Data response Example 1

Question 1

This question is taken from the Standard Level Specimen Paper for May 2009. The same question appeared on the Higher Level Specimen Paper.

Example 1. Answers and comments

(The minimum answer required on the markscheme for each mark is given in the centre first).

Reading both the graph and the data table it can be seen that the initial temperature was 23.03 oC. The highest temperature reached (again taken either from the graph or the data table) was 23.70 oC hence ∆T (the temperature difference) is 0.67 oC. Note that it is worth examining the graph to see if the temperature had started to go down once the reaction had finished due to losing heat to the surroundings. If this had been the case then the graph would need to be extrapolated back to when the reaction started to compensate for heat loss resulting in a higher value for ∆T.

The formula for sucrose is given (C12H22O11) so using Table 5 of the 2009 Data Booklet the molar mass of sucrose can be calculated. Use the values to two decimal places to give 324.34 [(12 x 12.01) + 22 x 1.01) + (11 x 16.00). The amount (in moles) is then the actual mass taken divided by the molar mass. Since the mass is to four significant figures and this is fewer than the molar mass (which has been calculated to five significant figures) the answer should be given to four significant figures.

(c)

The heat capacity of the system is given. This means that it would take 10.114 kJ of energy to raise the temperature of the system by 1oC (or 1 degree Kelvin). When 0.4385 g of sucrose was burnt the temperature rose by 0.67 K and the energy given out was 10.114 x 0.67 kJ. However 0.4385 g of sucrose is much less than one mole of sucrose so a lot more energy will be given when one mole of sucrose is combusted. Since 0.4385 g of sucrose is 1.281 x 10-3 mole the amount of energy (in kJ) given out by one mole will be (10.114 x 0.67) divided by 1.281 x 10-3. The answer can only be given to two significant figures as the temperature rise is only given to two significant figures. Clearly if the answers to (a) and (b) had been calculated wrongly the answer to (c) (i) would be wrong hence the need to take account of error carried forward (ECF).

The value given in Table 12 of the (2009) Data Booklet is -5640 kJ mol-1. Since the value in (c) (i) is given to two significant figures this value too should be to two significant figures (5600 kJ mol-1). The experimental error is simply the difference between the value obtained and the expected value divided by the expected value expressed as a percentage.

The hypothesis was based on three assumptions. The enthalpy of combustion of sucrose is larger than that of TNT so this would not explain why TNT is explosive. Similarly in the equations given the amount of gas evolved for one mole of TNT is 4 moles (2 moles of TNT gives 5 moles of steam and three moles of nitrogen) whereas one mole of sucrose gives 23 moles of gas so this cannot be the explanation. The reason why TNT is explosive is because of the rapid rate of the reaction. This is actually the true reason but in fact cannot be deduced from the information given (see general comments below).

General comments on Example 1

This question was a specimen question and as such did not go through the rigorous system of checking that a genuine examination question would go through. It is interesting to look at this question critically. There are some inconsistencies which are examples bad practice but of much more importance there is some wrong chemistry. How many examples of bad practice and errors can you spot?

1. Three relatively minor inconsistencies are in the way in which the data is presented in the table.

(i) There is inconsistent use of significant figures in the temperature. In some cases four significant figures are given e.g. 23.03 oC in other cases only three are given e.g. 23.7 oC. You would penalise students in their internally assessed practical work if they did this and the value of 23.7 should be expressed as 23.70 oC.

(ii) The headings on the columns give the units in brackets - (s) and (oC). Although not wrong the units are normally given in IB Chemistry papers with a slash e.g. time / s and temperature / oC. This in fact is the case in this question for the data given in part (d).

(iii) The units are altered during the question. In the data table they are given in degrees Celsius whereas Kelvin is used in the value of the specific heat capacity. Although not a major error it is good practice to be consistent.

2. A much more serious error is that the value for the enthalpy of combustion of TNT in part (d) is given as 3406 kJ mol-1. Enthalpies of combustion are always exothermic and the correct value is – 3406 kJ mol-1.

3. The product water in both reactions is given as a gas. When standard enthalpies of combustion are given the products (and reactants) should be in their normal states at 298 K and I atmosphere pressure hence water should be given in the liquid state. This fact should be known for Higher Level but not for Standard Level.

4. The real error in this question however lies in the equation given for the ‘enthalpy of combustion’ of TNT. This equation is the decomposition of TNT. It is not the enthalpy of combustion as the TNT is not burning in oxygen. The true equation for the enthalpy of combustion is:

4C7H5N3O6(s) + 38O2(g) → 28CO2(g) + 5H2O(l) + 12NO2(g)

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