CHEM1901/ J-8 June 2013

CHEM1901/3 2013-J-8 June 2013 The atmosphere of Venus contains 96.5 % CO 2 at 95 atm of pressure, leading to an average global surface temperature of 462 C. The energy density of solar radiation striking Venus is 2625 J m 2 s 1. The radius of Venus is 6052 km, and the average albedo (the fraction of solar radiation reflected back into space) of its surface is 0.90. Calculate the magnitude of the greenhouse effect on Venus. 5 An albedo of 0.90 means that 90% of the solar energy is reflected and only 10% is available to heat the planet. If Venus has a radius R, the energy input from the Sun is: E in = (2625 πr 2 0.10) J The energy output from the surface area 4πR 2 is: E out = (4πR 2 5.67 10-8 T 4 ) J At equilibrium, the energy input and output are equal and so the temperature can be calculated: 2625 πr 2 0.10 = 4πR 2 5.67 10-8 T 4 T 4 = 0.10 2625 / (4 5.67 10-8 ) K so T = 184 K As the actual temperature on Venus is 462 C = 735 K, the magnitude of the greenhouse effect is: greenhouse effect = (735 185) K = 551 K Answer: 551 K The main absorption bands of CO 2 lie in the energy range 600 750 cm 1. What range of wavelengths (in nm) corresponds to this energy range? 600 cm -1 corresponds to a wavelength of (1/600) cm = 0.00167 cm. As 1 cm = 0.01 m and 1 nm = 10-9 m, this corresponds to 16700 nm. 750 cm -1 corresponds to a wavelength of (1/750) cm = 0.00133 cm. As 1 cm = 0.01 m and 1 nm = 10-9 m, this corresponds to 13300 nm. ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2013-J-8 June 2013 Sketch the emission spectrum of Venus on the axes below. Note the wavelength of maximum intensity, and point out any other important features. The wavelength corresponding to the intensity maximum is given by Wien s displacement law: T λ max 2.898 10 6 K nm Using T = 735 K, λ max (2.898 10 6 K nm) / (735 K) = 3940 nm λ max = 3940 nm Intensity y CO 2 absorption 4000 13,000 17,000 wavelength

CHEM1901/3 2013-J-12 June 2013 Differential scanning calorimetry (DSC) is an experimental technique that measures the temperature of a sample as a function of the heat supplied to it. Negative or positive peaks on a DSC curve therefore indicate endothermic or exothermic processes respectively. The figure below shows a series of DSC curves collected for methane at different pressures. The scales of all the heat flow curves are the same, but they have been offset from zero for clarity. Clearly identify the type of phase change associated with every peak in the DSC curve. 8 s l s l s l l g l g s g On the 0.1, 1 and 10 bar curves, there are two phase transitions observed so melting and boiling occurs. From the size of the peak is consistent with the single peak in the 100 bar being melting. The size of the peak in the 0.01 bar curve is not consistent with melting or boiling so it must correspond to sublimation. ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2013-J-12 June 2013 Use the DSC data shown to sketch a pressure-temperature phase diagram on the graph below (note that pressure is on a log scale). Label all the important regions of the phase diagram. solid supercritical fluid liquid critical point gas triple point

CHEM1901 2012-J-8 June 2012 Explain the following phenomena. (a) When O 2 dissolves in water at room temperature, the total entropy of the system decreases. 6 The decrease in entropy of O 2 going from (g) (aq) outweighs the increase in entropy of pure water going to water with O 2 dissolved in it. (b) When MgSO 4 is dissolved in water, there is a very small but measurable decrease in volume. The hydration spheres around the dissolved ions (especially the small Mg 2+ ion) create a denser structure than that of random H 2 O molecules in pure water. (c) A crystal (e.g. of NaCl) in its lowest energy configuration (thermodynamic ground state) will always contain defects at finite temperatures. Defects increase enthalpy, but they also increase entropy. So a small number of defects will lower the total free energy (except at T = 0 K). THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY

