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Introductory Chemistry B CH4751 Lecture Notes 11-20 Dr. Erzeng Xue

CH4751 Lecture Notes 11-20 (Erzeng Xue) . Introductory Chemistry B CH4751 Lecture Notes 11-20 Dr. Erzeng Xue. CH4751 Lecture Notes 11 (Erzeng Xue) . Chemical Reactions. Chemical Reaction - Observation. Reaction (1) CH 4 + 2O 2 ® CO 2 + 2H 2 O

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Introductory Chemistry B CH4751 Lecture Notes 11-20 Dr. Erzeng Xue

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  1. CH4751 Lecture Notes 11-20 (Erzeng Xue) Introductory Chemistry B CH4751Lecture Notes 11-20Dr. Erzeng Xue

  2. CH4751 Lecture Notes 11 (Erzeng Xue) Chemical Reactions Chemical Reaction - Observation Reaction (1) CH4 + 2O2® CO2 + 2H2O Reaction (2) CH4+ CO2® 2CO + 2H2 When carrying out these reactions we found that • at 400K (123°C), the reaction (1) will proceed and reaction (2) will not • at 1000K(723°C), reactions (1) & (2) both proceed; but • rxn (1) can go complete (until CH4 orCO2consumed completely) • rxn (2) won’t complete (with a feed CH4=CO2=1 & CO=H2=0, max. conv.=63% at 1000K) • The reaction (1) will give out heat, but the reaction (2) will require heat. Why?

  3. CH4751 Lecture Notes 11 (Erzeng Xue) Chemical Reactions Chemical Reaction Thermodynamics • Each molecule contains certain types and quantity of chemical energy • There is always energy change In a chemical reaction because of • breaking / reformation of chemical bonds • out-giving or in-taking heat • There are different energies associated with a substances & a reaction (A systematic study of various forms of energy & their changes is called Thermodynamics) We will learn some of these energies • The meanings • How to get values / do simple calculate • How to use them as a tool to study chemical reactions

  4. CH4751 Lecture Notes 11 (Erzeng Xue) refers to standard pressure (1 atm.) Temperature Chemical Reactions Chemical Reaction Thermodynamics • The heat of formation, H, (also called Enthalpy of Formation or Enthalpy) • H is an energy associated with heat • H is specific for each substance and is dependent of temperature & pressure e.g. at 1000K: H°CH4=-89, H°O2=0, H°CO2=-394, H°H2O=-241, H°CO=-111, H°H2=0 (kJ/mol) (H values for various substances can be found in physical chemistry/Chem Eng handbooks) • In a reaction we are interested in the enthalpy change, DH, which is calculated using For Rxn(1) CH4 + 2O2® CO2 + 2H2ODH°1000=-801 kJ/mol Rxn (2) CH4+ CO2® 2CO + 2H2DH°1000=+260k J/mol • The meaning • When DH<0, a reaction releases heat Þ reaction is exothermic, as in rxn (1) • When DH>0, a reaction requires heat Þ reaction is endothermic, as in rxn (2)

  5. CH4751 Lecture Notes 11 (Erzeng Xue) reaction can proceed (but we don’t know how fast it will be!) reaction at equilibrium (no further change possible-‘dead’ state) reaction will NOT proceed (or can proceed backward!) for a reaction at constant T, P, Chemical Reactions Chemical Reaction Thermodynamics • The Gibbs Free Energy, G, (also called Free Energy) • G is a thermodynamic function related to a reaction. It is a function of H, T & S (entropy) • G is specific for each substance & is a function of H, T & S (entropy) e.g. at 1000K: G°CH4=-+30, G°O2=0, G°CO2=-395, G°H2O=-192, G°CO=-200, G°H2=0 (kJ/mol) (G values for various substances can be found in physical chemistry/Chem Eng handbooks) • The Gibbs Free energy change, DG, in a reaction can be calculated using For Rxn(1) CH4 + 2O2® CO2 + 2H2ODG°400 »DG°1000=-801 kJ/mol Rxn (2) CH4+ CO2® 2CO + 2H2DG°400=+145, DG°1000=-24 kJ/mol • Use of DG - Rxn(1) DG <0 at 400 & 1000K-spontaneous, Rxn(2) DG <0 at 400K, will not proceed

  6. CH4751 Lecture Notes 11 (Erzeng Xue) Chemical Reactions Chemical Reaction Thermodynamics Example of DH°T calculation CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)CH4+ CO2® 2CO + 2H2 Coeff. 1 2 1 2 1 1 2 2 H°400( -77 0 -393 -242 -77 -393 -110 0 kJ/mol H°1000 -89 0 -394 -248 -89 -394 -111 0 kJ/mol Equation to use DH°400 =[1x(-393)+2x(-242)]-[1x(-77)+2x(0)]= -800 kJ/mol DH°1000 =[1x(-394)+2x(-248)]-[1x(-89)+2x(0)]= -801 kJ/mol Reaction (1) DH°400 =[2x(-110)+2x(0)]-[1x(-77)+1x(-393)]=+250 kJ/mol Reaction (2)DH°1000=[2x(-111)+2x(0)]-[1x(-89)+1x(-393)]= +260 kJ/mol Note: The heat of formation of single element gases (O2, H2, N2 etc) is defined as zero.

