CHAPTER 11 - DOC 1 by maclaren1

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									         CHAPTER 11 - ENTHALPY OF CHEMICAL REACTIONS

11.1 Introduction
All substances possess a certain amount of kinetic energy; their constituent particles may
be moving with vibrational, rotational or translational energy (or all three) depending on
whether the substance is in the solid, liquid or gaseous state. We say that substances have
„thermal energy‟ and the amount of this energy can be measured using the „temperature‟
scale. As we have noted earlier the two main temperature scales we use are the „Celsius‟
scale and the „Kelvin‟ scale, where 0 C = 273 K. In this chapter we shall be looking at
the changes in energy which may occur during a chemical reaction and how we may
quantify these changes by measuring the „change in enthalpy‟ of the reaction.

11.2 The concept of enthalpy:  H notation
The enthalpy (symbol H) or „heat content‟ of a chemical is a measure of its thermal
energy. When a chemical reaction proceeds there is almost always a change in the total
energy of the reactants compared to the total energy of the products. This change in
overall energy can be easily measured as a change in temperature where, if the reactants
have more energy than the products there will be a net energy loss and the temperature
will increase. If however, the reactants have less energy than the products the extra energy
needed will come from the surroundings and the overall temperature will decrease. Due
to the ease with which it can be measured, therefore, chemists talk about the change in
enthalpy from reactants to products, where:

          change in enthalpy = enthalpy of the products - enthalpy of the reactants

                                   H = Hproducts - H reactants

11.3. Heats of reaction
A reaction in which energy (or „heat’) is given off is known as an exothermic reaction -
the enthalpy of the reactants was greater than that of the products and the H value is
negative. As has been noted previously, an exothermic reaction is accompanied by an
increase in the temperature of the surroundings. Similarly, a reaction in which energy
(heat) must be taken from the surroundings is known as an endothermic reaction and the
H value is positive.

These most important of concepts may be represented on a „Heat change diagram‟.

Fig. 1

 Exothermic (H = -ve)                      Endothermic (H = + ve)

     Hr                                                              Hp
                        H                         H

                             Hp                  Hr
These simple diagrams do not attempt to show what is happening as the reactants change
into products - merely the end result of the transformation. The amount of energy that
must be supplied to actually initiate the reaction is known as the activation energy and
can also be represented on heat change diagrams:

Fig.2

 Exothermic (H = -ve)                      Endothermic (H = + ve)

                        Activation
                        energy                                       Hp
         Hr
                                                                                Activation
                                                 Hr                             energy
                            Hp


The larger the activation energy (or „hump‟ on the diagram) the more energy that must be
supplied to initiate the reaction. Some reactions, such as that of nitroglycerine reacting
with oxygen from the air in an explosion, have very small activation energies. Others,
such as the reaction of nitrogen and oxygen gases to produce the noxious gas nitrogen
monoxide, only occur to any appreciable extent at high temperatures - have high
activation energies. In industry it is often desirable to make a reaction proceed more
easily, and hence at lower cost, by using a suitable catalyst.

The function of a catalyst is to provide an alternate reaction pathway with a smaller
activation energy. By doing this the rate of the reaction is increased.

Fig. 3

Reaction pathway without catalyst              Reaction pathway with catalyst

                                                                   Activation
                        Activation                                 energy reduced
                        energy
                                               Hr
         Hr
                             Hp                                        Hp



Note that the magnitude and sign of the H value is not affected by the catalyst. By
reducing the activation energy the reaction is able to proceed at a faster rate - the position
of equilibrium and the equilibrium constant remain unchanged.
11.4 Specific Heat Capacity
Could the well-known phrase “A watched pot never          Table 11.1 Specific heats of
boils!” be alluding to the fact that water requires an    selected substances
extraordinarily large input of energy before its            Substance    Specific Heat content
temperature increases significantly? When supplied                       (J. C-1.g-1)
with an equivalent amount of heat energy, water (due           Water              4.184
to its strong hydrogen bonding between molecules)            Ammonia              2.124
                                                              Ethanol             1.413
will have a lesser increase in temperature than almost
                                                             Alumina              0.778
any other common substance, as is demonstrated in              Iron               0.448
Table 11.1. Once more we see that water possesses             Copper              0.386
physical and chemical properties that set it apart from        Gold               0.128
most other substances.

