LECTURE 4 Variation of enthalpy with temperature

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Cornelia Booker
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1 LECTURE 4 Variation of enthalpy with temperature So far, we can only work at 25 C. Like c v we define a constant pressure heat capacity, c p, as the amount of heat energy needed to raise the temperature of 1 mole of substance by 1 K at constant pressure. c p dq p dh = = dt dt p p or dh = c p dt Thus, H must increase by (c p dt) if heated by dt. T dh = c dt T 2 1 p the Kirchoff Equation First approximation: treat c p as independent of T for small temp. ranges or use the average value over the range of T H H = c ( T T ) T T p Particularly useful: recall that standard enthalpies of formation are defined at 25 C. Hence, we can write H f, T = H f, c p ( T )

2 Enthalpies of Reaction A model reaction: R P The enthalpy of R at 25 C is H f, 298 (R). The enthalpy at some other temperature T is [ H f, 298 (R) + c p (R) (T )]. H T (reaction) = [ H f, T (P) ] [ H f, T (R)] = [ H f, 298 (P) + c p (P) (T )] [ H f, 298 (R) + c p (R) (T )] = [ H f, 298 (P) - H f, 298 (R)} + {[ c p (P) - c p (R)] (T )} or H T (reaction) = H 298 (reaction) + c p (T ) where c p = ν i p ν i p c (products) - c (reactants)

3 The complete combustion of ethane releases kj mol -1 at 25 C. Calculate H (combustion) at 100 C. c p / J K -1 mol -1 : C 2 H 6(g) 52.6; O 2(g) 29.4; CO 2(g) 37.1; H 2 O (l) C 2 H 6(g) + 3½ O 2(g) 2 CO 2(g) + 3 H 2 O (l) H 298 = kj mol -1 c p = [2 c p (CO 2 ) + 3 c p (H 2 O)] - [c p (C 2 H 6 ) + 3½ c p (O 2 )] = [2 (37.1) + 3 (75.3)] - [ ½ (29.4)] = J K -1 mol -1 H 373 = H c p T = x 10 3 (J mol -1 ) + (144.6 J K -1 mol -1 )(75 K) = kj mol -1

4 Other types of enthalpy change Enthalpy of combustion - often referred to as the Heat of combustion the enthalpy change when 1 mol of a compound reacts completely with excess oxygen gas. e.g. the combustion of 1 mol of methane gas at 25 C evolves kj. CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (l) H 298 (comb.) = kj mol -1 H 298 (combust) are easy to measure and often used to determine H f,298 values. e.g. combustion of methane CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O (l) H (comb) = [2 H f (H 2 O) + H f (CO 2 )]-[ H f (CH 4 ) + 2 H f (O 2 )] = [2 (-285.8) + (-393.5] - [ H f (CH 4 ) + 2 (0)] = H f (CH 4 ) H f (CH 4 ) = 74.8 kj mol -1

5 The bomb calorimeter - already seen for U(combustion) at constant volume H(combustion) = U(combustion) + (pv) (combustion) If the gases behave ideally (pv) = (pv) prod - (pv) react = RT {(n gas ) prod - (n gas ) react } (pv) = n gas R T H(comb.) = U(comb.) + n gas R T where n gas is the number of moles of gas involved in the reaction. Why do we only consider gases? Volume of 1 mol of ideal gas: 25 C, 1 atm = cm 3. At 25 C: water occupies ~18 cm 3 mol -1 ; hexane 132 cm 3 mol -1, benzene 89 cm 3 mol -1 and naphthalene 120 cm 3 mol -1. Volumes of solids and liquids are negligible.

6 An example Following the earlier calculation, calculate the enthalpy of combustion of naphthalene at 25 C. From earlier, U(comb.) for naphthalene at 25 C = kj mol -1 C 10 H 8 (s) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O (l) To convert H to U using Eqn. 2.a, we need n gas. n gas = n gas (products) - n gas (reactants) = = -2 mol H(comb.) = J mol -1 + (-2 mol) (8.314 J K -1 mol -1 ) (298.2 K) H(combustion) = J mol -1 = kj mol -1 Exercise: Take the values from Example 2.8 and calculate the error in neglecting the volumes of naphthalene and water.

7 Note that the difference between the internal energy and enthalpy is relatively small in this case. Enthalpies of phase change Consider S, L, G for now. solid to liquid: melting or fusion. Liquid molecules have greater motion than solid - energy has to be added to the solid - fusion is an endothermic process. The enthalpy of fusion, H fus is the energy required at 1 bar pressure to melt 1 mole of a pure component at its melting point, T m. An earlier name, still in common usage is that H fus is the latent heat of fusion. It essentially represents the energy needed to overcome the intermolecular forces in the solid, Also, H(freezing) = - H(fusion)

8 liquid vapour - vaporization. H vap is the energy required at 1 bar pressure to vaporize 1 mole of a pure liquid at its boiling point, T b. Vaporisation is also endothermic. H vap is considerable larger than H fus since the molecules are completely separated in vaporization whereas a considerable degree of intermolecular interaction remains in liquids. Solid vapour - sublimation. Some compounds do not display a liquid phase an example is solid carbon dioxide which, at atmospheric pressure, sublimates at -78 C. H(sublimation) = H(fusion) + H(vaporization)

9 Examples Calculate the energy required to turn 100 g of ice at 0 C into steam at 100 C. The process required can be conveniently split into three parts g ice 100 g water at 0 C H 1 = H fus for 100 g g water, 0 C 100 g water, 100 C H 2 = c p T for 100 g g water, 100 C 100g steam, 100 C H 3 = H vap for 100 g 100 g of water = 100 / 18 mol = 5.56 mol H total = H 1 + H 2 + H 3 = 5.56 mol [ x ( ) x 10 3 ] = kj (100g) -1 Note that the factor of 10 3 occurs in H 1 and H 3 since they are measured in kj mol -1 while c p is in J K -1 mol -1.