CHEM1901/3 2010-J-10 June 2010 Iron oxide that has been exposed to the atmosphere for any length of time will generally contain a mixture of magnetite, Fe 2 O 3, and haematite, Fe 3 O 4. This mixture can be converted to pure Fe 3 O 4 by heating it under an excess of flowing hydrogen gas: 3Fe 2 O 3 (s) + H 2 (g) 2Fe 3 O 4 (s) + H 2 O(g) H = 11.0 kj mol 1 If 0.364 kj of heat is liberated during the conversion of a 7.18 g sample of iron oxide (consisting only of magnetite and haematite) to pure Fe 3 O 4, what was the initial mass percentage of haematite in the sample? 4 From the chemical equation, 1 mol of H 2 liberates 11.0 kj of heat. Hence, the number of moles of H 2 that must be reacting to produce 0.0364 kj of heat is: number of moles of H 2 = (0.0364 / 11.0) mol = 0.00331 mol From the chemical equation, 1 mol of H 2 reacts with 3 mol of Fe 2 O 3 and so: number of moles of Fe 2 O 3 = (3 0.00331) mol = 0.00993 mol The molar mass of Fe 2 O 3 is (2 55.85 + 3 16.00) g mol -1 = 159.7 g mol -1. Hence, this amount of Fe 2 O 3 corresponds to: mass of Fe 2 O 3 = number of moles molar mass = (0.00993 mol) (159.7 g mol -1 ) = 1.59 g The mass percentage of Fe 2 O 3 in the initial 7.18 g sample is therefore: mass percentage of Fe 2 O 3 = (1.59 / 7.18) 100% = 22.1 % The mass percentage of haematite is therefore (100.0 22.1)% = 77.9%. Answer: 77.9% Fe 3 O 4 can be further reduced to FeO under flowing hydrogen. Fe 3 O 4 (s) + H 2 (g) 3FeO(s) + H 2 O(g) Based on the following table of thermodynamic data, what is the minimum temperature at which this would be a spontaneous reaction? Fe FeO(s) Fe 3 O 4 (s) H 2 (g) H 2 O(g) f H (kj mol 1 ) 272 1118 242 S (J K 1 mol 1 ) 27 61 146 131 189 Using rxn H = f H (products) - f H (reactants): rxn H = [3 f H (FeO) + f H (H 2 O] [ f H (Fe 3 O 4 )] = ([(3-272) + (-242)] [-1118]) kj mol -1 = 60. kj mol -1 ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2010-J-10 June 2010 Using rxn S = S (products) - S (reactants): rxn S = [3S (FeO) + S (H 2 O] [S (Fe 3 O 4 ) + S (H 2 )] = ([(3 61) + (189)] [(146 + 131)]) J K -1 mol -1 = 95 J K -1 mol -1 The reaction is spontaneous if G < 0. As G = H - T S, this condition becomes: Hence: H - T S < 0 (60. 10 3 J mol -1 ) T (95 J K -1 mol -1 ) < 0 T > (60. 10 3 J mol -1 ) / (95 J K -1 mol -1 ) T > 630 K Answer: T > 630 K

CHEM1901/3 2010-J-11 June 2010 Would the resulting FeO be stable under those conditions, or would it be reduced further to Fe metal by the following reaction? Explain, with calculations as appropriate. FeO(s) + H 2 (g) Fe(s) + H 2 O(g) 3 For this reaction, using rxn H = f H (products) - f H (reactants): rxn H = [ f H (H 2 O] [ f H (FeO)] = ([-242] [-272]) kj mol -1 = 30. kj mol -1 Using rxn S = S (products) - S (reactants): rxn S = [S (Fe) + S (H 2 O] [S (FeO) + S (H 2 )] = ([27 + 189] [61 + 131]) J K -1 mol -1 = 24 J K -1 mol -1 At 630 K, G = H - T S = (30. 10 3 ) (630 K) (24 J K -1 mol -1 ) = 15000 J mol -1 As G > 0, the reaction is not spontaneous and FeO is stable.

