Chemistry Practice - Bonding Energy Kinetics - Lab Report Example

CHEMISTRY BONDING/ENERGY/KINETICS PART-A Due in week commencing 23.02.08 For the pairs of atoms listed below, whether they will tend to form ionic bonds, covalent bonds or polar covalent bonds. Explain the reason for your choice. (a) Potassium and chlorine Potassium is a metal while chlorine is a non-metal. The difference in electronegativity values of potassium and chlorine is high (the electronegativity values of potassium and chlorine are respectively 0.8 and 3.0 (Brown, LeMay and Bursten, 1991)). Potassium will have a tendency to form the cation, K+. Chlorine will have the tendency to form the anion, Cl-. There will be no sharing of electrons, hence giving rise to an ionic bond. (b) Iodine and iodine Iodine is in Group VII and has 7 electrons in its outer shell, one of them being unpaired. A single bond will be formed between two iodine atoms, with an equal sharing of these two outer shell electrons. The bond is covalent in nature. (c) Oxygen and hydrogen There is a sharing of electrons between these two atoms, giving rise to a covalent bond. The electronegativity value of H is 2.1 while that of O is 3.5 (Brown, LeMay and Bursten, 1991). There will hence be a distortion of the electron cloud, thus causing the molecule to be polar. The bonding between oxygen and hydrogen is polar covalent in nature. 6 marks 2. A number of burettes were filled with different liquids which were run out of the jet. A charged rod was held near to the jet of liquid. (a) Copy and complete the following table to show in which cases the jet of liquid would be attracted to a charged rod and what the strongest intermolecular force is. The first has been done for you Liquid Formula Attracted Yes/No Strongest Intermolecular force Water H2O Yes Hydrogen bond Heptane C7H16 No Van der Waals Tetrachloromethane CCl4 No Van der Waals Methanol CH3 OH Yes Hydrogen bond 6 marks (b) What causes molecules of some liquids to be attracted to a charged rod 1 mark Polar molecules are attracted to a charged rod. In these molecules, the centres of positive and negative charges do not cancel each other out, giving rise to permanent dipoles. When a positive rod is used, the negative end of the dipoles are attracted towards the rod while the positive end of the dipoles are attracted to a negative rod. 3. Using the 'electron pair repulsion theory', state and explain the shape of the following compounds and in each case sketcha diagram to show the arrangemant of the atoms in space, labelling the bond angles. (a) AlCl3 In AlCl3, the three outer electrons of the aluminium atom are bonded to the outer electrons of three chlorine atoms. Since there are no lone pair of electrons, the structure will not be distorted. According to the electron pair replusion theory, the three bonded pairs repel each other as far apart as possible. AlCl3 will hence be trigonal planar in shape with a bond angle of 120 (Brown, LeMay and Bursten, 1991). (b) SiH4 In SiH4, the four outer electrons of the silicon atom are bonded to the outer electrons of four hydrogen atoms. There are no lone pair of electrons. According to the electron pair replusion theory, the four bonded pairs repel each other as far apart as possible. SiH4 will hence be tetrahedral in shape with a bond angle of 109.5. (c) PCl5 In PCl5, the five outer electrons of the phosphorus atom are bonded to the outer electrons of five chlorine atoms. There are no lone pair of electrons. According to the electron pair replusion theory, the five bonded pairs repel each other as far apart as possible. The shape of PCl5 will be trigonal bipyramid, with bond angles of 120 and 90. 12 marks 4. Enthalpy changes for the reactions of carbon and aluminium with oxygen are given in the following equations, where all substances are in their standard states. C(s) + O2(g) CO2(g) H = -394 kJ mol -1 4Al(s) + 3O2(g) 2Al2O3(g) H = -3352 kJ mol -1 (a) What is the value for the standard enthalpy of combustion of carbon The standard enthalpy of combustion of a compound is the enthalpy change when 1 mole of the substance is completely burnt in oxygen under standard conditions. In the case of carbon, it is the enthalpy change associated with the following reaction: C(s) + O2(g) CO2(g) According to the data given above, the standard enthalpy of combustion of carbon is -394 kJ mol -1. 