  7. CH4751 Lecture Notes 11 (Erzeng Xue) Chemical Reactions Chemical Reaction Thermodynamics Example of DH°T calculation CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)CH4+ CO2® 2CO + 2H2 Coeff. 1 2 1 2 1 1 2 2 G°400( -42 0 -394 -224 -42 -394 -146 0 kJ/mol G°1000 +19 0 -396 -193 +19 -396 -200 0 kJ/mol Equation to use DG°400 =[1x(-394)+2x(-224)]-[1x(-42)+2x(0)]= -800 kJ/mol DG°1000 =[1x(-396)+2x(-193)]-[1x(+19)+2x(0)]= -801 kJ/mol Reaction (1) DG°400 =[2x(-146)+2x(0)]-[1x(-42)+1x(-394)]=+144 kJ/mol Reaction (2)DG°1000=[2x(-200)+2x(0)]-[1x(19)+1x(-396)]= -23 kJ/mol Note: The Gibbs Free Energy of single element gas (O2, H2, N2 etc) is defined as zero.

  8. CH4751 Lecture Notes 11 (Erzeng Xue) Chemical Reactions Chemical Reaction Thermodynamics • The values of DG°T and DH °T Equations In both cases G° and H° values for the reactants and products have to be those at the reaction temperature, indicated by the subscript. • For common substances, G° and H° values are given as a function of T in handbooks - okay • For some less common substances, you may only find values at 298K, G°298 and H°298 How do you convert values of G°298 and H°298 to those of G°T and H°T ? Here is the equations you can use to calculate the values of G°T and H°T from G°298 and H°298 in which, S°T is the entropy and Cp is the heat capacity at constant pressure

  9. CH4751 Lecture Notes 11 (Erzeng Xue) reaction can proceed (but we don’t know how fast it will be!) reaction at equilibrium (no further change possible-‘dead’ state) reaction will NOT proceed (or can proceed backward!) for a reaction at constant T, P, Chemical Reactions Chemical Reaction Thermodynamics • Summary • Will a reaction proceed in the direction specified? • Check DG°T value of the reaction. The DG°T value of a reaction can be calculated by The G° values of reactants / products can be found in literature. Remember • Is a reaction exothermic or endothermic? • Check DH°T value of the reaction. The DH°T value of a reaction can be calculated by The H° values of reactants / products can be found in literature

  10. CH4751 Lecture Notes 12 (Erzeng Xue) [NH4]2 [NH4]1 [NH3]1 [NH3]2 t Chemical Reactions Chemical Reaction Equilibrium • Until now we assume reaction A + B ® C + D goes to complete • Meaning a reaction only stops when either A or B is consumed completely • Experimental observations • Some reactions will ‘cease’ without complete consumption of limiting reactant • Without altering reaction conditions (T, P, [ ] etc.) the ratio of conc. remains constant • After a change (T, P, [ ] etc.), the ratio of conc’s changes to another constant value Example 1: NH3(aq)+ H2O(l) D NH4+(l) + OH-(aq) Follow concentrations of each component with time at constant T and P, at t® ¥ • After changing T, a new constant ratio is established • If [NH3] is reduced the amount [NH4] decrease accordingly while the ratio above remains constant • We say the reaction has reached equilibrium state

  11. CH4751 Lecture Notes 12 (Erzeng Xue) NO2 NO O2 t Chemical Reactions Chemical Reaction Equilibrium • Experimental observations Example 2: 2NO(g) + O2(g)D 2NO2(g) Again at constant T and P, when t® ¥ • Reaction equilibrium is achieved when t® ¥ • We say the reaction has reached equilibrium state • The ratio of product concentration to reactants, with the stoichiometry coefficient as the power index, is called reaction quotient • When reaction quotient = constant value Þ reaction reaches equilibrium • The value of reaction quotient at equilibrium, is called equilibrium constant, Keq • At equilibrium, reactants may or may not be consumed completely e.g. A feed gas mixture: NO=500ppm, O2=10%, N2=89.95%, achieves equilibria at the following T’s Temperature / °C 50 325 500 NO remaining at equil / ppm 0 96.5 390 NO conversion at equil / % 100 80.3 22

  12. CH4751 Lecture Notes 12 (Erzeng Xue) NO2 NO O2 t Chemical Reactions Chemical Reaction Equilibrium • Important concepts of reaction equilibrium 2NO(g) + O2(g) D 2NO2(g) • Is the reaction between reactants still going on? YES. Reaction goes forward as well as reverses. At equilibrium: Rforward = Rreverse though there is no NET change of all conc.’s • The equilibrium constant, Keq, has a meaning of Keq=Rforward / Rreverse • Changing T causes both Rforward & Rreverseto change, leading to a new Keq. • If Keq >>1, which means Rforward >> Rreverse, thereaction tends to go forward If Keq <<1, which means Rforward << Rreverse, thereaction tends to go backward • Will there be an equilibrium constant for reactions that go complete? YES. There is Keq for all reactions (at a constant P) once T and conc’s are fixed. For a reaction that tends to go complete, Rforward >> Rreverse (Keq >>1).