CHEM1901/3 2009-J-8 June 2009 A new process has been developed for converting cellulose from corn waste into the biofuel butanol, C 4 H 9 OH. A bomb calorimeter with a heat capacity of 3250 J K 1 was used to determine the calorific value by burning 5.0 g of butanol in excess oxygen. Write a balanced reaction for the combustion of butanol in oxygen. 5 C 4 H 9 OH(l) + 6O 2 (g) 4CO 2 (g) + 5H 2 O(g) Calculate the heat released from this combustion if the temperature of the calorimeter increased from 23.0 to 78.6 C during the test. The heat, q, required to change the temperature of the bomb calorimeter by T is given by: q = C T The temperature increases from 23.0 to 78.6 C corresponding to T = 55.6 C = = 55.6 K. Hence: q = (3250 J K -1 )(55.6 K) = 181000 J = 181 kj Answer: 181 kj Use this value to determine the calorific value and molar enthalpy of combustion of butanol. The calorific value is the energy content per gram. As 5.0 g of butanol releases 181 kj, the calorific value is: calorific value = = 36 kj g-1. The molar enthalpy of combustion is the energy released by combusting one mole. The molar mass of C 4 H 9 OH is: molar mass = (4 12.01 (C) + 10 1.008 (H) + 16.00 (O)) g mol -1 = 74.12 g mol -1 5.0 g of butanol thus corresponds to: number of moles =.. = 0.067 mol As 0.067 mol generates 181 kj when combusted, the molar enthalpy of combustion is negative and is given by: H =. = -2.7 103 kj mol -1

CHEM1901/3 2008-J-8 June 2008 Write a balanced equation for (i) the explosive decomposition, and (ii) the combustion in air, of TNT, C 7 H 5 N 3 O 6 (s). (i) 2C 7 H 5 N 3 O 6 (s) 12CO(g) + 3N 2 (g) + 5H 2 (g) + 2C(s) (ii) 2C 7 H 5 N 3 O 6 (s) + 21 /2O 2 (g) 14CO 2 (g) + 5H 2 O(g) + 3N 2 (g) What is the essential difference between these two processes? In the combustion reaction, the oxidant comes from an external source (air). In the explosion, the oxidant is contained in the explosive material. What is the increase in the number of moles of gas (per mole of TNT consumed) for each of these two processes? For process (i), the change in the number of moles of gas, n, is: n = n(moles of gaseous products) n(moles of gaseous reactants) = ((12 + 3 + 5) (0)) mol = 20 mol As (i) involves 2 mol of TNT, n = 10 mol per mole of TNT Answer: (i) 10 mol For process (ii), the change in the number of moles of gas, n, is: n = n(moles of gaseous products) n(moles of gaseous reactants) = ((14 + 5 + 3) ( 21 /2)) mol = 23 /2 mol As (i) involves 2 mol of TNT, n = ½ 23 /2 mol per mole of TNT Answer: (ii) 23 /4 mol Which of these two processes releases more energy into the surroundings? Data: f H (TNT) = 6.9 kj mol 1 f H (CO(g)) = 111 kj mol 1 f H (CO 2 (g)) = 393 kj mol 1 f H (H 2 O(g)) = 242 kj mol 1 Using r H = Σ m f H (products) - Σ m f H (reactants), the enthalpy change for reaction (i) can be written as r H = [12 f H (CO(g))] [2 f H (C 7 H 5 N 3 O 6 (s) )] = ([12-111] [2 6.9]) kj mol -1 = --1346 kj mol -1 ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2008-J-8 June 2008 Similarly. the enthalpy change for reaction (ii) can be written as r H = [14 f H (CO 2 (g)) + 5 f H (H 2 O(g))] [2 f H (C 7 H 5 N 3 O 6 (s) )] = ([14-393 + 5-242] [2 6.9]) kj mol -1 = -6726 kj mol -1 f H (N 2 (g)), f H (H 2 (g)) and f H (C(s) all represent formation of the elements from their standard states and are all zero. The combustion reaction releases more energy. Answer: combustion reaction