1 mark (b) What is the value for the standard enthalpy of combustion of aluminium The standard enthalpy of combustion of aluminium is the enthalpy change associated with the following reaction: Al(s) + 3/2O2(g) 1/2Al2O3(g) According to the data given above, the standard enthalpy of combustion of aluminium is therefore -838 kJ mol -1. 1 mark (c) What is the value for the standard enthalpy of formation of aluminium oxide 1 mark The standard enthalpy of formation of a substance is the change in enthalpy when 1 mol of a substance is formed from its elements under standard conditions. The standard enthalpy of formation of aluminium oxide corresponds to the enthalpy change associated with the following reaction: 2Al(s) + 3/2 O2(g) Al2O3(g) The standard enthalpy of formation of aluminium oxide is equal to half the standard enthalpy of combustion of aluminium, that is -1676 kJmol-1. 5. (a) State the First Law of Thermodynamics. 2 marks The First Law of Thermodynamics states that the internal energy of a system is constant unless it is changed by doing work or heating. It hence depends only on the initial and final states of the system. This law can be expressed mathematically as: U = q + w where U is the change in internal energy of the system, q is the energy transferred as heat and w is the work done on the system (Atkins, 1990). (b) State Hess's Law of Constant Heat Summation and show that it is derived from the First Law of Thermodynamics. 2 marks Hess's Law of Constant Heat Summation states that the energy change in converting the initial reactants to the final products is the same regardless of the route taken provided that the initial and final conditions are identical. The statement of Hess's Law given above implies that the change in internal energy of a system depends only on its initial and final states. This is what is stated in the First Law of Thermodynamics, showing that Hess's Law is derived from this fundamental law of conservation of energy. (c) Given the following data, construct a Hess's Law cycle and calculate the standard enthalpy of formation of ethyne Hf [C2H2]. 2C(s) + H2(g) C2H2 (g) Hc carbon = -394 kJ mol-1 Hc hydrogen = -286 kJ mol-1 Hc ethyne = -1300 kJ mol-1 4 Marks Hess's Law Cycle: 2 C(s) + 5/2 O2 (g) + H2 According to the diagram, 2 Hc (C) + Hc (H2) = Hc (C2H2) + Hf (C2H2) -1074 = -1300 + Hf (C2H2) Standard enthalpy of formation of ethyne = 226 kJ mol-1 6. Given the following data, construct a Hess's Law cycle and calculate the standard enthalpy of hydration of ethene C2H4(g) + H2O(l) C2H5OH (l) ethene ethanol Hf ethene = +52 kJ mol-1 Hf water = -286 kJ mol-1 Hf ethanol = -278 kJ mol-1 4 Marks Hess's Law cycle: According to the diagram, Hf (C2H5OH) = Hf (C2H4) + Hf (H2O) + Hhydration (C2H4) -278 = -234 + Hhydration (C2H4) Standard enthalpy of hydration of ethane = -44 kJ mol-1 7. 1-bromobutane was hydrolysed by aqueous sodium hydroxide solution. C4H9Br + NaOH C4H9OH + NaBr 1-bromobutane butan-1-ol During the reaction the sodium hydroxide is used up. Samples of the reaction mixture were drawn off at regular time intervals and analysed by titration with standard hydrochloric acid to find out the concentration of NaOH. The results are given below: Time / secs concentration of OH- / mol dm-3 0 0.500 100 0.350 200 0.250 300 0.180 400 0.125 500 0.090 600 0.063 700 0.040 800 0.030 (a) Plot a graph of these results with "time" on the horzontal axis and concentration on the vertical axis. 3 Marks (b) Select 4 suitable points along the curve and draw tangents to find the rates. Contruct a table of "Rate" against "Concentration". 4 Marks From the tangents, the following values can be obtained: Initial conc of NaOH /mol dm-3 Initial rate of NaOH disappearance / mol dm-3 s-1 0.500 0.00150 0.180 0.00055 0.090 0.00027 0.040 0.00010 (c) Plot a graph of Rate (vertical axis) against Concentration (horzontal axis) 3 Marks (d) From your graph in (c) deduce the Order of Reaction. 1 mark Since the graph of rate vs concentration of NaOH is a straight line, the order of reaction is first order with respect to NaOH (e) Write a rate equation with respect to concentration of OH-. 2 marks Rate = k[OH-1]1 (f) Calculate rate constant k from the gradient of the graph drawn in (c). 