → 3 H2O(l)                                       ΔrH° = –857.4kJ/mol-rxn

    2 Fe(OH)3(s) → Fe2O3(s) + 3 H2O(l)                                 ΔrH° = -288.6 kJ/mol-rxn


    2 Fe(s) + 3/2 O2(g) → Fe2O3(s)                                       ΔHf = -824.2 kJ/mol

    Exercise \(\PageIndex{4}\)

    Determine the standard enthalpy of formation of calcium carbonate from the thermochemical equations given below.

    Ca(OH)2(s) → CaO(s) + H2O(l)                                ΔrH° = 65.2 kJ/mol-rxn

    Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l)              ΔrH° = −113.8 kJ/mol-rxn

    C(s) + O2(g) → CO2(g)                                           ΔrH° = −393.5 kJ/mol-rxn

    2 Ca(s) + O2(g) → 2 CaO(s)                                    ΔrH° = −1270.2 kJ/mol-rxn

    Answer

    −1207.6 kJ/mol-rxn

     

    Enthalpy of Reaction from Standard Enthalpies of Formation

    Exercise \(\PageIndex{5}\)

    Calculate ΔrH° for the combustion of ammonia, using standard molar enthalpies of formation:

    4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(l)

    Molecule             ΔfH° (kJ/mol-rxn)

    NH3(g)                      –45.9

    NO2(g)                      +33.1

    H2O(l)                       –285.8

    Answer

    –1398.8 kJ/mol-rxn

    ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants)

               = [(4 * ΔH°f(NO2)) + (6 * ΔH°f(H2O))] - [(4 * ΔH°f(NH3)) + (7 * ΔH°f(O2))]

               = [(4 * 33.1) + (6 * -285.8)] - [(4 * -45.9) + (7 * 0)]

               = -1398.8 kJ/mol-rxn

    Exercise \(\PageIndex{6}\)

    What is ΔrH° for the following phase change?

    LiF(s) → LiF(l)

    Substance           ΔH°f (kJ/mol-rxn)

    LiF(s)                     –616.93

    LiF(l)                      –598.65

    Answer

    18.28 kJ/mol-rxn

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