Enthalpy Worksheet - Answer Key

 

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  1. What is Enthalpy?  What is its symbol?

    A thermodynamic potential consisting of the internal energy of the system plus its pressure and volume.

    Enthalpy is represented by, H. 

    2.  

     

  2. What is the equation for ΔH?

    The overall equation is

    ΔH = ΔE + Δ(PV)

    If the reaction is under constant volume circumstances then the equation is rewritten

    ΔH = ΔE + VΔP                   

    If the reaction is under constant pressure circumstances then the equation is rewritten
    ΔH = ΔE + PΔV

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  3. Under constant pressure circumstances what does ΔH equal?

    ΔH = qp. 

    4.  

     

  4. Is enthalpy a state function?

    ΔH is a state function. 

    5.  

     

  5. Can enthalpy tell us whether a reaction is exothermic or endothermic?  If so, how?

    Yes – if the reaction is run under constant pressure.
    If ΔH is negative the reaction is exothermic.
    If ΔH is positive the reaction is endothermic.

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  6. For the following, predict the relationship between ΔE and ΔH for the following reactions at constant pressure:

      7.  

    • 2HF(g)→ H2 (g) + F2(g)

      ∆H = ∆E.

      •  

    • N2(g) +3H2 (g)→ 2NH3 (g)

      ∆H < ∆E.

      •  

    • 4NH3 (g) + 5O2 (g)→ 4NO(g) + 6H2O (g)

      ∆H > ∆E.

     

     

  7. Hess’ Law capitalizes on what property of enthalpy?

    Hess’ Law capitalizes on the state function status of enthalpy.  You can use the enthalpy values of known reactions to find the ∆H of a new reaction.

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  8. How do the following changes to a reaction affect ΔH values?

      9.  

    • Reverse of reaction → Reverses sign of ∆H

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    • Multiplies reaction → Multiply ∆H by the same constant

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    • Add up reactions → add up the ∆H values for all reactions

     

     

  9. Given

    C6H4(OH)2 (aq)    → C6H4O2 (aq) + H2 (g)         ΔH =177.4 kJ

     

    H2 (g) + O2 (g)     → H2O2 (aq)                             ΔH = -191.2 kJ

     

    H2 (g) + ½O2 (g)    → H2O (g)                                      ΔH = -241.8 kJ

     

    H2O (g)    → H2O (l)                                           ΔH = -43.8 kJ

    Determine ΔH for:

    C6H4(OH)2 (aq) + H2O2 (aq)    → C6H4O2(aq)+ 2H2O (l)

    ΔH = –202.6 kJ

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  10. This symbol  ΔH°f represents standard enthalpy of formation.

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  11. What is heat of formation?

    This is the enthalpy that accompanies the formation of a compound from its elements in their standard states (how they are naturally found).

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  12. What are the requirements to write out the heat of formation reaction equation for a molecule?

      13.  

    • Only one mole of the compound is formed.

      •  

    • The reactants must be pure elements in their standard state. 

     

     

  13. Which of the following represent a heat of formation reaction equation?
    If it is not correct, why?

      14.  

    • 6C(s) + 6H2O(l)→ C6H12O6 (s)

      No.  H2O is a compound, not an element.  Only elements are allowed on the reactant side.

      •  

    • Na(s) + Cl(g)→ NaCl(s)

      No.  Cl(g) is not the standard state for chlorine, Cl2 (g) is.

      •  

    • 2H2 (g) + O2(g)→ 2H2O(l)

      No.  Only one mole of H2O(l) is formed in a standard enthalpy of formation – not two.

      •  

    • 1/2 N2 (g) +  3/2 H2 (g)→ NH3 (g)

      Yes.

     

     

  14. An element in its standard state has a ΔH°f equal to zero.  Why?

    Because it does not take any energy to get them in their standard state.  It’s simply the way they come.  Like O2 (g) or Hg (l).

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  15. The equation to derive the ΔH°rxn from ΔH°f is:

    ∆Horxn = ∑∆Hof (product) -  ∑∆Hof (reactant)

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  16. Calculate ΔH° rxn for the following:

      17.  

    • 4 Na(s)   +   O2 (g)   →   2 Na2O(s)

       

      ΔHfo = –416 kJ

      ΔHo = –832 kJ

      •  

    • 2 Na(s)  +   2 H2O(l)     →  2NaOH(aq)   +   H2(g)

       

                         ΔHfo = –286 kJ         ΔHfo = –470 kJ           

      ΔHo = –368 kJ

      •  

       

    • 2 Na(s)   +     CO2 (g)        →   Na2O(s)            +      CO(g)

       

      ΔHfo = –393.5 kJ      ΔHfo = –416 kJ       ΔHfo = –110.5 kJ

      ΔHo = –133 kJ

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