2 marks The slope of the above line is 0.003. Hence the rate constant has a value of 0.003 s-1 PART-B 1. (a) (i) Draw a dot & cross diagram to show the bonding in hydrogen sulphide, H2S. State and explain the shape of H2S using the 'electron pair repulsion theory', estimating the bond angle. 4 Marks Hydrogen and sulphur have 1 and 6 outer electrons, respectively. For sulphur to complete its octet, it forms two bonded electron pairs with two hydrogen atoms. The central atom has two bonded electron pairs and two non-bonded electron pairs. Based on the electron pair repulsion theory, one can conclude that H2S has a bent shape. The bond angle is estimated to be around 104.5o (this angle puts the non-bonded electron pairs furthest possible). (ii) Hydrogen sulphide (H2S) is a gas at room temperature whereas water (H2O) is a liquid boiling at 100oC. Suggest an explanation for this. 3 Marks The covalent bonds in water molecules are more polar than those in hydrogen sulphide. Contrary to hydrogen sulphide molecules that are held together by van der Waals forces of attraction, water molecules are held together by hydrogen bonding as well as van der Waals forces of attraction. Due to the hydrogen bonding operating in water, molecule are closer to each other and hence water exists as a liquid. (b) (i) Draw a dot & cross diagram to show the bonding in methanal, HCHO. State and explain the shape of HCHO using the 'electron pair repulsion theory', estimating the (H-C-H) bond angle. 4 Marks Carbon is the central atom in methanal. The carbon atom is linked to two hygrogen atoms by two single bonds and one one oxygen atom by a double bond. To determine the shape of the molecule, the double bond is counted as a single bonded pair. With three bonded pairs and no lone pairs, the shape of methanal is given to be trigonal planar. However, due to the higher electronic-charge density on the double bond of central atom, the double bond repels the other two single bonds with a greater force. As result the H-C-H angle is below 120o, the reported value is in fact 116 o (Brown, LeMay and Bursten, 1991). (ii) Methanal (HCHO) is a gas at room temperature whereas methanol (CH3OH) is a liquid. Suggest an explanation for this. 3 Marks Due to the large difference in the electronegativity values between oxygen and hygrogen, the O-H bond in methanol is polar. The methanol molecules can participate in hydrogen bonding,whereas methanal molecules cannot. The hydrogen bonding makes the methanol molecules closer to each other, and hence it is a liquid at room temperature. As there is no hydrogen bonding between the methanal molecules, the molecules are less tightly held together and methanal exists as gas at room temperature. 2. Predict the shapes of (i) phosphine, PH3 The central atom has 3 bonded pairs and one nonbonding pair, hence the shape is trigonal pyramidal (ii) sulphur trioxide, SO3 The central atom has 3 bonded pairs and no non-bonding pair, hence the shape is trigonal planar (iii) the sulphite ion, SO32- The central atom has three bonded pairs and one nonbonding pair, hence thr shape is trigonal planar. (iii) the amide ion, NH2- The central atom has two bonded pairs and two nonbonding pair, hence the amide ion has a bent shape. 4 Marks 3. Use the values for average bond enthalpies (E) from the table below to calculate the enthalpy changes in each of the reactions: (a) and (b) Bond C-C C=C C-H C=O C-Cl H-Cl O-H O=O E/kJ mol-1 346 611 412 743 339 431 463 497 (a) CH4 (g) + 2O2(g) CO2(g) + 2H2O(g) 3 Marks Enthalpy change = -(2 x E(C=O)) + -2(2 x E(O-H)) + (4 x E(C-H)) + 2(1 x E(O=O)) = -(2 x 743) + -2(2 x 463) + (4 x 412) + 2(1 x 497) = -696 kJmol-1 (b) CH2=CH2(g) + HCl(g) CH3CH2Cl(g) 3 Marks Enthalpy change = -(5 x E(C-H)) + -(1 x E(C-C))+ -(1 x E(C-Cl)) + (4 x E(C-H)) + (1 x E( C=C)) + (1 x E(H-Cl)) = -(5 x 412) + -(1 x 346) + -(1 x 339) + (4 x 412) + (1 x 611) + (1 x 431) = -55 kJmol-1 (c) How would your answer to (a) compare to the data book value for the standard enthalpy of combustion of methane Explain your answer. 2 Marks The absolute value is lower than the value quoted in the data book. In the Above equation, water is in the gaseous state, but under standard conditions (the value quoted in the data book) water is a liquid. As in the above case some energy is used to vaporise water, the net energy released is thus lower than the one obtained under standard conditions. 