  13. CH4751 Lecture Notes 12 (Erzeng Xue) reaction is spontaneous reaction at equilibrium (no further change possible) reverse reaction is spontaneous for a reaction at constant T, P, Chemical Reactions Chemical Reaction Equilibrium • Equilibrium constant and Gibbs Free Energy For reaction vAA + vBB DvCC + vDD Remember: The value of DG° determines the direction of reaction No more change possible º reaction in equilibrium º DG°= 0 • Is the DG value related to the equilibrium constant? YES. DG° and Keq are related by the equation below (calculate one from the other) DG°T= - RTln(Keq)

  14. CH4751 Lecture Notes 12 (Erzeng Xue) for liquid phase rxn for gas phase rxn Chemical Reactions Chemical Reaction Equilibrium • Equilibrium constant - for different type of rxns General form: vAA + vBB DvCC + vDD • gas phase 2NO(g) + O2(g) D 2NO2(g) • gas-solid phase CaCO3(s) D CaO(s)+CO2(g) • liquid phase NH3(aq)+H2O(l) D NH4+(l)+OH-(aq) • liquid-solid Cu(OH)2(s) D Cu2+(aq)+2OH-(aq) • gas-liquid NH3(g)+H2O(l) D NH4OH(aq) • For reactions that have gas components, we normally use pressure to represent the conc’s • For reactions involves gas+liquid or gas+solid, only gas terms appear in the Keq expression

  15. CH4751 Lecture Notes 12 (Erzeng Xue) Chemical Reactions Factors Affecting Reaction Equilibrium • Le Chatelier’s Principle “When a system in equilibrium is subjected to an external stress, the system will establish a new equilibrium, when possible, so as to minimise the external stress” Stresses: Changes in [ ], temperature or pressure Example: N2(g) + 3H2(g) D 2NH3(g) + heat (exothermic) a) Effect of ([ ]). Increasing [ ] of substance shifts equil. in the direction of the long arrow N2 + 3H2 2NH3 + heat N2 + 3H2 2NH3 + heat N2 + 3H2 2NH3 + heat b) Effect of heat. Addition or removal of heat at constant temperature Addition of heat: N2 + 3H2 2NH3 + heat c) Effect of Pressure. (only affects reactions that have volume change before & after). Increase in pressure: N2 + 3H2 2NH3 + heat 4 volumes ® 2 volumes

  16. CH4751 Lecture Notes 12 (Erzeng Xue) Chemical Reactions Chemical Reaction Equilibrium • Write the equilibrium expression for the following reactions and determine the units for Keq: 1)2O3(g)D 3O2(g) 2)Ag+(aq) + 2NH3(aq)D Ag(NH3)2+(aq) 3)2NaN3(s)D 3Na(s) + 3N2(g) 4)2Na(s) + Cl2(g)D2NaCl(s) 5)2NaCl(s)D 2Na(s) + Cl2(g) 6)N2(g) + 3H2(g)D 2NH3(g) Note 1. The Keq expression depends on how the rxn equation is written (compare rxns 4&5). 2). The unit of Keq depends on the way how the rxn eqn is written & the unit used of each.

  17. CH4751 Lecture Notes 12 (Erzeng Xue) Chemical Reactions Chemical Reaction Equilibrium 1. Analysis shows that a mixture of N2 (2.46 atm), H2 (7.38 atm) and NH3 (0.116 atm) at 472°C in reaction (N2(g) + 3H2(g) D 2NH3(g)) is in equilibrium state. Calculate: 1) Keq; 2) DG°; 3). The total pressure. 4) Will the rxn be push to the product by decreasing the reaction pressure? Give reason why? 5) Will the removal of NH3 from reaction mixture promote the product formation? Explain why. 1) 2) DG°T= - RTln(Keq)=-8.314x(472+273)ln(2.79x10-5)=65 kJ/mol 3) Ptotal= PN2+PH2+ PNH3=2.46+7.38+0.116=9.956 atm 4) A decrease reaction P favours the reverse rxn because Volreactant> Volproduct. 5) Yes. As Keq=P2NH3/(PN2xP3H2)=constant, the removal of NH3 will reduce PNH3, to compensate the change, more N2 and H2 will be converted to NH3 in order to keep the same Keq (reaction quotient).

  18. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction • What we know about a chemical reaction so far • Reaction equation - quantitative representation of a chemical reaction • Reaction stoichiometry coefficients and balancing reaction equations • We can judge if a give chemical reaction would proceed in the direction specified • A reaction has a tendency if DG of a reaction is smaller than zero • We can decide if a given reaction gives out or takes up heat • If DH < 0, reaction is exothermic; if DH > 0, reaction is endothermic • For reactions that are feasible, to what extent they will complete • Chemical reaction equilibrium, equilibrium constant • Now we know a rxn would proceed in the direction specified, questions are: - how fast that reaction is going to proceed under give conditions? - how to quantitatively describe the rate of a reaction and make comparison? - how can we explain that some reactions occur faster than others? e.g. 2NO(g) + O2(g) = 2NO2(g) (slow); 2NaN3 = 2Na + 3N2 (very fast)

