THERMOCHEMISTRY ASSIGNMENT

Confidence Building Questions

  1. A person weighing 65 kg climbed 11 m up a rope; how much work (in kJ) was involved? (g = 9.8 ms-2)

    Hint: 7.0 kJ

    Skill:
    Calculate mechanical work or energy involved.
    In reality, more energy is required due to low efficiency of energy utilization.
  2. Assume the ideal gas law. What is the volume occupied by 1.00 mole (18.0 g) of water vapor at its boiling point, 373 K at a pressure of 1 atm? (R = 0.08205 L atm/(mol K))Hint: 30.6 L
    Discussion
    Molar volume is 30.6 L at 373 K, but 22.4 L at STP. Water is a liquid at STP, and its vapor pressure is very low.
  3. When water boils, the vapor expands against a constant atmospheric pressure of 1 atm. Calculate the work (J) done by one mole of water vapor if its volume is 30.6 L.
    (1 atm = 101.3 kPa, R = 8.314 J/(mol K), 1 L = 0.001 m3)
    Hint: -3100 J

    Skill:
    Calculate pressure-volume work.
  4. The heat of vaporization for one mole of water is 40.7 kJ (i.e., 40.7 kJ absorbed by 18 g of H2O

    H2O). If 3.1 kJ of work is done by the system, what is the increase in internal energy dE?

    Hint: 37.6 kJ

    Skill:
    Evaluate internal energy.
    When water evaporates, its vapor occupies 30.6 L at 373K, and work is done by pushing its way in the atmosphere of 1 atm. 0.0306m3×101.3kNm2=3.1kNm=3.1kJ 0.0306m3×101.3kNm−2=3.1kNm=3.1kJ 

  5. Let us consider the condensation of water vapor, or gas H2O

    H2O. A H2O

    H2O gas loses 40.7 kJ of heat when it condenses, and the environment performs 3.1 kJ of work when the volume contracts. Calculate the gain (or loss) (negative change) in internal energy (kJ).

    Hint: -37.6 kJ

    DIscussion
    This problem states the reversed process of the previous problem.
  6. A battery does 63 J of electrical work and loses 3.0 J of heat. Which of the following (all in unit J) ARE correct:

    1. dE = 66
    2. dE = -60
    3. dE = 63
    4. dE = -66
    5. = 63
    6. = -63
    7. = 60
    8. = -66
    9. = 3
    10. = -3
    11. = 60
    12. = -60
    Hint: d f & j are all true.

    Skill:
    Various skills are involved.

  7. It is known that 1.0 cal is equivalent to 4.184 J, and 40.7 kJ is required to evaporate 1.0 mol of H2O

    H2O (molar mass, 18.0). Calculate the heat (in cal) required to evaporate 1.0 g H2O

    H2O.

    Hint: 540 cal

    A reminder of unit conversion:

    40700Jmol×1cal4.184J×1mol18g=540cal

    40700Jmol×1cal4.184J×1mol18g=540cal 

  8. What is the sign of the enthalpy change for the following reaction?

    2CO+O22CO2

    2CO+O2→2CO2 

    Positive or negative?

    Hint: negative

    Skill:
    Explain exothermic and endothermic reactions in terms of enthalpy of reaction.
  9. Given the following standard enthalpies (kJ) of formation at 25 °C,

    NH3

    NH3, -46.1;   CO2

    CO2, -393.5;   C2H4

    C2H4, 52.3;   NO2

    NO2, 33.2

    which molecule will require the most energy to form?

    Hint: C2H4

    C2H4 

    Skill:
    Explain the meaning of enthalpy.
  10. Given the follwoing enthalpies (kJ) of formation,

    H2O

    H2O, -285.8;   H2S

    H2S, -20.6;   SO2

    SO2, -296.8;   SO3

    SO3, 395.7;   CO2

    CO2, -393.5; and  CO

    CO, -110.5

    calculate the enthalpy change for the reaction:

    2H2S+SO23S+2H2OdH=?

    2H2S+SO2→3S+2H2OdH=? 

    All reactants and products are in their standard states.

    Hint: -233.6 kJ

    Skill:
    Evaluate the standard enthalpy of reaction from standard enthalpies of formation of reactants and products.Further hint
    Writing all the equations out presents the evaluation in a logical way.

    2H2+O2H2O2H2S2H2+2SSO2S+O2dH=571.6kJ(2times)dH=41.2dH=296.8(11)(12)(13)

    (11)2H2+O→2H2OdH=−571.6kJ(2times)(12)2H2S→2H2+2SdH=41.2(13)SO2→S+O2dH=296.8 

  11. Given the following enthalpies (kJ) of formation,

    H2O

    H2O, -285.8;   H2S

    H2S, -20.6;   SO2

    SO2, -296.8;   SO3

    SO3, 395.7;   CO2

    CO2, -393.5; and   CO

    CO, -110.5

    calculate the enthalpy change for the reaction:

    2CO+O22CO2

    2CO+O2→2CO2 

    All reactants and products are in their standard states.

    Hint: -566 kJ

    Skill:
    Evaluate the standard enthalpy of reaction from standard enthalpies of formation of reactants and products.
  12. The hydration of lime, CaO

    CaO (molar mass 56), gives off 65.2 kJ per mole:

    CaO+H2OCa(OH)2,dH=65.2kJ

    CaO+H2O→Ca(OH)2,dH=−65.2kJ 

    How much heat (in kcal) is given off when 5.0 kg of CaO

    CaO is hydrated? Conversion factor: 1 kcal = 4.184 kJ.

    Hint: -1391 kcal

    Further interest:
    Ca(OH)2 Ca(OH)2 

    is called slaked lime. It is used in the construction industry.

  13. The enthalpy of formation of CO2

    CO2 from graphite is -393.5 kJ. The enthalpy of formation of diamond is +1.9 kJ. Calculate the heat of combustion of diamond to form CO2

    CO2.

    Hint: -395.4 kJ

    Skill:
    Know that dHof for graphite is zero (the standard state of carbon). Calculate enthalpy of combustion. Diamond is unstable with respect to graphite.

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