CHEM1901/3 2008-J-10 June 2008 A calorimeter containing 300.0 ml of water at 25 C was calibrated as follows. A 1000.0 W heating coil was run for 10.0 s, after which time the temperature had increased by 7.5 C. Calculate the heat capacity of the empty calorimeter. The specific heat of water is 4.184 J K 1 g 1. 6 With a power of 1000.0 W = 1000.0 J s -1, the amount of heat generated by the coil is: q = (1000.0 J s -1 ) (10.0 s) = 1.00 10 3 J As the density of water at 25 C is 0.997 g ml -1, 300.0 ml of water corresponds to: mass of water = m = density volume = (0.997 g ml -1 ) (300.0 ml) = 299 g The heat required to heat this quantity of water by 7.5 C is: q = T = (4.184 J K -1 g -1 ) (299 g) (7.5 K) = 9380 J The remaining heat, (1.00 10 3 9380) J = 620 J, is used to heat the calorimeter. This also increases in temperature by 7.5 C so the heat capacity of the calorimeter is: c calorimeter = =. = 82 J K-1 Answer: 82 J K -1 A solution containing 0.040 mol Ag + (aq) was mixed with a second solution containing 0.050 mole Br (aq) in this calorimeter, causing AgBr(s) to precipitate. The temperature increased by 2.4 C. Given the solubility product constant is K sp (AgBr) = 5 10 13 M 2, calculate the equilibrium concentrations of Ag + (aq) and Br (aq) present in the final solution of volume 320 ml. To calculate the equilibrium constants of Ag + (aq) and Br - (aq), a reaction table can be used, with x representing the number of moles which do not precipitate: Ag + (aq) Br - (aq) AgBr(s) initial (mol) 0.040 0.050 0 final (mol) x 0.010 + x 0.040 - x The concentrations in the final solution of volume 320 ml are therefore: [Ag + (aq)] =. and [Br -. (aq)] =. Hence, the solubility product is: ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2008-J-10 June 2008 K sp = [Ag + (aq)][br - (aq)] =. = 5 10-13.. K sp is very small, x is tiny and so 0.010 x ~ 0.010. This approximation gives: or. = 5 10-13.. [Ag + (aq)] =. = (5 10-13 ).. M = 1.6 10-11 M [Br - ] =.. M = 0.031 M [Ag + (aq)]: 1.6 10-11 M [Br (aq)]: 0.031 M Calculate the enthalpy of solution of AgBr(s). As the final solution has a volume of 320 ml, its mass is: mass = density volume = (0.997 g ml -1 ) (320 ml) = 319 g The heat produced causes the temperature of this mass of solution and the calorimeter to rise by 2.4 C: q = ( T) + ( T) = ((4.184 J K -1 mol -1 ) (319 g) (2.4 K)) + ((82 J K -1 ) (2.4 K)) = 3400 J This corresponds to the heat given out in the precipitation of 0.040 mol of AgBr(s). As the precipitation is exothermic, the heat of solution is endothermic: heat of solution = + = +85 kj mol-1. Answer: +85 kj mol -1

CHEM1901/3 2008-J-11 Use the figure below to help answer the following. June 2008 4 Write a balanced equation for the smelting of one of these metal oxides with coke in which a major product is CO 2. Give the approximate temperature range over which this reaction is spontaneous and state what happens outside this temperaturee range. Between 400 950 K, the Ni NiO lines is below the C CO 2 line and hence the oxide will be reduced by coke to produce CO 2 : 2NiO(s) + C(s) 2NI(l) + CO 2 (g) Below 400 K, it appears thatt the Ni NiO will be above the C CO 2 line so this reduction will not occur. Above 950 K, the C CO line is higher than the C CO 2 line so the reduction produces CO 2. (Any CO 2 produced would be reduced by C to produce CO.) Over what temperature range can ZnO be reduced by Fe? What other metal could be used instead to increase the temperature range in which metallic Zn was produced? The Zn ZnO line falls below the Fe FeO line at approximately 1450 K and reduction of ZnO by Fe will thus occur above this temperature. The Zn ZnO line is below the Ca CaO, Al Al 2 O 3 and Mn MnO lines over the entire temperature range of the diagram and so Ca, Al or Mn could be used to reduce ZnO at these temperatures. THIS QUESTION CONTINUES ON THE NEXT PAGE.

CHEM1901/3 2008-J-12 June 2008 Estimate the partial pressure of CO that would be expected at equilibrium in the smelting of ZnO by coke at 1500 K. 5 At 1500 K, C + ½O 2 CO Zn + ½O 2 ZnO lnk p (1) ~ 21 lnk p (2) ~ 14 Hence, for: ZnO + C Zn + CO K p = p(co) = K p (1) / K p p(2) or lnk p = ln(k p (1)/K p (2)) = lnk p (1) lnk p (2) = 21 14 = 7 K p ~ 10 3 and hence p(co) ~ 1000 atm Metallic copper is produced by smelting chalcopyrite, CuFeS 2 (s), directly in oxygen to produce iron oxides and SO 2. Write a balanced equation for this reaction, and sketch the lnk p versus temperature curve for Cu-CuO on the diagram on page 24. Clearly label the curve you have drawn. CuFeS 2 (s) + 5 /2O 2 (g) Cu(s) + FeO(s) + 2SO 2 (g) The lnk p curve for Cu CuO lies below Fe FeO at all temperatures. A sketch (in bold / red) is shown below.