  19. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction Kinetics • Chemical reaction kinetics study the rate of chemical reactions • Definition of chemical reaction rate • The number of moles of a reactant converted (consumed) in a reaction per unit time for a reaction A + B ® C + D (mol/s) • When we say rate we always refer to ONE of the components in the reaction • The minus sign refers to that the concentration of reactant decreases in reaction • Reaction rate equation (or kinetic equation) • Many forms exist • The most common one where k reaction rate constant A0 pre-exponential factor Ea reaction activation energy a,b,l,d reaction orders with respect to A, B, C, D, respectively R gas constant T reaction temperature in Kelvin scale Arrhenius equation Kinetic parameters

  20. CH4751 Lecture Notes 13 (Erzeng Xue) Ea reactant energy product reaction process Chemical Reactions Chemical Reaction Kinetics • Meanings of kinetic parameters, k, A0, Ea, a, b, l, d • Reaction rate constant, k • It tells how fast a reaction can occur • It is a constant dependent on temperature but independent of concentrations • Pre-exponential factor, A0 • It refers to the frequency of collision between molecules, the higher frequency, the faster rxn • Reaction activation energy, Ea • It can be understood as the energy barrier for a reaction to overcome • The higher Ea value is, the more difficult for a reaction to occur • Reaction orders w.r.t. each component, a, b, l, d • The magnitude of these values reflects the effectiveness of each component in the reaction • The values of a, b, l, d can be positive or negative or zero • The values of a, b, l, d can be integrals or fraction • All these kinetic parameters have to be determined experimentally

  21. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction Kinetics • Factors affecting the rate of a chemical reaction • Reaction temperature, T • An increase T will lead to increasing k, thus reaction rate. • The dependence of k on T is given by differentiating k expression, the higher Ea value is, the more significant of the effect of increasing T on the reaction rate • Concentration of reactants / products • The effect of increasing a concentration is positive if the respective order is positive • The larger the value of order is the stronger the effect of increasing conc on the rate • When order equals to zero, there is no effect of concentration on the rate. • The presence of a catalyst • A catalyst can alter reaction rate (speeding up desired rxns or slowing down undesired rxns) Note: we assume rate of mass transfer (to meet) is sufficient high comparing to rA.

  22. CH4751 Lecture Notes 13 (Erzeng Xue) mass transfer mass transfer reaction reaction Chemical Reactions Chemical Reaction Kinetics • The reaction rate and mass transfer rate When we discuss the reaction rate, it only makes sense if there are sufficient number reactant molecules can be delivered to the reaction site. • We say a reaction is kinetic control when the rate of mass transfer > the rate of rxn rA. • This means that molecules being transported to the reaction site are ‘queuing’ for reaction • If the mass transfer rate is slower than the reaction rate, the overall rate we observed will be the rate of mass transfer, not the reaction rate - diffusion control • This means that molecules are waiting to be delivered before reacting ‘queuing’ for reaction • The concept of rate determining step (r.d.s.) • The slowest step in a reaction process determine the overall rate of a reaction

  23. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction Kinetics • Catalysis and catalysts Catalyst is a substance which can alter reaction rate without itself being destroyed or consumed(many other definitions and this is one of them) • 95% of chemical industries apply one or more catalysts in their processes e.g. polymerisation, air/water depolution, ammonia synthesis, cracking heavy oil to LPG, etc • A catalyst can be an acid, a base; can be a liquid or a solid. Most industrial catalysts are metals, metal oxides or a mixture of them formulated & made in special ways • Use of catalysts in industry • Speeding up desired reactions thus increase the process output • Slowing down undesired reaction thus reduce the unwanted waste products • Altering reaction route by changing the relative speed of certain steps in a reaction network therefore realising certain products which would not be possible without catalysts. • Allowing some rxns to occur under mild conditions e.g. working with heat sensitive materials • Enzymes are catalysts that participate in bio-active processes • etc

  24. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction Kinetics • Calculation of reaction rate Example: a gas phase reaction 2N2O5=4NO2+O2 occurs at 300°C. The concentrations of N2O5 found in the reaction mixture at different time intervals are given below: t h 0 1 2 3 5 7 9 [N2O5] mol/L 1.40 1.07 0.80 0.58 0.34 0.18 0.09 Calculate the rxn rates w.r.t. N2O5, NO2 & O2 1) betw. 0-1h; 2) betw. 3-5h; 3) average betw 0-9h. N2O5 consumption rate NO2 formation rate O2 formation rate Eqn’s to use: 1) 0-1 h 2) 3-5h 3) aver.

  25. CH4751 Lecture Notes 13 (Erzeng Xue) Chemical Reactions Chemical Reaction Kinetics • More about reaction rate 2N2O5 = 4NO2 + O2 1) 0-1 h rN2O5=0.33 rNO2=0.66 rO2=0.165 mol/Lh 2) 3-5h rN2O5=0.12 rNO2=0.24 rO2=0.06 mol/Lh 3) aver. rN2O5=0.146 rNO2=0.291 rO2=0.073 mol/Lh • For the same reaction, the reaction rate expressed by different components varies with their stoichiometry coefficients • rN2O5 : rNO2 : rO2 = 2 : 4 : 1 or • Given reaction rate for one of the rxn components you should be able to calc others. • For the same reaction the reaction rate may vary with the time • because of change of reactant conc’s with time & rate in general is proportional to [ ]’s. • When rxn orders w.r.t. reactant > 0 (usually they are) the rxn rate rbeginning > rlater • As reaction rate is a function of temperature, the determination of reaction of reaction rate must be done at a constant temperature (you may need to determine the rate at a different T, or you may need to vary temperature to determine such as Ea • Don’t forget to put the correct unit to the reaction rate you determined

