Catatan Popular

  • Enthalpy of Formation, ∆Hf
    • The change of heat when 1 mole of a   compound is formed from its elements at their   standard states.              H 2 (g) + ½ O 2(g) ...
  • Enthalpy of Solution, Hsoln
    The enthalpy of solution , enthalpy of dissolution , or heat of solution is the enthalpy change associated with the dissolution of a subst...
  • Enthalpy of Sublimation, Hsubl
    The heat (or enthalpy) of sublimation is the amount of energy that must be added to a solid at constant pressure in order to turn it direct...
  • Enthalpy of Combustion, ∆Hc
    • The heat released   when 1 mole of substance is burned completely in excess oxygen.     C (s) +   O 2(g)    → CO 2(g)                  ∆ ...
  • Enthalpy of Neutralization, ∆Hn
    The enthalpy of neutralization ( Δ H n o ) is the change in enthalpy that occurs when an acid and base undergo a neutralization reaction...
  • Enthalpy of Hydration, Hhyd
    • The heat change when 1 mole of gaseous ions is hydrated in water . • e.g:   Na + (g) ¾ ® Na + (aq)     D H hyd = - 406 kJ mol -1   C...
  • Born-Haber Cycle
    Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. ...
  • Calorimetry
    Calorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction. Calorimetry is a way of deter...
  • Enthalpy of Atomisation, Ha
    • The heat change when 1 mole of gaseous atoms is formed from its element      • D H a is always positive because it involves only breakin...
  • thermochemistry
    Thermochemistry is the study of the energy and heat associated with chemical reactions and/or physical transformations. A reaction may r...

question and answer

Problems
Example 1
Between Br2(l) and Br2(g) at 298.15 K which element would have a standard enthalpy of formation?
Solution:
We find that Br2(l) has the most stable form between Br2(l) and Br2(g), which means it has the lower enthalpy and
thus Br2(l) has ?Hf = 0. The answer is that Br2(g) has a standard enthalpy of formation.
Note: that the element phosphorus is a unique case. The reference form in phosphorus isn't the most stable form.
The solid white phosphorus is the reference form in this case. Even though red phosphorus is the most stable form,
In this case white phosphorous ?Hf = 0 even though its not the most stable form.
Again, standard enthalpies of formation can be either positive or negative.


Example 2
The enthalpy of formation of carbon dioxide at 298.15K is ?Hf = -393.5 kJ/mol CO2(g). Write the chemical equation.

Solution: In writing this equation you have to make sure you set it up for one mole of CO2(g). You must find the
stable forms of elements to. In this case they would be O2(g) and graphite for carbon.
?Hf = -393.5 kJ
The general equation for the standard enthalpy change of formation is:
?Hreactiono = ??Hfo (Products) - ??Hfo (Reactants)
When we plug in our equation for the formation of CO2:
?Hreactiono = ?Hfo[CO2(g)] - (?Hfo[O2(g)] + ?Hfo[C(graphite)]
As O2(g) and C(graphite) are in the most elementally stable form, they each have a standard enthalpy of formation
equal to 0:
?Hreactiono = -393.5 kJ = ?Hfo[CO2(g)] - ((1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol))
?Hfo[CO2(g)]= -393.5 kJ


Example 3
Using the values in the above table of standard enthalpies of formation, calculate the ?Hreactiono for the formation of
NO2(g).
NO2(g) is formed from the combination of NO(g) and O2(g) in the reaction:
To find the ?Hreactiono, we'll need to use the formula for the standard enthalpy change of formation:
?Hreactiono = ??Hfo (Products) - ??Hfo (Reactants)
The standard enthalpy of formation values we find from the data table are as follows:
O2(g): 0 kJ/mol
NO(g): 90.25 kJ/mol
NO2(g): 33.18 kJ/mol
When we plug these values into the formula for the standard enthalpy change of formation:
?Hreactiono = (1 mol)(33.18 kJ/mol) - ((2 mol)(90.25 kJ/mol) + (1 mol)(0 kJ/mol))
?Hreactiono = -147.32 kJ

Example 4
The lattice energy of NaCl calculated from a physics view point using the Madlung constant of the NaCl structure type is 788 kJ/mol. The estimated enthalpy of hydration for sodium and chloride ions are 406 and 363 kJ/mol respectively. Estimate the enthalpy of solvation for NaCl.
Solution
Using the cycle shown above and the formula, we have
Hhyd = Hcrystallization + Hsolv
Hsolv = -769 - (-788) kJ     = 19 kJ/mol.
Discussion
A positive value indicates an endothermic reaction. However, the value is small, and depending on the source of data, the estimated value may change. This value of 19 kJ/mol is too high compared to the value given earlier for NaCl of 3.88 kJ/mol, due to a high value of lattice energy used.