4. (a) Given the following data, construct a Hess's Law cycle and calculate the standard enthalpy of formation of ethane, Hf [C2H6]. 2C(s) + 3H2(g) C2H6 (g) Hc carbon = -394 kJ mol-1 Hc hydrogen = -286 kJ mol-1 Hc ethane = -1560 kJ mol-1 4 Marks Hf (C2H6) + Hc (C2H6) = 2Hc (C) + 3Hc (H2) Hf (C2H6) + -1560 = (2 x -394) + (3 x -286) Hf (C2H6) = -86 kJmol-1 (b) Use the values for average bond enthalpies (E) from the table below along with the standard enthalpy of atomisation of carbon to calculate the standard enthalpy of formation of ethane, Hf [C2H6] using the equation given in (a). Structural formula of ethane: H H | | H C C H | | H H Bond C-C C-H H-H Hat [C] = 717 kJ mol-1 E/kJ mol-1 346 412 436 3 Marks Hf [C2H6] = -(6 x E(C-H)) + -(1 x E(C-C)) + (2 x Hat [C]) + (3 x E(H-H)) Hf [C2H6] =-(6 x 412)) + -(1 x 346) + (2 x 717) + (3 x 436) Hf [C2H6] =-76 kJmol-1 (c) Comment briefly on the discrepancy between the two calculated values for the standard enthalpy of formation of ethane in (a) and (b). Stating, with a reason, which of the two values is likely to be more accurate. 2 marks In method (a), the exact enthalpy values (obtained at standard conditions) of the various elements/compounds have been used to calulate Hf [C2H6]. However, in method (b) average bond enthalpies (not exact values, particularly for C-H bond) have been used to calculate Hf [C2H6] (g). As in method (a) exact enthalpy values have been used, this method is expected to give a more accurate value for Hf [C2H6]. 5. (a) Draw a diagram of the energy distribution of gas particles in a system at one temperature T1. On the same diagram, show the shape the distribution at some higher temperature T2. 3 Marks (b) Relate the two curves in (a) to the change in the rate of a gas phase reaction with increased temperature. 2 Marks For a reaction to occur, a molecule must possess a minimum amount of energy known as the activation energy (Ea). When the temperature is increased, the maxima of the distribution curve is shifted towards the right and a larger number of molecules possess this activation energy. As at T2 the population of molecules possessing Ea is higher than that at T1, the reaction occurs at a faster rate at T2. (c) Draw a labelled energy profile showing the energy changes during an endothermic reaction. Use this and the diagram drawn in (a) to explain how catalysts increase the rate of reactions. 4 Marks A catalyst lowers the activation energy of a reaction, see the figure below. With this lower activation energy, more molecules can participate in a given reaction. Hence the rate of reactions are increased by the mediation of catalysts. 6. Two gases react according to the equation: X (g) + 2Y(g) XY2(g) Experiments were done at 700oC to determine the rate equation. The following results were obtained: Experiment number Initial [X] /mol dm-3 Initial [Y] /mol dm-3 Initial rate of formation of XY2 /mol dm-3 s-1 1 0.1 0.1 1 x 10-4 2 0.1 0.2 4 x 10-4 3 0.2 0.1 1 x 10-4 (a) State with reasons the order with respect to X. 2 Marks The order of reaction with respect to X can be determined by comparing the results from experiments 1 and 3, in which the concentration of Y is constant. As the initial rate of reaction is unchanged by doubling the concentration of X in experiment 3, the reaction rate is zero order with respect to X. State with reasons the order with respect to Y. 2 Marks The order of reaction with respect to Y can be determined by comparing the results from experiments 1 and 2. As the rate of reaction is quadrupled by doubling the concentration of Y while keeping the concentration of X constant, one can conclude that the rate of reaction is second order with respect to Y. (b) Write the rate equation for the reaction 1 Mark Reaction rate = k[Y]2, where k is the rate constant. (c) Using the results from experiment 1, calculate the value of the rate constant for the reaction and state its units. 3 Marks k= reaction rate/[Y]2 = 1 x 10-4 mol dm-3s-1/(0.1 mol dm-3)2 = 1 x 10-2 dm3 mol-1 s-1 References Atkins, P.W, 1990, Physical Chemistry, 4th edn, Oxford University Press, U.K Brown, T.L, LeMay, H.E & Bursten, B.E, 1991, Chemistry: The Central Science, 5th edn, Prentice Hall, Englewood Cliffs, NJ 07632 Read More