# Thermodynamics Experiment The Enthalpy of H 2 O 2 Decomposition in Aqueous Solution

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1 Thermodynamics Experiment The Enthalpy of H 2 O 2 Decomposition in Aqueous Solution OBJECTIVES 1. To illustrate the use of calorimetry in the experimental measurement of thermodynamic data. 2. To illustrate the use of thermodynamics and specifically Hess Law calculation of enthalpies of reaction, using H 2 O 2 decomposition as a model system. INTRODUCTION This experiment uses a calorimeter comprising a dewar vessel fitted with an electrically driven stirrer and a mercury thermometer. The calorimeter is well insulated from the surroundings. When an exothermic reaction is carried out inside the calorimeter, the temperature rises. Provided the reaction and associated temperature rise are rapid then correction for heat exchange between the calorimeter and the surroundings is unnecessary. In a typical experiment the temperature inside the calorimeter is monitored for a period of five minutes. The reaction is then initiated, causing a temperature rise. The temperature inside the calorimeter is then monitored for a further five minutes. A schematic temperature vs time graph is shown below. The readings are divided into two periods: pre-reaction and post-reaction. Ideally, during the pre-reaction and post-reaction periods the temperature should remain constant, but in practice there is often a slight drift owing to heat generated by the stirring and temperature equilibration between the calorimeter and its contents. The pre- and post-reaction period readings should be linearly extrapolated to five and a half minutes. The difference Tf Ti then gives the correct temperature rise. Figure 1: A schematic temperature vs time graph.

2 In this experiment a reaction of known molar enthalpy change is carried out inside the calorimeter. Measurement of the temperature rise enables determination of the heat capacity of the calorimeter. Then, A reaction of unknown molar enthalpy change is carried out inside the calorimeter. From the measured temperature rise and the known heat capacity of the calorimeter the unknown molar enthalpy may be calculated. When an exothermic reaction is carried out in the calorimeter, the heat liberated by the reaction is not transferred from the calorimeter to the surroundings, but instead is used to heat up the calorimeter and its contents. If the temperature increase is T = Tf Ti, then the heat, q, required to heat the system by T is: q = C p T [1] where Cp is the isobaric heat capacity of the whole system. This heat comes from the enthalpy change of the reaction: q = r H (Note: remember the reaction is exothermic i.e. H is negative) We therefore have: r H = C p T [2] We use a reaction of known rh to calibrate the calorimeter (i.e. determine C p ). We use the equation: C p = r H / T [3] N.B. In this experiment, r H is negative because the reaction is exothermic, and so C p is positive. The calorimeter is then used to measure the molar enthalpy change for the unknown reaction, we reapply the equation: r H = C p T [2] where T is the rise in temperature for the unknown reaction. EXPERIMENTAL PROCEDURE Equipment: Isobaric calorimeter with stirrer and precision thermometer 100 ml volumetric flask 250 ml measuring cylinder 10 ml pipette Chronometer Pipette filler Wash bottle Spatula Weighing boats Chemicals: Sodium nitrite NaNO 2 Sulphamic acid NH 2 SO 3 H 100 volume hydrogen peroxide solution H 2 O 2

4 Figure 2: Schematic data layout. Drag a box around the complete set of data and insert or draw a scatter graph (do not connect the points). Add a linear trendline to each of the two data sets, for the pre-reaction temperatures forecast the line forwards and for the post-reaction temperatures forecast the line backwards. Draw a vertical line at five and a half minutes and where this line intersects your extrapolations will give you T i and T f. Your graph should resemble figure 1. T is simply then T f T i. Calorimeter calibration (i) (ii) The sulphamic acid was used in excess, hence the number of moles of reaction is equal to the number of moles of NaNO 2 reacting. Calculate the quantity of NaNO 2 added to the calorimeter in moles. Given that r H m for the calibration reaction is kj mol 1, calculate r H using your quantity of sodium nitrate and hence determine C p using equation [3]. Hint: bear in mind the units, 1 kj = 1000 J. Enthalpy of decomposition of H 2 O 2 (i) The calorimeter contains the same volume of solution in both calibration and decomposition experiments so we can assume the same heat capacity. From the observed temperature rise and C p, calculate r H for the H 2 O 2 decomposed in the calorimeter using equation [2]. (ii) The hydrogen peroxide used was 100 volume, meaning it releases 100 times its own volume of oxygen gas upon decomposition. Its concentration was determined, by titration against standard potassium permanganate solution, as mol dm 3. Given this concentration, calculate the quantity of H 2 O 2 added to the calorimeter in moles. (iii) From the number of moles of H 2 O 2 added to the calorimeter, calculate r H m, in units of kj mol 1, for the decomposition.

5 (iv) The experiment measures the enthalpy of decomposition of H 2 O 2 in aqueous solution, i.e. r H for the reaction: H 2 O 2 (aq) H 2 O (l) + ½ O 2 (g) For the dissolution of pure liquid H 2 O 2 in a large excess of water: H 2 O 2 (l) + water H 2 O 2 (aq) r H = 2.1 kj mol 1 Use Hess Law to calculate the enthalpy for the decomposition of liquid H 2 O 2. H 2 O 2 (l) H 2 O (l) + ½ O 2 (g) Calculation of the O O bond dissociation energy in H 2 O 2 (i) Write out the equation for the decomposition of pure H 2 O 2 (l). Above the arrow, write in the value you obtained for the reaction enthalpy in the previous part. Leave a little space and then underneath the equation write elements in their standard states. Draw an arrow from there up to each substance. The arrows represent the enthalpies of formation of the species. Bearing in mind the enthalpy of formation of O 2 (g) is zero and given the enthalpy of formation of liquid water is kj mol 1 it is now possible to determine the enthalpy of formation of liquid hydrogen peroxide using a Hess Law cycle. (ii) The enthalpy of vaporisation of H 2 O 2, vap H, is r H for the process: H 2 O 2 (l) H 2 O 2 (g) Given the value of vap H is kj mol 1, once again write elements in their standard states under this equation, draw arrows and using Hess Law calculate the enthalpy of formation of gaseous hydrogen peroxide. (iii) The dissociation energy of the (O O) bond in hydrogen peroxide, written D(HO OH), is r H for the process H 2 O 2 (g) = 2HO (g) Given the value of f H for OH (g) is kj mol 1, once more, write elements in the standard states under this equation, draw arrows and using Hess Law calculate D(HO OH) Finally, the dissociation energy of the O O bond in the oxygen molecule, D(O 2 ), is kj mol 1. This should be significantly higher than the value you have just calculated for the dissociation energy of the O O bond in H 2 O 2, D(HO OH). Whilst there is no extra credit for doing so, can you think of a reason why this might be? You will be assessed on your calculated values for the enthalpy of decomposition of liquid hydrogen peroxide and, D(HO OH), the bond dissociation energy of hydrogen peroxide.

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