Industrial Chemistry Past Paper PDF

INDUSTRIAL CHEMISTRY IMPORTANT REMINDER  ALL REACTIONS ARE REVERSIBLE IN THIS CHAPTER. PLEASE USE THE REVERSIBLE SYMBOL ⇌ IN CHEMICAL EQUATIONS. CHAPTER 51 Calculation of the Order of Reaction Experiment [A] / mol dm-3 [B] / mol...

INDUSTRIAL CHEMISTRY IMPORTANT REMINDER  ALL REACTIONS ARE REVERSIBLE IN THIS CHAPTER. PLEASE USE THE REVERSIBLE SYMBOL ⇌ IN CHEMICAL EQUATIONS. CHAPTER 51 Calculation of the Order of Reaction Experiment [A] / mol dm-3 [B] / mol dm-3 Initial Rate 1 0.01 0.01 10 2 0.02 0.01 20 3 0.01 0.03 90 Let Rate = k[A]x[B]y From Expt. 1, 10 = k (0.01)x (0.01)y … (1) From Expt. 2, 20 = k (0.02)x (0.01)y … (2) (2) ÷ (1) 2 = 2x x=1 From Expt. 3, 90 = (0.01) (0.03)y … (3) (3) ÷ (1) 9 = 3y y=2 ∴ The order of reaction with respect to A is 1, while the order of reaction with respect to B is 2. The overall order of reaction is 3. The rate equation is Rate = k[A][B]2 (Initial rate is used because the initial concentration of reactants is known.) Rate Constant Units Since the rate of reaction has a unit of mol dm-3 s-1 and the concentration of the reactants have their own units, the unit of the rate constant exists to balance the unit of the equation so that the final unit of the rate of reaction is mol dm-3 s-1. Zeroth order: mol dm-3 s-1 First order: s-1 Second order: dm3 mol-1 s-1 Third order: dm6 mol-2 s- 1 Calculation Experiment [A] / mol dm-3 [B] / mol dm-3 Initial Rate 1 0.01 0.01 10 Rate = k[A][B]2 From Expt. 1, 10 = k (0.01) (0.01)2 k = 100000 dm6 mol-2 s- 1 The graph of log (rate) vs log (concentration) Equation: Rate = k[A]n  log (rate) = n log [A] + log k Slope: order of reaction y-intercept: log (rate constant) Information from this graph: BOTH order of reaction & rate constant  CHAPTER 52 Activation Energy  The minimum amount of energy required for a chemical reaction to occur. Effective Collision  Collisions between reactant particles in the right orientation and possess energy levels higher than the activation energy. Calculation of the Activation Energy 𝐸 − 𝑎 𝑘1 𝐸𝑎 1 1 k=A𝑒 𝑅𝑇  log =− ( − ) 𝑘2 2.3𝑅 𝑇1 𝑇2 Experiment Rate Constant Temperature (°C) 1 10 25 2 100 35 Gas constant R = 8.314 J K-1 mol-1 𝑘1 𝐸𝑎 1 1 log =− ( − ) 𝑘2 2.3𝑅 𝑇1 𝑇2 10 𝐸𝑎 1 1 log =− ( − ) 100 2.3 × 8.314 25 + 273 35 + 273 𝐸𝑎 = 175511 J mol-1 𝐸𝑎 = 175.511 kJ mol-1  The temperatures in the Arrhenius Equation are in Kelvin (K). Therefore, if the given temperatures are in Celsius, 273 must be added to the value before calculation.  The unit of gas constant R is in J K-1 mol-1. Therefore, the value of the calculation in the third line will be in J mol-1. The final answer must be converted into kJ mol-1. Energy Profile Activated Complex / Transition State  Exists at the point of maximum energy  Contains partially broken and partially formed bonds  Highly unstable and cannot be isolated Intermediate  The mixture of chemicals formed between steps of reactions (the low point between the two peaks in the red graph)  Can be isolated Rate Determining Step  The step of the reaction that has the highest Ea (the first peak of the reaction)  The Ea of the second step of the reaction is measured from the energy level of the intermediates to the peak, so the rate determining step is the first step, although the second peak is higher. Having a catalyst lowers the activation energy of the reaction, so the peak of the red graph is lower than that of the black graph. Maxwell-Boltzmann Distribution Curve The area under the two curves are the same, since the number of particles in the two systems are the same. The total kinetic energy in the system increases when the temperature increases, so the number of particles with kinetic energy higher than the activation energy also increases. Therefore, the number of effective collisions and the rate of reaction increase. When there is a catalyst, the activation energy decreases, and the orange dotted line moves left. More particles now have kinetic energy higher than the activation energy, so the number of effective collisions and the rate of reaction increase. The graph of log (rate constant) vs 1 / temp 𝐸𝑎 Equation: k = A 𝑒 −𝑅𝑇  𝐸𝑎 1 log 𝑘 = − × + log 𝐴 2.3𝑅 𝑇 𝐸𝑎 Slope: − 2.3𝑅 y-intercept: log A (useless) Information from this graph: Activation Energy 1 Remarks: k ∝ 𝑡𝑖𝑚𝑒  CHAPTER 53 Catalyst  A catalyst provides an alternative pathway for a reaction with a lower activation energy.  A catalyst remains chemically unchanged at the end of the reaction.  A catalyst remains unchanged in quantity at the end of the reaction.  A catalyst is specific in action.  There are negative catalysts that make the activation energy higher.  Catalysts increase the rates of BOTH forward and backward reactions to the same extent. Thus, the yield of product will not increase.  Catalysts have no effect on BOTH equilibrium position and equilibrium constant.  Homogeneous catalysts have the same physical state as the other reactants.  Heterogeneous catalysts have different physical states as the other reactants.  