NJC H2 CHEM P2 Solution Prelim

NJC Preliminary Examination H2 Chemistry Paper 2 Solutions 1 Planning (P) A student was provided with a spirit burner . He was asked to determine the enthalpy change of combustion under laboratory conditions, Hc, for ethan ol using the following setup. Knowing that ther e could be significant heat loss arising from the above experimental setup, he decided to calibrate the calorimeter by burning 0.60 g of methanol (CH 3OH) to determine the calorimeter’s heat capacity, C calorimeter, which accounts for both the water and the copper can. Heat capacity is defined as the number of J oules of heat needed to raise the temperature of the calorimeter by one Kelvin or one degree Celsius. The temperature of the calorimeter rose from 25.0°C to 33.8°C. The same calorimeter was then used to measure the enthalpy of combustion of ethanol. (a) Given the enthalpy change of combustion of methanol is − 715 kJ mol –1, use the information above to calculate the heat capacity of the calorimeter, C calorimeter, stating its units. [2] Heat evolved from methanol = ( ) × 715000 = 13400 J Heat gained by calorimeter = CT = C × 8.8 Heat evolved from methanol = Heat gained by calorimeter 13400 = (8.8)C C = 1520 J K–1 calorimeter (copper can containing 50 cm3 of water) draught shield wick spirit burner containing selected alcohol stirrer lid thermometer

(b) Given the enthalpy change of combustion of ethanol is approximately −1370 kJ mol–1, calculate the minimum mass of ethanol required to give the same temperature change as that in the calibration. [2] For the combustion of ethanol to give the same temperature change as that in the calibration, it means: Heat evolved from ethanol = Heat gained by calorimeter ethanol × (1370 × 103) = 13400 ethanol = ( ) = 9.871 × 10–3 mol Minimum mass of ethanol = (9.871 × 10–3) × 46.0 = 0.450 g (c) Write a plan to determine the enthalpy change of combustion, Hc, of ethanol that the student will carry out. You may assume that you are provided with:  a thermometer with divisions of 0.2 °C division;  the apparatus normally found in a school or college laboratory. Your plan should include details of  the procedure to determine the enthalpy change of combustion of ethanol;  the readings recorded using appropriate table(s), including units;  precautions taken to ensure reliability of the experiment  an outline of how the results would be used to determine the enthalpy change of combustion of ethanol based on the plan that you have written using arbitrary values. [7] Procedure 1. Use the same calibration apparatus setup for the experiment. 2. Rinse the spirit burner with ethanol and allow the wick to dry off/ use a new wick with same length. 3. Weigh the spirit burner containing 1.0 g of ethanol using an electronic balance. 4. Measure the initial temperature of the water using a 0.2 oC division thermometer. 5. Light the wick of the spirit burner. 6. Monitor the temperature of water using the thermometer with constant stirring using the stirrer. 7. Extinguish the flame with a cap when the rise of temperature of the water reaches about 8.8oC. 8. Measure the final temperature of the water.

3 9. Weigh the spirit burner with the remaining ethanol again. 10. Repeat the experiment until the % difference of T/m < 5%. Tabulation of results Initial temperature of water / oC T1 Final temperature of water / oC T2 T / oC T2 – T1 = T Initial mass of spirit burner with ethanol / g x Final mass of spirit burner with remaining ethanol / g y Mass of ethanol burnt, m / g x – y = z T/m Calculation Heat gained by water and calorimeter = C × T = (1520)T J = (1.52)T kJ Amount of ethanol = mol Hc (ethanol) = – (1.52)T/ ( ) kJ mol–1 (d) Identify one potential safety hazard in this experiment and state how you would minimise this risk. [1] Methanol is toxic. Wear glove. Methanol is toxic and volatile. Conduct the experiment in a fume hood. Methanol and ethanol are highly flammable. Alcohols are placed away from flame when not in used in experiment. [Total: 12]

2 The simplest chemical reactions are those that occur in the gas phase in a single step, such as the transfer of a chlorine atom from ClNO2 to NO. ClNO2(g) + NO(g) NO2(g) + ClNO(g) (a) (i) An equimolar mixture of C lNO2(g) and NO(g), at total initial pressure of 3 atm, is allowed to react in a closed vessel at 1000 K. When equilibrium is attained at the 5th minute, the partial pressure of ClNO2 is found to be 0.57 atm. Calculate the value for the equilibrium constant, Kp , of this system. ClNO2(g) + NO(g) NO2(g) + ClNO(g) Initial/ atm 1.5 1.5 0 0 Change/ atm 0.93 0.93 +0.93 +0.93 Eqm/ atm 0.57 0.57 0.93 0.93 ( ) ( ) ( ) ( ) ( ) ( ) Kp = 2.67 (no units) (ii) At the 10 th minute, more C lNO2 gas was pumped into the vessel at 1000 K, increasing the partial pressure of ClNO2 to 1 atm. Suggest how the position of the equilibrium would change. By LCP, the position of the equilibrium will shift forward to consume some of the extra ClNO2 in the reaction to partially decrease its partial pressure. (iii) Hence illustrate clearly, in a pressuretime graph below, the changes in the partial pressures of ClNO2 and NO2 when (I) the above gaseous system first reaches equilibrium at the 5th minute, (II) more ClNO2 gas was added into the vessel at the 10th minute and a new equilibrium is attained at the 15th minute. Time Pressure 0 ClNO 1.5 0.57 1.0 5 10 15 0.93 NO2

5 (iv) Suggest whether the addition of an inert gas into the vessel would affect the position of the equilibrium. The position of the equilibrium is not affected by the addition of the inert gas as the partial pressure of all gaseous reactants and products remained unchanged [7] (b) ClNO2 can behave as an ideal gas under certain experimental conditions. (i) State the two assumptions of kinetic theory of ideal gas. There are no/ negligible intermolecular forces of attraction or repulsion between the gaseous particles. A gas is composed of tiny gaseous particles that have a negligible volume compared to the volume of the gas container. (ii) Predict whether ClNO2 behaves ideally under high pressure. Under high pressure, the gaseous particles are very close together in a small volume.  They experience significant intermolecular forces of repulsions/ attractions between them. OR  The volume of the gas particles is no longer negligible compared to the volume of the container. So it is not behaving ideally under high pressure. [4] (c) (i) Draw the dot-and-cross diagrams of NO and C lNO. Hence, suggest the shape and bond angle of ClNO. Shape : bent Bond angle : Stating a value that is within the range – 110  x < 120 (ii) Hence, suggest why the formation of ClNO from NO is favoured. NO contains a single unpaired electron. Formation of N -Cl bond allows N to achieve noble gas/ octe