The „specific heat capacity‟ is defined as the amount of heat energy required to raise the
temperature of 1 gram of a substance by 1 C or 1K.

Example 11.1
A calorimeter is filled with 100 ml of water and its temperature is recorded as 18.28 C.
A current of 2.41A at a potential difference of 5.92V is then passed through the water for
exactly 2 minutes. What temperature will the water reach, presuming the calorimeter is
well insulated?

Solution
First, we must calculate the amount of energy supplied to the water.
                    E = V.I.t
                  E = 5.92 x 2.41 x 120
                  E = 1712 J
The specific heat capacity of water is 4.184 J. C-1.g-1
        ie. to raise 1 g of water by 1 C requires the input of 4.184 J of energy
                  to raise 100 g of water requires 100 x 4.184 J = 418.4 J
                                              1712
                  Temperature increase =            = 4.09 C
                                              418.4

11.5 The calorimeter
Calorimeters are devices used to measure the thermal energy released or absorbed as a
chemical reaction proceeds. They can be used in a wide variety of applications, including
reactions involving aqueous solutions, gases and even in the combustion of foods.

        11.5.1 Components of calorimeters
All calorimeters basically consist of a well insulated outer container enclosing an inner
chamber where the reaction will take place. They have an electrically heated coil to allow
for calibration, an accurate thermometer to measure any change in temperature and a
stirring device to ensure even mixing of the reactants. If the reactants involve gases, the
inner chamber will contain a sealed glass vessel (or „bomb‟) into which the gaseous
reactants are placed. When the reaction is to take place, the glass bulb is broken to allow
contact of all reactant species.

Fig. 4 - Solution and bomb calorimeters




        11.5.2 Calibration of calorimeters
Before any calorimeter can be used to measure the enthalpy of a reaction it must be
calibrated
          ie. we must know how much energy is required to raise the calorimeter and its
 contents by 1 C. In most cases this is achieved by passing an electric current at a known
    voltage through a heating coil within the calorimeter and measuring the increase in
     temperature that results. The formula for calculating the calorimeter constant is:

                                                  E   VIt
                       Calorimeter constant =                        Units: J.C-1
                                                 T T


Example 11.2
A student wishes to calibrate a calorimeter in order to perform an experiment involving
the heat change of a particular reaction. To do this she fills the calorimeter with exactly
100.0 ml of deionised water and then passes a current of 1.88A through the heating coil
for exactly 2 minutes. The potential difference across the coil is measured at 4.13V and
the increase in temperature that occurs is 3.81 C.

Solution
                                   E   VIt 413  188  120
                                            .     .
       Calorimeter constant =                             245 J.C-1
                                  T T           .
                                                 381

11.6 Thermochemical equations
A thermochemical equation is one which gives the relevant H value for that particular
equation exactly as it is written. For example, the thermochemical equation for the
combustion of methane in air can be written as:
       CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O (l)           H = -890 kJ mol-1
This equation literally means that “when 1 mole of methane reacts with 2 moles of
oxygen to produce 1 mole of carbon dioxide and 2 moles of water that 890 kJ of energy is
released”. If the equation were doubled, the H value would also double. If the equation
were reversed, the H value would also change sign:
       2CH4 (g) + 4O2 (g) ----> 2CO2 (g) + 4H2O (l)          H = -1780 kJ mol-1

       CO2 (g) + 2H2O (l) ----> CH4 (g) + 2O2 (g)             H = +890 kJ mol-1

Thermochemical equations can be used to perform a range of stoichiometric calculations
involving heats of reaction.

Example 11.3
Calculate the amount of energy released when 500 ml of methane gas at S.L.C. reacts
with excess air according to the equation
       CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O (l)       H = -890 kJ mol-1

Note: VM of any gas at S.L.C. = 24.5L

Solution
                   V    0.500
       n(CH4) =               0.0204 mol
                  VM     24.5
From the equation, 1 mol of yields 890 kJ
                0.0204 mol yields „x‟ kJ
       By ratio, x = 890 x 0.0204 = 18.16

The energy released by the combustion reaction is 18.16 kJ.