CHEM1901/3 2008-J-14 June 2008 State the Second Law of Thermodynamics, and explain how an exothermic process in a closed system changes the entropy of the surroundings. 4 In any spontaneous process, the entropy of the universe always increases. A closed system can exchange heat or work with its surroundings. An exothermic reaction releases heat, q, into the surroundings: surr S = q / T > 0 The enthalpy of reaction defines the heat released into or absorbed from the surroundings at constant temperature and pressure. The standard enthalpy and entropy of solution of poly(oxyethylene) in water are H = 7.8 kj mol 1 and S = 31 J K 1 mol 1. Use these data to predict whether the solubility of poly(oxyethylene) in water increases or decreases when the solution is warmed. The surroundings receive 7.8 kj mol -1 from this exothermic reaction. Hence: surr S = (7.8 10 3 kj mol -1 ) / T As sys S = 31 J K 1 mol 1, the total change in entropy is: univ S = ( 31 + 7800/T) J K 1 mol 1 Spontaneous process occur when univ S is positive: At low T, the second term is large and dominates the first term. This leads to univ S > 0 and dissolution being favoured. At high T, the second term is small and the first term dominates. This leas to univ S < 0 and dissolution being unfavoured. Hence, the solubility decreases with temperature.

CHEM1901/3 2007-J-7 June 2007 Write a balanced equation for the combustion of methylhydrazine, CH 3 NHNH 2 (g). 6 CH 3 NHNH 2 (g) + 5 /2O 2 (g) CO 2 (g) + N 2 (g) + 3H 2 O(g) Using bond enthalpies, estimate the enthalpy of combustion of methylhydrazine. Bond enthalpies: Bond H / kj mol 1 Bond H / kj mol 1 C N 285 O H 464 N N 159 O=O 498 C H 416 C=O 806 N H 391 N N 945 H H H C N N H H H + 5 /2 O O O C O + N N + 3 H O H atom H(CH 3 NHNH 2 ) = 3 H(C-H) + H(C-N) + 3 H(N-H) + H(N-N) = (3 416) + (285) + (3 391) + (159) = 2865 kj mol -1 atom H(O 2 ) = H(O=O) = 498 kj mol -1 atom H(CO 2 ) = 2 H(C=O) = (2 806) = 1612 kj mol -1 atom H(N 2 ) = H(N N) = 945 kj mol -1 atom H(H 2 O) = 2 H(O-H) = (2 464) = 928 kj mol -1 comb H = atom H(reactants) atom H(products) = = [ atom H(CH 3 NHNH 2 ) + 5 /2 atom H(O 2 )] [ atom H(CO 2 ) + atom H(N 2 ) + 3 atom H(H 2 O)] = [(2865) + ( 5 /2 498)] [(1612) + (945) + (3 928)] = 1231 kj mol -1 Answer: 1231 kj mol -1 ANSWERS CONTINUES ON THE NEXT PAGE

CHEM1901/3 2007-J-7 June 2007 Liquid methylhydrazine and liquid oxygen can be used as a rocket fuel. Calculate the calorific value (in kj g 1 ) of this fuel. The molar mass of O 2 is (2 16.00) = 32.00. The molar mass of CH 3 NHNH 2 is: (12.01 (C)) + (6 1.008 (H)) + (2 14.01 (N)) = 46.078 To burn one mole of methylhydrazine requires 5 /2 moles of liquid O 2 to be carried so the combined mass is (46.078) + ( 5 /2 32.00) = 126.078. The calorific value is therefore: calorific value = H 1231 comb = = 9.8kJ g molar mass 126.078 1 Answer: 9.8 kj g -1 When methylhydrazine is used as a rocket fuel, the usual oxidant is dinitrogen tetroxide rather than liquid oxygen. Why? N 2 O 4 has a boiling point of 21 C, so there are no problems associated with handling cryogenic liquids. N 2 O 4 and methylhydrazine are hypergolic - the reaction occurs spontaneously on contact.