  26. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 1.How many grams of water are produced in the oxidation of 1.0g of glucose, C6H12O6? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O Step 1: Use molar mass of glucose to convert g to moles 1 mole C6H12O6=6x12(C)+12x1(H)+6x16(O)=180g/mol number of moles C6H12O6=1.0g x (1mol/180g)=5.55x10-3 mol Step 2: Use balanced equation to determine no. of moles of H2O produced 1 mole C6H12O6 produces 6 moles H2O the no. of moles of H2O produced: 5.55x10-3 moles C6H12O6x6=0.033 moles H2O Step 3: Convert moles of H2O to grams using molar mass 1 mole of H2O=2x1(H)+1x16(O)=18 g/mol Grams of H2O produced: 0.033 mol of H2Ox(18g/mol)x = 0.6g H2O (answer) Note: You cannot use the weight directly in the calculation. It has to be converted to moles.

  27. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 2. In a reactor one put 180g of glucose (C6H12O6) and 160g of O2. Can you produce 108g of H2O. Why? What is the maximum amount of H2O which can be produced? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O Step 1: Convert all components from grams to moles: No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6 No.moles of O2= 160/32g/mol=5 mole of O2 No.moles of H2O= 108/18g/mol=6 mole of O2 Step 2: Find out how much glucose AND O2 you need to produce 108g H2O. To produce 108g which is 6 moles of H2O, you will need 1mole glucose AND 6 moles of O2. Do we have enough glucose? - Yes. Do we have enough O2? - No. Step 3: Every 6 molecules of O2 will burn 1 molecule of glucose, this will proceed UNTIL one of the reactant consumed completely, in this case O2. When all O2 is consumed the reaction will stop and the max. amount of H2O which can be produced can be calculated from O2 available: 5moles of O2 gives 5moles (or 90g) of H2O Note: When one of reactants is consumed completely the reaction will stop.

  28. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 3. As in Question 2, one puts 180g of glucose (C6H12O6) and 160g of O2. When O2 is completely consumed, what is the glucose left & what is the percentage conversion of glucose? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O Step 1: Convert all components from grams to moles: No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6 No.moles of O2= 160/32g/mol=5 mole of O2 No.moles of H2O= 108/18g/mol=6 mole of O2 Step 2: Find out how much glucose left after all O2 has consumed. The molar ratio of glucose and O2 in the reaction=1:6. For a consumption of 5 moles of O2, the amount glucose reacted will be 1x5/6=5/6 moles or 0.83 moles, or 0.83x180=150g. The amount glucose left over=1-0.83=0.167moles or 180-150=30g Step 3: The percentage conversion of glucose at complete conversion of O2 (try the weight base) Note: The conversion (%) calculated based on moles is the same as that based on weight .

  29. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 4. The brown gas NO2 can form colorless gas N2O4, 2NO2D N2O4. At 25°C the concentrations of NO2 & N2O4 are 0.018 M & 0.055 M respectively when at equilibrium. 1) Calculate the equilibrium constant Keq at 25°C. 2) If in another equilibrium system of the same gases at the same temperature, the NO2 concentration is found to be 0.08 M, what is the concentration of N2O4? Step 1: Determine the equilibrium constant From equilibrium constant definition: Step 2: When at equilibrium

  30. CH4751 Lecture Notes 14 (Erzeng Xue) ln k slope= -12750 1/T Chemical Reactions Chemical Reaction Calculations Question 5. The rate constants of a reaction are determined to be 3x10-5 mol/L.h at 200°C and 4x10-4 mol/L.h at 250°C. Estimate the reaction activation energy. Arrhenius eqn relates the rate constant to activation energy Mthd.1: lnk=lnA0+(-Ea/R)(1/T), A plot of lnk against 1/T will produce a straight line, the slope of which is -Ea/R. So that Ea=slope x R Ea=-12750 x 8.314=106,000 J/mol = 106 kJ/mol Mthd 2: Let A0,1=A0,2 T1=273+200 K, T2=273+250 K, k1=3x10-5 & k2=4x10-4 mol/L.h, R=8.314 JK/mol

  31. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 6. Two catalysts A & B are compared for their catalytic activity for reaction R®P. When A is present it takes 10s for R to change from 2 to 0.5 moles and when B is present it takes 20s for R to decrease from 5 to 2 moles at the same temperature and with the quantities of catalyst. Which catalyst is more active for the reaction concerned? Answer: The activity of two catalysts can be compared based on the average reaction rate when A & B presence separately. The A catalyst is more active for the reaction concerned.

  32. CH4751 Lecture Notes 14 (Erzeng Xue) 1/[A] slope=k t Chemical Reactions Chemical Reaction Calculations Question 7. Verify that the rate constant of a reaction following second order rate law rA=-k[A]2 can be determined from the slope of a line obtained by plotting 1/[A]t against reaction time t, where [A]t is the concentration of A measured at time t. Answer: Second order rate law: (1) rearrange: (2) Define boundary conditions: at t=0, [A]=[A]0 and at t=t, [A]=[A]t integrate eqn (2), t from 0-t and [A] from [A]0 to [A]t (3) compare eqn (3) with linear eqn Y=aX+B, which is a straight line with slope a Let A plot of vs. t will give a straight line with slope=k.