Example 11.4
What mass of propanol must be burnt in excess air to produce 10,000 kJ of energy,
according to the reaction
       2C3H7OH (l) + 9O2 (g) ----> 6CO2 (g) + 8H2O (l) H = -4034 kJ mol-1

Solution
       From the equation, 2 mol of propanol yields 4034 kJ of energy
                         „x‟ mol yields 10,000 kJ
              By ratio, 2 x 10,000 = „x‟ x 4034
                         x = 4.96 mol
              M (C3H7OH) = 60 g.mol-1
                         mass of propanol = 297 g
Example 11.5
When hydrogen sulfate is mixed with water to produce sulfuric acid a substantial amount
of heat is evolved. If 100 ml of hydrogen sulfate (density = 1.834 g.cm-3) is added to 1
litre of water at 18.5 C, calculate the temperature that the water will reach after
dissolution. The relevant equation is :
                 H2SO4 (l) water  H2SO4 (aq)
                                                     H = -7.2 kJ mol-1

Solution
                   m
       d(H2SO4) =       m (H2SO4) = 1.834 x 100 = 183.4 g
                   V
                    m 1834   .
       n(H2SO4) =             = 1.87 mol
                   M        .
                          981
       From the equation, 1 mol of H2SO4 yields 7.2 kJ
                      1.87 mol of H2SO4 yields „x‟ kJ
                      x = 13.46 kJ

       The specific heat capacity of water is 4.184 J. C-1.g-1
                       To increase the temperature of 1 litre of water by 1 C requires
                           4.184 x 1000 = 4184 J
                                                      13,460
                       Increase in temperature =              = 3.22 C
                                                       4184

Example 11.6
1.82g of a biscuit was dried and then burnt in a bomb calorimeter in the presence of
excess pure oxygen. The temperature of the water bath surrounding the reaction chamber
rose from 22.085 C to 23.117 C. To calibrate the calorimeter, a current of 3.77A at a
potential difference of 5.29V was passed through the heating coil for a period of 72.1
seconds; a rise in temperature of 3.326 C being recorded. Calculate the heat content of
the biscuit.

Solution
First, we must calculate the calorimeter constant:
                                   E    VIt 5.29  377  721
                                                      .    .
        Calorimeter constant =                              432.3 J.C-1
                                  T T             .
                                                   3326

       The change in temperature caused by the combustion of the biscuit
              T = 23.117 - 22.085 = 1.032 C
        Energy released by 1.82g of biscuit = 432.3 x 1.032 = 446.2 J
                                    446.2
        Heat content of biscuit =        = 245 J g-1
                                      .
                                     182
11.7 Summary/ Objectives
At the end of this Chapter you should
        - understand that chemical reactions involve a change in the energy of the
reactants as they change into products and that this change is known as the „enthalpy‟ of
the reaction
        - appreciate that it is much easier to consider a change in enthalpy as a reaction
proceeds, such that
                          H = Hproducts - H reactants
        - recognise that heat change diagrams are most useful devices for representing
changes in enthalpy and the effect of catalysts on the activation energy of a reaction
        - be able to interpret accurately and fully these heat change diagrams
        - recall that endothermic reactions have a negative H value as they release heat
into the environment during the course of a reaction
        - recall that exothermic reactions have a positive H value as they absorb heat
from the environment during the course of a reaction
        - understand the role of a catalyst as providing an alternative pathway for a
reaction of lower activation energy
        - be able to display catalytic function on a heat change diagram and/or interpret
such a diagram
        - recall the definition of „specific heat capacity‟ of a substance as „the amount of
heat energy required to raise the temperature of 1 gram of a substance by 1 C or 1K‟
        - recall that water has a very high specific heat capacity and understand the
reasons for this (in terms of hydrogen bonding)
        - be familiar with the components of both „solution‟ and „bomb‟ calorimeters and
appreciate the function of these components in terms of the overall purpose of a
calorimeter
        - recall how to calibrate a calorimeter from both a theoretical standpoint and in
practice, having been supplied with the relevant data
        - recognise the experimental errors inherent in calorimetric determinations and
discuss how these errors may be minimised
        - appreciate how to use thermochemical equations in stoichiometric calculations
to determine energy changes and/or H values, having been supplied with the relevant
data
                                  Chapter 11 Questions