CHEM1901/3 2007-J-8 June 2007 Estimate the average temperature of Mercury given the solar power density at its surface of 9150 J m 2 s 1, and assuming an average albedo of 6% and zero Greenhouse effect. 2 With an albedo of 6%, 94% is not reflected. Hence, E out = 4πr 2 5.67 10-8 T 4 J s -1 E in = πr 2 9150 0.94 J s -1 These balance at a temperature: T = 4 2 ( π r ) (9150) (0.94) 2 8 (4π r ) (5.67 10 ) π = 441 K Answer: 441 K

CHEM1901/3 2007-J-10 June 2007 Use the figure below to help answer the following. 4 Write a balanced equation for the smelting of NiO by coke. In what temperature range will this process be spontaneous? NiO(s) + C(s) Ni(l) + CO(g) The reaction is spontaneous at temperatures where the C-CO line is above the Ni-NiO line. As marked on the diagram, this occurs at temperatures above ~620 K. Why are (a) aluminium and (b) tungsten not recovered from their oxides by smelting with coke? What alternative processes are used and why? Tungsten forms tungsten carbide if smelted with coke. The oxide is usually reduced using another reductant such as H 2. Aluminium is usually recovered electrolytically as smelting only occurs at temperatures well above 1800 K, which are not readily achieved.

CHEM1901/3 2007-J-13 June 2007 State the Second Law of Thermodynamics, and explain how the enthalpy of reaction is related to the entropy change of the surroundings. 5 In any spontaneous reaction, the entropy of the universe always increases. The enthalpy of reaction defines the heat released into or absorbed from the surroundings at constant temperature and pressure. The entropy change of the surroundings arises from the heat released. By definition, surr S = qsurr T As the heat from the reaction is lost to the surroundings, q surr = rxn H at constant P. Hence, surr S = H rxn T Formic acid HCOOH, can dimerise in the gas phase according to the reaction 2HCOOH (HCOOH) 2 with a standard enthalpy and entropy of dimerisation of H = 62 kj mol 1 and S = 150 J K 1 mol 1 respectively. Predict the temperature-dependence of the dimerisation reaction. universe S = sys S + surr S = sys S H rxn T Assuming that S and H do not change greatly with temperature, this becomes: universe S = (-150) 3 ( 62 10 ) = -150 + T T 3 (62 10 ) J K -1 mol -1 Dimerization is spontanteous if universe S > 0. This occurs when T < 3 (62 10 ) (150). Hence, dimerization is spontaneous at T < 413 K and the reverse process is spontaneous at T > 413 K. More dimer is formed at low temperature. Draw a structure that shows the bonding in the dimer. H-bond H O C O H H O C O H

CHEM1901/3 2006-J-7 June 2006 Write the equation whose enthalpy change represents the standard enthalpy of formation of hydrazine, N 2 H 4 (g). 4 N 2 (g) + 2H 2 (g) N 2 H 4 (g) Write the equation whose enthalpy change represents the enthalpy of combustion of hydrazine, N 2 H 4 (g) to produce water vapour. N 2 H 4 (g) + O 2 (g) N 2 (g) + 2H 2 O(g) Given the following data, calculate the standard enthalpy of formation of N 2 H 4 (g). H f (H 2 O(g)) = 242 kj mol 1 H comb (N 2 H 4 (g)) = 580 kj mol 1 Using o o o rxn f f H = m H (products) n H (reac tan ts), the enthalpy change for the combustion reaction is: o o o comb f 2 f 2 4 H = [2 H (H O(g))] [ H (N H (l))] as the remaining Hence: compounds are elements in their standard states. o 1 f 2 4 [2 242] [ H (N H (l))] = 580kJ mol o 1 f H (N2 H 4(l))] 96kJ mol = + Answer: +96 kj mol -1