10 A gas mixture consisting of 25% nitrogen and 75% hydrogen (by volume) was passed over a catalyst at a rate of 112 dm 3 min -1 (Gas volumes corrected to 0 C and 1 atm pressure). The reaction took place at 450 C and 1 bar pressure and complete conversion to ammonia was achieved. Given the following data, calculate the rate of heat evolution or absorption. H f, 298 (NH 3 ) kj mol -1 Mean c p / J K -1 mol -1 : N ; H ; NH The enthalpy change at 25 C can be calculated N 2(g) + 3 H 2(g) 2 NH 3(g) H 298 = [ 2 x H f, 298 (NH 3 )] - [ H f, 298 (N 2 ) + 3 H f, 298 (H 2 )] = [ ] - [ 0 ] kj mol -1 H 298 = kj mol -1 Correcting the enthalpy change to 450 C makes use of the Kirchoff Eqn. H T (reaction) = H 298 (reaction) + c p (T ) c p = [2 x c p (NH 3 )] - [ c p (N 2 ) + 3 x c p (H 2 )} = [2 x 39.7] - [ (3 x 29.3)] = J K -1 mol -1 H 723 = Jmol -1 + ( J K -1 mol -1 ) ( )K = Jmol -1 + ( Jmol -1 ) H 723 = kj mol -1 Hence, at 450 C, kj is evolved for each mole of the equation as written. From the stoichiometry, this means that kj is evolved per mole of nitrogen which reacts. Since, for a gas, the volume is proportional to the number of moles, the 1:3 mixture is in the stoichiometric amount. We know that at 0 C and 1 atm pressure, 1 mole of an ideal gas

11 occupies 22.4 dm 3. Thus, the total amount of gas passing over the catalyst amounted to 5 moles per minute, of which 1.25 moles were N 2. Rate of heat evolution = (moles of N 2 min -1 ) (heat evolved per mole N 2 ) = ( 1.25 mol min -1 ) ( kj (mol N 2-1 ) = kj min -1 The negative sign indicates an exothermic reaction so that heat is evolved.

12 Using the data from a text, estimate the standard enthalpies of formation for (a) gaseous cyclohexane, C 6 H 12 and (b) gaseous benzene, C 6 H 6. Compare the true values in a data table. (a) The formation of cyclohexane can be written 6 C (graph) + 6 H 2(g) C 6 H 12(g) H 1 = H f (C 6 H 12(g) ) To form cyclohexane, we need to make 6 C C bonds and 12 C H bonds. From the definition of bond energies, this refers to the reaction 6 C (g) + 12 H (g) C 6 H 12(g) H 2 = 6 (-344) + 12 (-415) = kj mol -1 the values being negative since this is the reverse of bond dissociation. Prior to this step, we need 6 C (graph) + 6 H 2(g) 6 C (g) + 12 H (g) H 3 = 6 H (graph.) + 6 H (H H) H 3 = 6 (716.7) + 6 (436) = kj mol -1. It is clear that H 1 = H 2 + H 3 = = kj mol -1 The actual value is kj mol -1. The agreement is therefore very good considering that average bond energies have been used throughout. (b) This calculation is performed in an analogous manner to part (a) except that benzene contains 6 C H, 3 C C and 3 C=C bonds. 6 C (graph) + 3 H 2(g) C 6 H 6(g) H 1 = H f (C 6 H 12(g) ) 6 C (g) + 6 H (g) C 6 H 6(g) H 2 = 3 (- 615) + 3 (-344) + 6 (-415) = kj mol -1

13 6 C (graph) + 3 H 2(g) 6 C (g) + 6 H (g) H 3 = 6 H (graph.) + 6 H (H H) H 3 = 6 (716.7) + 3 (436) = kj mol -1. It is clear that H 1 = H 2 + H 3 = = kj mol -1 The actual value is 83 kj mol -1. Clearly, there is poor agreement for benzene. This is due to the assumption of the alternating single and double bonds in benzene whereas the actual structure is a resonance stabilised hybrid of these. Note: This illustrates that care must be taken to consider the chemical properties of the molecules involved when using average bond enthalpy data. At best approximate results can be expected. The complete combustion of g of sucrose (C 12 H 22 O 11) in a bomb calorimeter with a heat capacity of JK C resulted in a temperature increase of C. Calculate H combustion for sucrose. at The energy released on combustion is given by (11140 JK K) = kj Since the bomb calorimeter is a constant volume device, this gives U combustion. The experiment used 5.85 x 10-3 moles. Hence, U combustion = / 5.85 x 10-3 = kj mol -1 The chemical equation for the reaction is C 12 H 22 O 11(s) + 12 O 2(g) 12 CO 2(g) + 11 H 2 O (l) so the number of moles of gas is the same before and after the reaction. Hence, n gas = 0 and H combustion = U combustion H combustion = kj mol -1

Thermochemistry: Heat and Chemical Change

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