Q1. Draw and clearly label heat change diagrams to represent
      (i) an exothermic reaction with a relatively small activation energy
      (ii) an endothermic reaction with a relatively large activation energy
      (iii) the effect of a catalyst on the activation energy of both of the above reactions

Q2. State which of the following reactions are exothermic and which are endothermic:
      (a) I2 (s) + Cl2 (g) ----> 2ICl (g)                        H = +35 kJ mol-1
      (b) N2 (g) + 3H2 (g) ----> 2NH3 (g)                        H = -92.3 kJ mol-1
      (c) BCl3 (l) + 6H2 (g) ----> B2H6 (g) + 6HCl (g)           H = +315 kJ mol-1
      (d) Ag+ (aq) + 2NH3 (aq) ----> Ag(NH3)2+ (aq)              H = -111 kJ mol-1
      (e) N2 (g) ----> 2N (g)                                    H = +946 kJ mol-1

Q3. Calculate the amount of heat energy absorbed from the surroundings when 25.0 g of
sodium sulfate decahydrate crystallises from water, given the equation:
  Na2SO4.10H2O (s) water  Na2SO4 (aq) + 10H2O (l)
                                                             H = +79.0 kJ mol-1

Q4. How much energy is required to convert 1 litre of boiling water into steam?
      The heat of vaporisation of water is given below:
             H2O (l) ----> H2O (g)          H = + 44.0 kJ mol-1

Q5. Anhydrous copper sulfate is a hydroscopic substance - it absorbs water from the
surroundings according to the equation:
                             
              CuSO4 (s) water  CuSO4 (aq)

When 5.73 g of this substance dissolves in water it produces 2.388 kJ of thermal energy.
Calculate the H for the dissolution reaction above.

Q6. Calculate the energy released when 10.0 g of carbon reacts with excess oxygen
according to the equation:
       C (s) + O2 (g) ----> CO2 (g)        H = -393.4 kJ mol-1

Q7. What mass of brown coal, with an average specific heat capacity of 25 kJ g-1 would
be required to heat 600 ml of water from 20 C to 100 C?
                                      [specific heat capacity of water = 4.184 J C-1g-1]

Q8. The heat of combustion of hexane is 4158 kJ mol-1. Calculate H for the reaction:
      2C6H14 (l) + 19O2 (g) ----> 12CO2 (g) + 14H2O (l)

Q9. Calculate the heat of combustion of butan-1-ol given the following:
 2C4H9OH (l) + 12O2 (g) ----> 8CO2 (g) + 10H2O (l)             H = -5354 kJ mol-1
Q10. Octane, the principle component of petrol, undergoes combustion with a plentiful
supply of oxygen according to the equation:
 2C8H18 (l) + 25O2 (g) ----> 16 CO2 (g) + 18H2O (l)         H = -10,900 kJ mol-1

Given the density of octane is 0.698 g cm-3 calculate the thermal energy released when 50
litres of octane undergoes complete combustion.

Q11. Much research is currently being undertaken to investigate the feasibility of using
ethanol (or an ethanol/petrol mixture) as a substitute for the fossil-fuel based petrol
currently employed in most cars. Given that ethanol has a density of 0.785 g cm-3 and that
it undergoes combustion according to the equation:
   C2H5OH (l) + 3O2 (g) ----> 2CO2 (g) + 3H2O (l)                 H = -1364 kJ mol-1

calculate the thermal energy released when 50 litres of ethanol undergoes complete
combustion. By referring to Question 9, which of ethanol or octane (on this basis alone) is
the better fuel? What other factors should be taken into account?

Q12. When an electric current is passed through water it will decompose to its
constituent elements according to the equation:
               2H2O (l) ----> 2H2 (g) + O2 (g)       H = +572 kJ mol-1

Calculate the amount of electrical energy that would be required to produce 500 cm3 of
oxygen gas at S.T.P.