  33. CH4751 Lecture Notes 14 (Erzeng Xue) Chemical Reactions Chemical Reaction Calculations Question 8. Reaction R®P follows the second order rate law rR=-k[R]2. Verify that the time required for the reactant R to fall to a half of its initial value is t1/2=1/(k[R]). Answer: Second order rate law: (1) After integration of eqn (1) with the boundary conditions: at t=0, [R]=[R]0 & at t=t1/2, [R]=[R]t1/2=0.5[R]0

  34. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Homogeneous and Heterogeneous Reactions • In a chemical reaction, the reactants can be in various physical states • Homogeneous - All of reactants & products in the same phase & no phase boundary • Heterogeneous - Involving multi-phases & phase boundary crossing * When two immiscible liquid, such as oil and water is regarded as heterogeneous type.

  35. CH4751 Lecture Notes 15 (Erzeng Xue) gas condensation evaporation sublimation deposition liquid Energy level freezing melting solid water vapour liquid water evaporating Temperature melting ice Energy (heat) added Chemical Reactions Phase & Phase Change • A substance can exist in different physical states • Gas / vapour • Liquid • Solid Note: Some other states may sometime mentioned. such as liquid-crystal, super-critical state, gel. etc. • The temperatures at which phase changes occur vary with substances and circumstances (e.g. P) • The energy required for phase change varies with substances and type of phase change. • The energy possessed by molecules of the same substance at different state are different

  36. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Reactions In Liquid Phase - Solution • Solution = solute + solvent • Solute is a solid or a gas - solute is dissolved in solvent (e.g. NaCl + H2O, O2 + H2O) • Solute is another liquid - solute and solvent are miscible (e.g. C2H6O + H2O) • solute in a solution can exist as molecules or ions, or both (such as weak acid) • some solutes are dissolved in a solvent in any proportions (e.g. C2H6O in water); others are only dissolved in a solvent in certain proportion - solubility limitation (e.g. NaCl or N2 in H2O) • Concentration of a solution - the amount of solute in the solution • Molar concentration (molarity) - number of moles solute in ONE litre solution • Molarity is the most commonly used concentration unit in chemistry • Weight percentage (wt.%) of solute in solution definition: solute wt%=100% x (wt of solute)/(wt. of solute + wt. of solvent) e.g. A solution contains 20g solute & 30g solvent solute wt%=100% x 20/(20+30)=40wt% (the wt% of solvent =100%-40%=60%) solute - substance that is dissolved solvent - dissolving medium

  37. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Reactions In Liquid Phase - Solution • Some important notes on the chemical reaction in solution • When a solute dissolve in solution, solute can either be present as molecules or ions • as molecules. e.g. dissolving sugar in water - not electronic conductivity. • as ions. e.g. dissolving salt (NaCl) in water - Na+ & Cl- both conduct electricity. • Strong acids or base, when dissolved in H2O, form only ions in solution • Weak acid or base, when dissolved in H2O, form mixture of molecules and ions (partial dissociatn) • The solubility of some gas solutes, when dissolved in a solvent, depends on the pressure of the solute gas above the solution ideal solution: Pi=xiPi* or xi=Pi / Pi*in which • Solvent molecules may form weak bond with solute molecules (e.g. Hydrogen-bond), which may to a certain degree change the reactivity of solute. • The presence of solute may affect certain properties of the resultant solution e.g. boiling point elevation, freezing point depression, osmosis pressure, etc. Pi- vapour pressure of solute i xi - mole fraction of solute i in solution Pi* - equil. vapour pressure of pure solute i

  38. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Gas Phase Reactions • Distinctive features of molecules in gas phase in relation to reaction • Having high energy • Sometime a liquid or even a solid substance is heat to gas phase to react e.g. steam reforming hydrocarbons, heavy oil cracking, • Moving freely within the space of reaction • High mass transfer rate - General magnitude of mass transfer rate in solid, liquid and gas solid 100 liquid 103 gas 105 • High heat transfer rate - This is very important for reactions involving heating/cooling • In practice, many reactions in which the reactants are liquid or solid at normal temperature are carried out at elevated temperatures in order to convert the reactants to gas • Compressible therefore sensitive to the reaction pressure • This has implication on the reactions involving the change of number of moles before and after reaction. • The main disadvantages of the gas phase reactions are • Usually high volume (large reactor) • not suitable for heat sensitive substances if heating to high temperature is required.

