Calorimetry calculations and conclusions

1. Zinc and copper(II) sulfate solution

The first thing to consider is the excess. The zinc was in excess - this means that there will be some left over at the end of the reaction. Not all of it reacts. This means that, provided that there is enough time for the reaction to finish, all of the copper(II) sulfate solution should have reacted. This is why we use the number of moles of copper(II) sulfate in the calculation, and not the number of moles of zinc.

Temperature change = 18.0°C to 27.5°C = 9.5°C

Heat absorbed = 4.18 ´ 25 ´ 9.5 = 993 J

993 ÷ 1000 = 0.993 kJ

Number of moles of copper(II) sulfate = 0.2 ´ 25.0 ÷ 1000 = 0.005 mol

DH = - 0.993 ÷ 0.005 = - 199 kJ mol^-1 [The accepted value is - 219 kJ mol^-1]

Did you remember to include the minus sign for this exothermic reaction? Failure to do so will lose you a mark in your exam. Often students get the sign wrong. The vast majority of chemical reactions are exothermic and have a negative sign. The fact that an exothermic reaction gets hotter and the temperature of the surroundings increases tends to make people think that it should have a positive sign. However, as chemists, we are not particularly interested in the surroundings. We are more interested in the chemicals that are left behind. If the surroundings have gained energy the chemicals must have lost energy - hence the negative sign for DH.

2. Citric acid and sodium hydrogencarbonate

Temperature change = 20.5°C to 4°C = 16.5°C

Heat absorbed = 4.18 ´ 25 ´ 16.5 = 1724 J

1724 ÷ 1000 = 1.724 kJ

Number of moles of citric acid = 1.0 ´ 25 ÷ 1000 = 0.025 mol

DH = + 1.724 ÷ 0.025 = + 69 kJ mol^-1

The accepted value is + 70 kJ mol^-1

Missing out the plus sign is very common, students even argue that they are right to do so! However, you will lose credit in your exam if you do so.

3. Conclusions

You will see that the values we obtained were smaller (ignoring the signs) than the accepted values. The main reason for this is heat loss or gain. We only allowed for the heat absorbed or lost from the water in the cup. However, more heat will be exchanged with the other chemicals, the thermometer, the plastic cup and the surrounding air. This is not allowed for in our calculations and so our answer is smaller than expected.

There are also some measurement errors in this experiment. The accuracy of the 25 cm³ measuring cylinder is ± 0.3 cm³ so the error is 0.3/25 × 100 = 1.2%. The thermometer is accurate to around ± 0.5 ºC, giving an error in the first experiment of 0.5/9.5 × 100 = 5.3%. This gives a total error of ± 6.5%. This does not include the error due to the heat loss or gain which is difficult to give a percentage value to. This means even without this major error our answer may lie within the range -186 to -212 kJ mol^-1. It would be better not to suggest it is more accurate than it is by rounding off the answer to two sig figs, - 200 kJ mol^-1. Though even rounding off to the nearest ten is quite optimistic as to the accuracy of our experiment.