Q13. As part of the process of manufacturing sulfuric acid the following reaction takes
place:
       2SO2 (g) + O2 (g) ----> 2SO3 (g)            H = -197 kJ mol-1

Calculate the heat energy released in the production of 1 tonne of SO3 gas.

Q14. An important reaction which occurs in the blast furnace is the reduction of carbon
dioxide to carbon monoxide, according to the equation:
       CO2 (g) + C (s) ----> 2CO (g)        H = +173 kJ mol-1

Calculate the amount of heat energy absorbed as 5000 litres of CO2 is reduced by excess
carbon to CO at 1.226 atm pressure and 1100 C.

Q15. By using the specific heat capacity data from Table 11.1 calculate the amount of
energy
       (i) required to raise the temperature of 20 ml of water from 11 C to 85 C
       (ii) given off as 45 ml of ethanol [d = 0.785 g.cm-3] cools from 37 C to 4 C
       (iii) absorbed by 1000 kg of iron as it is heated from 25 C to 1000 C
Q16. A 1.00 kg block of gold at a temperature of 488 C is plunged into a container
filled with 20 L of water at a temperature of 18.5 C. The gold block is removed when its
temperature has been lowered to 40 C. What is the temperature of the water now?
[specific heat of gold = 0.128 JC-1g-1]

Q17. A solution calorimeter with an internal volume of 100 ml has a constant of
419 J C-1. A student carefully adds 100 ml of 2 M HCl to the calorimeter and measures
the temperature as 19.60 C. He then adds 0.449g of powdered magnesium, stirs the
solution to ensure complete reaction and measures the temperature the solution has
reached as 25.42 C.
Calculate the heat of reaction for the equation
               Mg (s) + 2HCl (aq) ----> MgCl2 (aq) + H2 (g)

Q18. In order to determine the energy content of a sample of bread a food scientist
measures a 1.50g sample, dries it fully and places it into the combustion chamber of a
bomb calorimeter. The device is calibrated by passing a current of 2.44 A at a potential
difference of 5.08 V for 185 seconds. The temperature within the calorimeter increased
by 1.69 C. The chamber is then filled with oxygen and the bread sample is ignited. The
temperature of the water surrounding the combustion chamber is found to increase from
17.22 C to 29.16 C.
        (i) Calculate the energy content of the bread.
        (ii) State the major sources of error which may have affected the result obtained

Q19. A student wishes to determine the heat of reaction for a strong acid reacting to
completion with a strong base. The ionic equation for this reaction is:
              H3O+ (aq) + OH- (aq) ----> 2H2O (l)

She first electrically calibrates the calorimeter by filling it with 100.0 ml of deionised
water and passing a current of 1.496 A at a potential difference of 4.07 V for 160
seconds. The temperature of the water rose from 16.81 C to 19.58 C. She then carefully
added 50 ml of 0.110 M NaOH and 50.0 ml of 0.101 M HCl to the calorimeter and
measured the increase in temperature as 0.81 C.

       (i) Calculate the calorimeter constant
       (ii) Calculate the heat of neutralisation for the reaction
       (iii) Suggest a reason for using a higher concentration of the NaOH compared to
               the HCl.
       (iv) List the main experimental errors the student may have encountered.
       (v) Would you expect the heat of neutralisation to have been larger, smaller or the
same if 50 ml of 0.101 M acetic acid had been reacted with 50 ml of 0.110 M ammonia?
Give an explanation for your answer.
                          Solutions to Chapter 11 questions

Q2. a) endothermic    b) exothermic    c) endothermic       d) exothermic

e) endothermic               Q3. 6.13 kJ      Q4. 2440 kJ         Q5. 66.5 kJ mol-1

Q6. 327.8 kJ          Q7. 8.03g        Q9. -8316 kJ mol-1 Q10. 1.669 x 106 kJ

Q11. 5.819 x 105 kJ Q12. 12.8 kJ       Q13. 1.23 x 106 kJ   Q14. 9413 kJ

Q15. (i) 6.19 kJ (ii) 1.65 kJ (iii) 4.37 x 105 kJ   Q16. 21.24 C

Q17. -132 kJ. mol-1 Q18. -10.8 kJ. mol-1     Q19. (i) 351.7 C-1 (ii) -56.1 kJ. mol-1

								
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