CHEM1901/3 2006-J-12 June 2006 State the Second Law of Thermodynamics, and explain how the enthalpy of reaction is related to the entropy change of the surroundings. 5 The entropy of the universe always increases: univ S = sys S + surr S > 0 The surroundings act as a reservoir of heat for the system: heat gained or lost in a reaction is taken or given to the surroundings, respectively. As sys H = -q surr at constant pressure: surr S = q surr T = - sys H T Give an example of a chemical reaction or a chemical process that corresponds to each of the following. S > 0, H > 0 As surr S = - sys H < 0, such processes will only occur if the temperature is T sufficiently high that surr S is smaller in magnitude than sys S. Examples include: Melting of ice at 300 K: H 2 O(s) H 2 O(l) Boiling of water at 400 K: H 2 O(l) H 2 O(g) NH 4 NO 3 (s) NH 4 + (aq) + NO 3 (aq) S > 0, H < 0 As surr S = - sys H > 0 and sys S > 0, such processes will occur at all T temperatures. Examples include: Burning of ethane: 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) CaCl 2 (s) Ca 2+ (aq) + 2Cl (aq) S < 0, H < 0 As surr S = - sys H > 0, such processes will only occur if the temperature is T sufficiently low that surr S is larger in magnitude than sys S. Examples include: Freezing of ice at 263 K: H 2 O(l) H 2 O(s) Condensing of steam at 300 K: H 2 O(g) H 2 O(l)

CHEM1901/3 2005-J-9 June 2005 In an experiment, 50.0 ml of 1.00 M HNO 3 was combined with 50.0 ml of 0.540 M NaOH in a calorimeter. Give an equation for the reaction that took place. 4 H 3 O + (aq) + OH (aq) 2H 2 O(l) The temperature of the solution was found to increase by 2.98 C. If the heat capacity of the calorimeter was 80.0 J K 1 and the heat capacity of the final solution was 426 J K 1, determine the molar heat of reaction. As q = C ΔT for both the calorimeter and the solution, q = ((80.0 2.98) J) + ((426 2.98) J) = 1510 J As the reaction is a 1:1 reaction and fewer moles of OH are present, it is rate limiting. The number of moles of OH is: number of moles of OH = concentration volume = 0.0500 L 0.540 mol L -1 = 0.027 mol This number of moles generates 1510 J of heat. The reaction is exothermic as the temperature increases. The enthalpy change for one mole is therefore: Δ r H = 1510 J 0.027 mol = 55800 J mol-1 = 55.8 kj mol -1 Answer: 55.8 kj mol -1 The average bond enthalpy of the O-H bond is 463 kj mol 1. Explain briefly why the heat of neutralisation calculated in the first part of this question differs significantly from this value. The bond enthalpy refers to half the enthalpy change for the reaction H 2 O(g) O(g) + 2H(g) rather than the transfer of a proton from H 3 O + to OH in aqueous solution.

CHEM1901/3 2004-J-9 June 2004 Aluminium metal is a very effective agent for reducing oxides to their elements. For example, it is used as a component of the solid fuel in the space shuttle, and in the thermite reaction shown in lectures: Fe 2 O 3 (s) + 2Al(s) Al 2 O 3 (s) + 2Fe(s) Write a balanced equation for the reduction of SiO 2 (s) to silicon by Al(s). 4 3SiO 2 (g) + 4Al(s) 2Al 2 O 3 (s) + 3Si(s) Given the following thermochemical data, evaluate the enthalpy change per gram of reactants for the SiO 2 and Fe 2 O 3 reactions above. Compound H f (kj mol 1 ) Fe 2 O 3-821 Al 2 O 3-1668 SiO 2-877 Using Δ rxn H = Σ mδ f H (products) - Σ mδ f H (reactants), for the SiO 2 reaction with Al: Δ rxn H = [2Δ f H (Al 2 O 3 )] [3Δ f H (SiO 2 )] = {[2-1668] [3-877]} kj mol 1 = -705 kj mol -1 This is the enthalpy change for 3 mol of SiO 2 and 4 mol of Al. The combined mass of the reactants in these amounts is: mass of reactants = (3 (28.09 (Si) + 2 16.00 (O)) + 4 26.98 (Al)) g mol -1 = 288.16 g mol -1 The enthalpy change per gram of reactants is therefore: 705 kj mol 1 enthalpy change = 288.16 g mol 1 = -2.45 kj g-1 For the Fe 2 O 3 reaction with Al: Δ rxn H = [Δ f H (Al 2 O 3 )] [3Δ f H (Fe 2 O 3 )] = {[-1668] [-821]} kj mol -1 = -847 kj mol -1 This is the enthalpy change for 1 mol of Fe 2 O 3 and 2 mol of Al. The combined mass of the reactants in these amounts is: mass of reactants = ((2 55.85 (Fe) + 3 16.00 (O)) + 2 26.98 (Al)) g mol -1 = 213.66 g mol -1 ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2004-J-9 June 2004 The enthalpy change per gram of reactants is therefore: 847 kj mol 1 enthalpy change = 213.66 g mol 1 = -3.96 kj g-1 Answer (SiO 2 ): -2.45 kj g -1 Answer (Fe 2 O 3 ): -3.96 kj g -1 Which set of reactants would make the better rocket fuel on the basis of most energy provided per mass of fuel (i.e. biggest bounce per ounce )? Fe 3 O 3 /Al is better fuel than SiO 2 /Al