The rate of reaction depends on the surface area of the solid catalyst. Catalytic Reactions in Industrial Chemistry 1. Haber Process (with finely divided iron) 2. Decomposition of hydrogen peroxide into H2O and O2 (with manganese (IV) oxide) 3. Contact Process – production of SO3 from SO2 and O2 (with vanadium (V) oxide) 4. Catalytic Converters in Motor Vehicles (with platinum) 5. Enzymes (Enzymes are sensitive to pH and temperature. If the conditions are not right, the enzyme may be denatured, and its catalytic properties will be lost.)  CHAPTER 54 Important Industrial Processes Haber Process Equation: N2 + 3H2 ⇌ 2NH3 Conditions: 450°C and 200 atm, with finely divided iron as catalyst Use: To make chemical fertilizers Extra point to note: The hot NH3 and the unreacted N2 and H2 are pumped through a heat exchanger to heat up the N2 and H2 to 450°C. This saves energy costs. Steam Reforming of Methane (Production of syngas) Equation: CH4 + H2O ⇌ CO + 3H2 Conditions: 700-1000°C and 30 atm (endothermic reaction) Extra point to note: The product mixture of CO and H2 is known as syngas. The production of hydrogen can be further increased by this reaction: CO + H 2O ⇌ CO2 + H2 Production of Methanol Equation: CO + 2H2 ⇌ CH3OH Conditions: 250°C and 50-100 atm, with Cu+ZnO+Al2O3 as catalyst Use: To make methanal (formaldehyde) and ethanoic acid (acetic acid) Production of Nitric Acid Equations: 4NH3 + 5O2 ⇌ 4NO + 6H2O / 2NO + O2 ⇌ 2NO2 / 4NO2 + 2H2O + O2 ⇌ 4HNO3 Conditions: 900°C and 8 atm, with red hot platinum alloy as catalyst Use: To make ammonium nitrate, a chemical fertilizer Reichstein Process (Production of Vitamin C) Equation: Glucose  Sorbitol  Sorbose  KGA  Vitamin C Extra point to note: Vitamin C is chemically produced because the natural supply of citrus fruits and Vitamin C cannot meet the global demand. Electrolysis of Brine using Flowing Mercury Cell Equation at the graphite/platinum anode: 2Cl-  Cl2 + 2e- Equation at the mercury cathode: Na+ + Hg + e-  Na/Hg Equation at the second cell: 2Na/Hg + 2H2O  2NaOH + H2 + 2Hg Electrolysis of Brine using Membrane Cell Equation at the anode: 2Cl-  Cl2 + 2e- Equation at the cathode: 2H2O + 2e-  H2 + 2OH- Advantages of Membrane Cell vs Flowing Mercury Cell 1. Membrane cell avoids the use of toxic mercury 2. Less energy is used to circulate the mercury Reasons for the Conditions Pressure  Higher pressure increases the rate of reaction in gaseous reactions.  If the number of moles of gas on the R.H.S. is lower, the equilibrium position will shift to the right when the pressure increases, so the percentage yield will be higher.  The construction and maintenance cost for equipment that can withstand high pressure is very high. Temperature  Higher temperature increases the rate of reaction.  If the forward reaction is endothermic, higher temperatures will favour the forward endothermic reaction, so the percentage yield will be higher.  If the forward reaction is exothermic, lower temperatures will favour the backward endothermic reaction, and the equilibrium position will shift to the right to counteract the change, so the percentage yield will be higher. Catalyst  Adding catalysts will not increase the yield of product.  However, the rate of reaction is higher, so the operating temperature can be lower.  Therefore, catalysts can lower the energy cost. Adding Reactants Constantly  Adding reactants constantly will cause the equilibrium position to shift to the right to counteract the change, so the percentage yield of the product increases. Removing Products Constantly  Removing products constantly will cause the equilibrium position to shift to the right to counteract the change, so the percentage yield of the product increases.  Reaching the equilibrium point takes a long time, so removing the products beforehand will be more efficient.  CHAPTER 55 Principles of Green Chemistry 1. Less steps, less waste 2. Higher atom economy (percentage of reactant atoms that is turned into products) 3. Fewer toxic / harmful / corrosive chemicals are used 4. Less energy, higher energy efficiency (use of catalyst) 5. Renewable raw materials 6. No pollution to the environment / intensify greenhouse effect Atom economy vs Percentage yield  Atom economy refers to the percentage of reactant atoms that is turned into products if the reaction is 100% complete.  Percentage yield refers to the actual yield of the product compared to the theoretical yield. The difference is due to the incompletion of the reaction. Manufacture of Ethanoic Acid Fermentation of Sugar Equation: Sugar solution + Yeast  CH3CH2OH + O2  CH3COOH Sugar is a renewable raw material, and no toxic chemicals are used. The fermentation is done in room temperature, so no energy is used. However, the rate of reaction is very slow. Monsanto Process (carbonylation of methanol) Equation: CH3OH + CO  CH3COOH Conditions: 150-200°C and 30-60 atm, with rhodium and hydrogen iodide as catalyst Methanol can be produced from renewable sources, and there is 100% atom economy. However, there are side reactions, producing more waste. In the CATIVA process, an iridium catalyst is used instead, so that the rate of reaction is further increased, and there are less side reactions, producing less waste.