Example 5
The enthalpy of crystallization for KCl is -715 kJ/mol. The enthalpies of hydration for potassium and chloride are -322 and -363 kJ/mol respectively. From these values, estimate the enthalpy of solvation for KCl.
Solution
The enthalpy of hydration for KCl is estimated to be
Hhyd= -322 + (-363)       = -685 kJ/mol
Thus, the enthalpy of solvation is
Hsolv= -685 - (-715)       = 30 kJ/mol


Practical Applications of the Heat of Sublimation
The heat of sublimation can be useful in determining the effectiveness of medicines.  Medicine is often administered in pill (solid) form, and the substances which they contain can sublime over time if the pill absorbs too much energy over time.  Often times you may see the phrase "avoid excessive heat"2 on the bottles of common painkillers (e.g. Advil).  This is because in high temperature conditions, the pills can absorb heat energy, and sublimation can occur3 . 


Practice Problems
  1. If the heat of fusion for H2O is 333.5 kJ/kg, the specific heat capacity of H2O(l) is 4.18 kJ/(kg*K), the heat of vaporization for H2O is 2257 kJ/kg, then calculate the heat of sublimation for 1.00 kg of H2O(s) with the initial temperature, 273K (Hint:  273K is the solid-liquid phase change temperature and 373K is the liquid-gas phase change temperature).
  2. Using the information given in question one, calculate the heat of sublimation for 1.00 mole H2O when the initial temperature of the solid is 273K.  (Hint: molar mass of H2O is ~18.0 g/mol or 0.018 kg/mol)
  3. Using the information given in question one, calculate the heat of sublimation for 1.00 kg H2O when the initial temperature is 200K.  The specific heat capacity for H2O(s) is 2.05 kJ/(kg*K).
  4. If the heat of fusion for Au is 1.24 kJ/mol, the specific heat capacity of Au(l) is 25.4 J/(mol*K), the heat of vaporization for Au is 1701 kJ/kg, then calculate the heat of sublimation for 1.00 mol of Au(s) with the initial temperature, 1336K (Hint:  1336K is the solid-liquid phase change temperature, and 3081K is the liquid-solid phase change temperature).
  5. If the heat of sublimation for Cu at 3081K is 313.3245 kJ/mol, the specific heat capacity of Cu(l) is .0245 kJ/(mol*K), the heat of vaporization for Cu is 300.3 kJ/mol, then calculate the heat of fusion at 1356K for 1.00 mol of Cu(s) with the temperature (Hint:  1356K is the solid-liquid phase change temperature, and 3081K is the liquid-gas phase change temperature).
Practice Problem Solutions
  1. ΔHsub for 1kg H2O (at Ti=273K)= (333.5 kJ/kg)(1.0 kg) + (4.18 kJ/kg*K)(373-273K) + (2257 kJ/kg)(1.0 kg) = 30008.5 kJ/kg
  2. ΔHsub for 1mol H2O (at Ti=273K)= (30008.5 kJ/kg)(0.018 kg/mol) =  54.153 kJ/mol
  3. ΔHsub for 1kg H2O (at Ti=200K)= 30008.5 kJ/kg + (2.05 kJ/K*kg)(1.0kg)(273-200K) = 3158.15 kJ/kg
  4. ΔHsub for 1mol Au (at Ti=1336K)= (1.24 kJ/mol)(1mol) + (.0254 kJ/mol*K)(3081-1336K) + (1701 kJ/kg)(0.197kg/mol) = 380.66 kJ/mol
  5. ΔHfus for Cu (at T=1356K) = 337.8735 kJ/mol - (0.0245 kJ/mol*K)(2839-1356K) - (300.3 kJ/mol)(1mol) = 1.24 kJ/mo
 
Example 6 
  In an experiment, 0.100 g of H2 and excess of O2 were compressed into a 1.00 L bomb and placed into a calorimeter with heat capacity of 9.08 x 104 J0C-1. The initial temperature of the calorimeter was 25.0000C and finally it increased to  25.155 0C.  Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.
 
Problems
  1. Define lattice energy, ionization energy, and electron affinity. 
  2. What is Hess' Law?
  3. Find the lattice energy of KF(s).
    Note: Values can be found in standard tables.
  4. Find the lattice energy of MgCl2(s).
Solutions
  1. Lattice energy: The difference in energy between the expected experimental value for the energy of the ionic solid and the actual value observed. More specifically, this is the energy gap between the energy of the separate gaseous ions and the energy of the ionic solid.
    Ionization energy: The energy change associated with the removal of an electron from a neutral atom or ion.
    Electron affinity: The release of energy associated with the addition of an electron to a neutral atom or ion.
  2. Hess' Law states that the overall energy of a reaction may be determined by breaking down the process into several steps, then adding together the changes in energy of each step.
  3. Lattice Energy= [-436.68-89-(0.5*158)-418.8-(-328)] kJ/mol= -695.48 kJ/mol
  4. Lattice Energy= [-641.8-146-243-(737.7+1450.6)-(2*-349)] kJ/mol= -2521.1 kJ/mol


Solution
Heat released   = Heat absorbed by the 
                               calorimeter
                                        
               q       = C∆T
                    = (9.08 X 104 J0C-1) X (0.1550C)
                    = 1.41 X 104 J
                    = 14.1 kJ
        
  H2(g)   +  ½O2(g)   →  H2O(c)
             
  mole of H2  = 0.100
     2.016    
  =  0.0496 mol


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