IB Chemistry: Calculation of Enthalpy Changes

IB Chemistry: Calculation of Enthalpy Changes Calculation of Enthalpy Changes Introduction The average kinetic energy of the molecules in a substance is measured by its temperature.Heat on the other hand is the measure of the total energy of the molecules in a system, which includes not only the kinetic energy, but also other energy sources such as Potential Energy. This lesson will focus on how you can quantify the temperatures, heat change etc. Calculate the heat energy change when the temperature of a pure substance is changed The formula for heat energy is: Heat energy (J) = mass (g) x specific heat capacity (kJkg-1K-1 )x change in temperature (°C) Or in short hand notation: q= mcΔT Because different substances require different amounts of energy to produce the same amount of temperature change, they have different specific heat capacities. You are not supposed to memorize any of these values, as the question usually comes with this specific value, but it is highly recommended you memorize the Specific Heat Capacity value for water, which is 4.18 kJkg-1K-1 Let’s do an example. How much energy would I need to raise the temperature of 50g of water from 20°C to 60°C? Let’s use the formula: q= mcΔT q=unknown ( We want to solve this) m= mass = 50g c= specific heat capacity of water = 4.18 kJ-1kg-1 K-1

ΔT = from 20°C to 60°C = 60°C-20°C= 40°C So. q= 50.0 g x 4.18kJ-1kg-1 K-1 x 40.0°C q= 7560 kJ Design suitable experimental procedures for measuring the heat energy changes of reactions ● You are supposed to know an experiment “procedure” used to measure the enthalpy (heat) change of a reaction. ● We can use a process called Calorimetry to measure this “change”. Here is how the process works: 1. Clamp the metal can but make sure there is enough space below for a Bunsen burner to be lighted. 2. Use the measuring cylinder to fill the metal can with 120cm3 of distilled water. 3. Record the initial temperature with a thermometer. 4. Add 100cm3 of alcohol into the spirit burner and then weigh it on a balance. Record your results so far on a table. 5. Carefully put the spirit burner on the heat proof mat, remove the cap and then light the wick. 6. Allow the alcohol below to slowly heat up the distilled water for around 3 minutes. Use the glass rod to constantly stir the distilled water. 7. Put the cap on the spirit burner to extinguish the flame. 8. Measure the temperature of the distilled water again. Record your findings on a table,and then work out the temperature change from the beginning and the end of the experiment. 9. Reweight the spirit burner and the cap and work out the mass difference. This will tellus the mass of alcohol used throughout the experiment. 10. Repeat the experiment again with different alcohols but with the same “Controlled Variables” stated in the “Variables” section above. 11. If time permits, try to collect results from each alcohol at least 2-3 times. 12. Use the change in temperature and mass to calculate the “Energy Released”. The formula for this is q=mcΔT Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Please Note: This question is not based on any empirical experiment conducted and that theresults here may not reflect on actual laboratory results. 1. When 50 cm 3 of 0.5 moldm-3 of HCl reacts with 50 cm3 of 0.5 moldm-3 of NaOH, the temperature rise is 5.0 °C. Calculate the enthalpy change per mole?

Let’s first write a balanced equation: HCl + NaOH –> NaCl + H2O Next, let’s calculate the number of moles of HCl If we recycle back to our previous equations: Moles = Concentration x Volume Moles of HCl= 0.5 x 0.05 ( 0.05 and not 50 as we have to convert to dm-3, so we have to divide by 1000) = 0.025 Moles of NaCl = 0.025 1:1 mole ratio if you look at the coefficients of HCl and NaOH, so number of moles of NaCl isalso 0.025 Now we can use our previous formula to calculate enthalpy change: ΔH= mcΔT ΔH=(0.1) x 4.18 x 5 = (0.1 x 5 x 4.18) = 2.09 kJ ΔH mol-1 = 2.09 kJ for 0.025 moles = 2.09 /0.025 (divide by 0.025 to get the value of ΔH for 1 mol) = -83.6 kJ mol-1 There is a temperature rise, as indicated by the 5 degree temperature increase , so the reaction is exothermic. Therefore, 83.6 will be a negative value. Evaluate the results of experiments to determine enthalpy changes When you calculate the enthalpy change, it is almost impossible to have “flawless” results. You are bound to encounter errors and here are a few that may be encountered:

○ Heat loss through lid. ○ Heat transferred not to the reacting products, but to the surroundings instead. ○ Human error in readings. There are quite a few ways this can be asked in the exams – first, they can ask why your calculated result is different to that of the standard literature value (most likely the IB databooklet).We can rectify the first error by using a well insulated lid such as a polystyrene cup – to minimize the amount of heat loss through the lid. Problem two can also be rectified by using a well-insulated calorimeter.