  39. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Solid Phase Reactions • Solid phase reactions are usually slow due to limited mobility of molecules • When a solid reacts with another reactant which is liquid or gas, the reaction starts from outer surface of the solid • In industry if a reaction involves a solid reactant (at ordinary temperature) what we usually do is • dissolving solid in solvent • heating it to above the melting point so that it takes part in reaction as a liquid • Many catalytic reactions use solid catalysts • The reactant in this case can be a liquid or a gas or liquid-gas mixed phase • Solid catalysts are very easy to separate from liquid, gas or a liquid-gas mixture • It is easy to handle solid catalysts from practical point of view (loading, discharge etc)

  40. CH4751 Lecture Notes 15 (Erzeng Xue) Chemical Reactions Reactions involve multi-phases • Many reactions involve multi-phase in one reactor • reactants and products can be presented as any combination of two or three phases. • When multi-phases present in a rxn following issues become important: • Relative rate of reactant and/or product molecules diffusion within each phase as well as through phase boundaries must match the rate of reaction • Solubility of solids and/or gases in liquid phase • When a porous solid is involved the liquid/gas molecules transport within the pore • Both pressures (for gas phase) and concentrations (for liquid) are interlinked in the reaction network therefore these have to be considered systematically. • When reaction involves heating or cooling, as most of reactions do, this has to be dealt with by considering both mass transfer and heat transfer within and between different phases. • The main advantage of multi-phase reactions is the easiness for separation

  41. CH4751 Lecture Notes 15 (Erzeng Xue) gas phase reactant molecule j gas phase k liquid phase l mn o pore porous solid r q p Chemical Reactions Example of Reactions involving multi-phases • The long journey for reactant molecules j. travel within gas phase k. cross gas-liquid phase boundary l. travel within liquid phase m. cross liquid-solid phase boundary n. reach outer surface of solid o. travel with pore p. reach reaction site q. be adsorbed on the site and activated r. react with other reactant molecules (either adsorbed or approached from surface above • Product molecules must follow the reverse process to return to gas phase • Heat transfer follows similar process

  42. CH4751 Lecture Notes 16 (Erzeng Xue) Chemical Reactions Acids and Bases • Acids & Bases are one of the most important classes of chemicals • Acids and bases have been know to human for a long time • Acids taste sour (in fruit), change colour of certain dye • Bases taste bitter and feel slippery (like in soap, lime water) • Acids and bases are widely present in nature, • especially in plants, electrolyte balance in life system cycle etc • Acids and bases are widely used in industry for various purpose • Dissolving chemicals, e.g. HF, aqua regia (HCl:HNO3=3:1) • Reagents for producing various chemicals • Catalysing various types of reactions • Titration in volumetric analysis • etc

  43. CH4751 Lecture Notes 16 (Erzeng Xue) H2O H2O Chemical Reactions Acids and Bases - Definition • Classical definition Acids - Substances that, when dissolved in water, increase the concentration of H+ ions e.g. HCl(g) H+(aq) + Cl-(aq) Note: H+, which is a proton only (no e- ), is actually bond with water molecule forming H3O+, the rxn is HCl(g) + H2O (l)® H3O+(aq) + Cl-(aq) For simplicity, we often use H+ instead of H3O+. Bases - Substance that, when dissolved in water, increase the concentration of OH- ions e.g. NaOH OH-(aq) + Na+(aq) NH3 + H2O ® NH4+ + OH- • Brønsted-Lowry definition Acid is proton donor and Base is proton acceptor (because H+ is a proton and OH- of a base reacts with H+ giving water)

  44. CH4751 Lecture Notes 16 (Erzeng Xue) remove H+ add H+ add H+ remove H+ HX(aq) + H2O (l)® X-(aq) + H3O+(aq) acid base conjugate base conjugate acid add H+ remove H+ NH3(aq) + H2O(l)®NH4+(aq) + OH-(aq) base acid conjugate conjugate acid base HCl(aq) + H2O(l)® Cl-(aq) + H3O+(aq) acid base conjugate conjugate base acid Chemical Reactions Conjugate Acid and Base Pairs • An acid & a base always work together to transfer proton (donate-accept). A substance can function as an acid only if another substance behaves simultaneously as a base. • When an acid or a base is dissolved in water, ions are released - this process involves proton transfer. To mark the process and link the ions with its original acid or base, conjugate acid-base pairs are defined. • Acid and conjugate base always appear in pair; likewise base and conjugate acid appear in pair • When an acid losses proton (H+) it becomes the conjugate base of that acid (e.g. HX to X-) when a base receives a proton (H+) it becomes the conjugate acid of that base (H2O to H3O+) • If an acid dissolves in water, H2O is a base; if a base dissolves in water, H2O becomes an acid.

  45. CH4751 Lecture Notes 16 (Erzeng Xue) acidbase HCl Cl- H2SO4 HSO4- HNO3 NO3- H3O H2O HSO4 SO42- H3PO4 H2PO4 HF F- HC2H3O2 C2H3O2- H2CO3 HCO3- H2S HS- H2PO4 HPO42- NH4 NH3 HCO3 CO32- HPO4 PO43- H2O OH- OH O2- H2 H- CH4 CH3- acid strength increase negligible weak strong base strength increase negligible weak strong Chemical Reactions Strengths of Acids and Bases • The strength of acids and bases • The strength of an acid is the ability to donate proton, or increase [H+] when acid is dissolved in water. • likewise, the ability to accept proton, or [OH-], determine the strength of a base • Common acids and their relative strengths • Strong acids, paired with bases with negligible basicity - Able to completely transfer their proton to water - Their conjugate bases are the weakest, with negligible tendency to accept proton • Weak acids, paired with week bases - These acids are partially dissociated to ions - Their conjugate bases are also weak, with limited ability of accepting proton • Acidswith negligible acidity, paired with strong bases - These class of acids, though carrying H, give out no [H+] - Their conjugate bases, however, are strong bases • Water can act as acid as well as base

