.
.
Monday February 26, 2024   Day 34
Constant Volume Calorimetry: The "Bomb" Calorimeter

Textbook Readings:

6.5: Constant Volume Calorimetry:
        Measuring ΔU for Chemical Reactions
Course Lecture

6.10 pdf  Video     Bomb Calorimetry
Calorimetry (Constant V vs. Constant P)

 		Bomb Calorimeter Problems
Objectives

1.    Describe the similarities and differences between constant pressure (coffee cup) and constant
       volume (bomb) calorimetry

2.    Given bomb calorimetry experimental temperatures, masses and calorimeter constant, 
       calculate the  ΔHrxn



Homework Problems

34.1   One of the main components of gasoline is octane, C8H18.  The balanced combustion reaction
       for octane is as follows...
 C8H18(l)     +      25/2  O2(g)    →     8  CO2(g)     +     9 H2O(g)

      If a 1.00 gram sample of octane is burned in a bomb calorimeter that contains 1,200. grams of water,
      the temperature rises from 25.00 oC to 33.20 oC.  

      a. Calculate the heat gained by the water.

      b. Calculate the heat gained by the calorimeter given that Ccal = 837 J/oC

      c.  Determine the total heat gained by the calorimeter and water.

      d.  Determine the total heat produced released by the reaction.
 
      e. Determine  ΔHcombustion for octane in units of   kJ/gram     AND   kJ/mol
          

34.2   1.00 g Sucrose (C12H22O11) is burned in a bomb calorimeter that contains
          1,500. grams of water.  The temperature changes from 25.00 oC to 27.45 oC

         Calculate the heat of combustion (per mole AND per gram) for sucrose
         given that Ccal = 920. J/oC

34.3  What is so special about a bomb calorimeter?
       
Click and drag the region below for correct answers

34.1    a. +41.17056 kJ      b. +6.8634 kJ    c. + 48.0339 kJ   
           d.  - 48.0339 kJ       e.  ΔHcombustion =  -48.0 kJ/g      f. ΔHcombustion  =  -5.49 x 103  kJ/mol

34.2    a.  ΔHcombustion  = - 17.63 kJ/g    and ΔHcombustion  =  - 6.0347 x 103 kJ/mol

34.3   Because the combustion reaction takes place in a rigid "bomb" enclosure there is no
          change in volume.  This means that no work is done by the reaction (ΔV = 0) and that
          the heat produced is all the energy there was available.

Notice that exothermic reactions, produce heat and therefore have negative ΔHcombustion values.
.
.
.
.
Tuesday February 27, 2024   Day 35
Hess's Law

Textbook Readings

4.13 Standard Heat of Formation

4.14 Calculating Heat of Reaction from Heat
         of formation

6.8: Relationships Involving Enthalpy of Reactions
Course Lectures

6.9  pdf  Video    Hess's Law:
                             Standard State Enthalpies

6.8  pdf  Video    Hess's Law

Objectives 		Hess's Law and Heats of Formation

Homework Problems

35.1. Consider the three equations below.  
               ...Multiply the first by 1/6 and  reverse it. 
                  ...Divide equations 2 and 3 by two.  
                      ...Add the reactions together

               a.   What is the overall reaction
               b.  What is ΔH  for the overall reaction?

                        Eq. 1       3A + 6B → 3D               ΔH = -403 kJ/mol
                        Eq. 2         E + 2F → A                  ΔH = -105.2 kJ/mol
                        Eq. 3         C → E + 3D                 ΔH = +64.8 kJ/mol

35.2. Calculate ΔH for the reaction: C2H4 (g) + H2 (g) → C2H6 (g),  from the following data.

        a.  C2H4 (g)   + 3 O2 (g)        → 2 CO2 (g) + 2 H2O (l)                ΔH = -1411. kJ
        b.   C2H6 (g) + 3½ O2 (g)      → 2 CO2 (g) + 3 H2O (l)                ΔH = -1560. kJ
        c.    H2 (g)        + ½ O2 (g)      →                      H2O (l)                ΔH = -285.8 kJ

35.3.  Calculate ΔH for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g), given:  

        a.    N2 (g)     + 3 H2 (g)            → 2 NH3 (g)                                 ΔH = -91.8 kJ
        b.    C (s)       + 2 H2 (g)           →   CH4 (g)                                  ΔH = -74.9 kJ
        c.     H2 (g) + 2 C (s) + N2 (g)    → 2 HCN (g)                                ΔH = +270.3 kJ

35.4. Calculate ΔH for the reaction 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s), given: 

        a. 2 Al (s)     +     6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g)                   ΔH = -1049. kJ
        b. HCl (g)                              → HCl (aq)                                       ΔH = -74.8 kJ
        c. H2 (g)        +      Cl2 (g)       → 2 HCl (g)                                      ΔH = -1845. kJ
        d. AlCl3 (s)                            → AlCl3 (aq)                                     ΔH = -323. kJ

35.5 Calculate the standard enthalpy of combustion for the following reaction:

                            C2H5OH(ℓ)    +    (7/2) O2(g)     →     2CO2(g)   +    3H2O(ℓ)

       Given the following heats of formation values:  

                                         C2H5OH(ℓ)    ΔH°f = -278.0 kJ/mol             
                                         O2(g)              ΔH°f =            kJ/mol (...you should know this value)
                                         CO2(g)           ΔH°f = -393.5 kJ/mol              
                                         H2O(ℓ)           ΔH°f =  -285.8 kJ/mol

35.6  Calculate the standard enthalpy of combustion for the following reaction: 

                          C6H12O6(s)    +    6O2(g)    →   6CO2(g)    +    6H2O(ℓ) 
   
        Given the following heats of formation values:  

                                         C6H12O6(s)    ΔH°f = -1275.0 kJ/mol
                                         O2(g)              ΔH°f =            kJ/mol (...you should know this value)
                                         CO2(g)           ΔH°f = -393.5 kJ/mol              
                                         H2O(ℓ)           ΔH°f =  -285.8 kJ/mol      


35.7  Calculate the standard enthalpy of formation for glucose (C6H12O6), given the following values: 
  
                                        ΔH°comb, glucose = -2800.8 kJ/mol        
                                        ΔH°f, CO2             = -393.5 kJ/mol             
                                        ΔH°f, H2O             = -285.8

35.8   Complete combustion of 1.00 mol of acetone (C3H6O) liberates 1790 kJ:  

       C3H6O(ℓ)     +      4O2(g)       →     3CO2(g)      +      3H2O(ℓ)         ΔH°comb, acetone = -1790 kJ/mol
 
   Using this information together with the data below (values in kJ/mol),
   calculate the enthalpy of formation, ΔH°f ,  for acetone.

                                        ΔH°f,  O2(g)    =    (...you should know this)      
                                       ΔH°f,    CO(g)    =   -393.5  kJ/mol         
                                       ΔH°f,    H2O(l)    =  -285.83 kJ/mol

Click and drag below for answers.

35.1. F + ½ C → A + B + D      ΔH = +47.0 kJ/mol

35.2   ΔH = -137. kJ                                 

35.3   ΔH = +256.0 kJ                     

35.4   ΔH = -6387. kJ

35.5   ΔHcomb =  -1367 kJ/mol

35.6   ΔHcomb =  -2801 kJ/mol

35.7   ΔHf =  -1275 kJ/mol

35.8   ΔHf =  -248 kJ/mol.

.
.
.