CHEM1901/3 2003-J-2 June 2003 Use the thermochemical data provided to calculate the heat of reaction of the following reaction: PCl 3 (l) + Cl 2 (g) PCl 5 (s) Data: P 4 (s) + 6Cl 2 (g) 4PCl 3 (l) H = 1280 kj mol 1 P 4 (s) + 10Cl 2 (g) 4PCl 5 (s) H = 1774 kj mol 1 2 The reactions given in the data correspond to the enthalpy of formation of PCl 3 (l) and PCl 5 (s). The reactions as written correspond to the formation of 4 moles of PCl 3 (l) and of 4 moles of PCl 5 (s). The f H o values for these two substances are therefore: f H o (PCl 3 (l)) = (-1280 ¼) kj mol -1 f H o (PCl 5 (l)) = (-1774 ¼) kj mol -1 As Δ r H = Σ mδ f H (products) - Σ mδ f H (reactants) and f H o (Cl 2 (g)) = 0, Δ r H = ((-1774 ¼) ((-1280 ¼)) kj mol -1 = -124 kj mol -1 ANSWER: 124 kj mol 1 Identify one property of a molecule necessary for it to be considered a greenhouse gas. 1 To be a greenhouse gas, a molecule must be able to absorb infrared light: it must undergo a vibration which causes a change in the dipole moment.

CHEM1901/3 2003-J-6 June 2003 The normal boiling point of chloroform is 61.7 C and its enthalpy of vaporisation is 31.4 kj mol 1. Calculate the entropy of vaporisation for chloroform at 1 atm. 2 The entropy change for vaporisation is: Δ vap S = vaph T vap = 31.4 103 J mol 1 61.7+273.0 K = 93.8 J K-1 mol -1 Answer: 93.8 J K -1 mol -1

CHEM1901/3 2003-J-10 June 2003 If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent decomposition by the reaction 6 Ag 2 CO 3 (s) Ag 2 O(s) + CO 2 (g) The enthalpy change, H, for this reaction is 79.14 kj mol 1 in the temperature range of 25 to 125 C. Given that the partial pressure of CO 2 in equilibrium with solid Ag 2 CO 3 is 8.20 10 6 atm at 25 C, calculate the partial pressure of CO 2 necessary to prevent decomposition of Ag 2 CO 3 at 110 C. Assume that S does not change over this temperature range. As Ag 2 CO 3 and Ag 2 O are present as solids, they are not involved in the equilibrium expression which is simply, K p = pco2 = 8.20 10-6 atm = (8.20 10-6 1.01325 10 5 ) Pa = 0.831 Pa at 298 K. This is related to universe S by: universe S = RlnK p = (8.314 J K -1 mol -1 ) ln(0.831) = -1.54 J K -1 mol -1 = rxn S - rxn H T Hence, rxn S = 264 J K -1 mol -1. o = { rxn S - 79.14 10 298 3 } J K -1 mol -1 If rxn H and rxn S are the same at 110 C (378 K), universe S becomes: universe S = rxn S - rxn H T = 57.4 J K -1 mol -1 o = {264-79.14 10 383 3 } J K -1 mol -1 The equilibrium constant at 110 C (383 K) is then: universe K p = exp( S ) R = exp( 57.4 J K 1 mol 1 8.314 J K 1 mol 1) = 992 Pa = 0.00979 atm = P CO 2 (K p increases with temperature for an endothermic reaction.) Answer: 0.00979 atm