  46. CH4751 Lecture Notes 16 (Erzeng Xue) Chemical Reactions Acid and Base Equilibrium • The extent of ionisation of an acid or a base in water • Some acids (or bases) ionise in water completely, leaving no molecules behind • Other acids (or bases) ionise partially in water, forming an equilibrium between molecules and ions e.g. HF(aq) + H2O (l) D F-(aq) + H3O+(aq) (1) NH3(g) + H2O (l) D NH4+(aq) + OH-(aq) (2) • The tendency of ionisation of an acid (or a base) varies with the type of acids, we can use the concept of reaction equilibrium to indicate the degree of ionisation. The ‘equilibrium constant’ used to describe the degree of ionisation of an acid is called acid-dissociation constant, Ka, which is defined as for equili. (1) for equili. (2) ions molecule The higher the Ka value, the higher ion conc., the higher acidity/basicity

  47. CH4751 Lecture Notes 16 (Erzeng Xue) Chemical Reactions Quantifying the Strength of Acids and Bases • [H+] and [OH-] are the measure of the strengths of acids and bases • We know that an acid when dissolved in water releases [H+] and a base gives [OH-] • We also know that the strengths of an acid or a base depend on the [H+] and [OH-] • It comes naturally that [H+] & [OH-] are used to indicate the strengths of acids/bases • The range of [H+] and [OH-] • Dilute aqueous solutions at 25°C always give, Kw=[H+][OH-]=1.0x10-14 • For an acid [H+]>[OH-], Kw=[H+][OH-]=1.0x10-14 • For a base [OH-]>[H+], Kw=[H+][OH-]=1.0x10-14 • For pure water, which is neutral [H+]=[OH-]=1.0x10-7, Kw=[H+][OH-]=1.0x10-14 • pH scale • For convenience the low value of [H+] and [OH-], we use the scale of log10 [H+] define pH= -log10[H+] Scale: 1-14. Acid pH=0-7 [H+]>[OH-]; strong acids have low pH Base pH=7-14 [OH-]>[H+]; strong bases have high pH Note: When using [OH-] (which is less used), we have pOH= -log10[OH-] (=14-pH) water can act as an acid as well as a base at equilibrium H2O D H+ + OH- Equili. constant at 25 °C is found to be Further examine other aqueous solution the same relation holds In pure water [H2O] is constant Known [H+], [OH-] can be calculated by this eqn.

  48. CH4751 Lecture Notes 16 (Erzeng Xue) Chemical Reactions Calculation of pH Example 1: Calculate pH of 0.05M HNO3 solution HNO3 + H2O ® H3O+ + NO3- HNO3 is a strong acid, HNO3 ionizes completely in water, i.e. [H3O+]= 0.05M Þ pH = - log10[0.05] = 1.3 Example 2: Calculate pH and pOH of 0.05M NaOH solution NaOH + H2O ® Na+ + OH- NaOH is a strong base, NaOH ionizes completely in water, i.e. [OH-]=0.05M, Kw = [H3O+][OH-] = 1 x10-14 M2 Þ [H3O+] = 1 x10-14 M2 / [OH-] = 1 x10-14 M2 / 0.05 M = 2 x 10-13 M Þ pH = - log10[2 x 10-13] = 12.7 Þ pOH = 14 - pH = 14 - 12.7 = 1.3 (why is this result the same as that of example1?)

  49. CH4751 Lecture Notes 16 (Erzeng Xue) Chemical Reactions Calculation of pH Example 3: What is the [OH-], in mol/L, in a solution whose pH is 9.72? Known: pH = - log10[H3O+] = 9.72 Þ [H3O+] = 1.9 x 10-10 (mol/L) for any aqueous solution Kw = [H3O+][OH- ] = 1.0 x 10-14 (mol/L)2 Þ [OH- ] = Kw / [H3O+] = 11.0 x 10-14 (mol/L)2 / 1.9 x 10-10 (mol/L) = 5.3 x 10-5 (mol / L) Example 4: The acid-dissociation constant, Ka, of hydrofluoric acid is 6.8x10-4. What is the [H3O+] in a 2M HF solution? What is the pH of the solution? HF(aq) + H2O (l)D F-(aq) + H3O+(aq) initial 2 0 0 at equili. 2 - x x x By definition Solve the eqn for x ( = [H3O+]) pH = - log10[H3O+] = - log10(0.0365) = 1.44

  50. CH4751 Lecture Notes 17 (Erzeng Xue) Chemical Equilibria Aqueous Equilibria and Some Applications • In chemistry many aqueous systems involve equilibria • Human body fluids are in electrolyte equilibria in order to function properly • Electrolyte: aqueous solutions that contain ions • Plants contain weak acids, which maintain right balance for plants to grow • Many properties of a solution that has ions are affected by its equilibrium state. • etc. (In a broad sense, harmony=balance=equilibria) • Many phenomena in chemistry can be studied by means of equilibria. We will look at: • The behaviour of an equilibrated electrolyte solution when other ions are added Applications • Buffer effect • Acid-base titration • Solubility of ionic substances and the factors affecting it

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