Endothermic Reactions - I am going to investigate the temperature drop induced by the reaction: NH4Cl + 100 H20 = NH4Cl (aq 100 H20)
Endothermic Reactions
I am going to investigate the temperature drop induced by the reaction:
NH4Cl + 100 H20 = NH4Cl (aq 100 H20)
Calculating the amount of solid required to react with a certain amount of water
NH4Cl + 100 H20 = NH4Cl (aq 100 H20)
mole NH4Cl = 14 + 4 + 35.5
= 53.5g
mole H20 = 2 + 16
= 18g/cm3 (assuming that 1g is the same as 1cm3)
It can therefore be assumed that one mole of water occupies 18cm3 of space.
Ratio: NH4Cl : H20
: 100
53.5 : 18*100
53.5/18 : 1800/18
2.97 : 100
Plan
I will take 100cm3 of water and 2.97g of NH4Cl and combine them in a polystyrene beaker whilst measuring the change in the temperature with an accurate thermometer (accurate to 0.1cm3). I will surround my beaker with cotton wool wadding and a sheet of aluminium foil shiny side in to reduce heat loss. I will also make a cardboard lid surrounded in a similar way to further reduce heat loss to the surrounding atmosphere. When I add the NH4Cl to the water, I will first take the temperature of the distilled water (so no impurities exist in the water) and record it as accurately as possible, then I shall add the NH4Cl quickly and replace the lid. I will record the lowest temperature reached inside the beaker and repeat the experiment 5 times and take an average. Once I have a value for the change in temperature, I will be able to calculate a value for DH using the formula:
DH = mcDT
Where: DH = Enthalpy change, m = mass of solution, c = specific heat capacity of water (4.18 kJ/kg), DT = temperature rise.
Prediction
When anything dissolves in water, an endothermic reaction takes place. I can therefore assume that there will be a substantial, rapid drop in temperature and consequently there will be a substantial DH value. I also predict that my value will be smaller than that predicted for DH as some heat shall be lost to the surrounding atmosphere.
Fair Test
In order to keep my test fair, ...
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DH = mcDT
Where: DH = Enthalpy change, m = mass of solution, c = specific heat capacity of water (4.18 kJ/kg), DT = temperature rise.
Prediction
When anything dissolves in water, an endothermic reaction takes place. I can therefore assume that there will be a substantial, rapid drop in temperature and consequently there will be a substantial DH value. I also predict that my value will be smaller than that predicted for DH as some heat shall be lost to the surrounding atmosphere.
Fair Test
In order to keep my test fair, I will attempt to remove any alterations in conditions that may affect my experiment. The main problem with doing a temperature change experiment is losing heat to the surrounding atmosphere, I will therefore attempt to reduce heat loss to an absolute minimum. In order to do this I will have to reduce the three main forms of heat loss, convection, conduction and radiation. In order to reduce radiation, I will wrap shiny aluminium foil around the cup so that radiation is reduced. In order to reduce conduction, I will wrap an insulator, cotton wool wadding, around the cup. This will also help reduce convection, however, it is difficult to entirely remove convection without a vacuum being created around the polystyrene cup, or even using a vacuum flask. However, by using cotton wool wadding convection will be kept to an absolute minimum. I shall also use a lid in order to prevent heat loss in this way. The lid will be similarly insulated to prevent as much heat loss as is possible. In this way I hope that heat loss to the surrounding atmosphere will be reduced to an absolute minimum.
In order to keep the test fair, I will also use exactly the same amounts of NH4Cl and water to do the experiment with, the same polystyrene cup, the same wadding and aluminium foil and the same thermometer. I will also measure the temperature after the reaction has taken place.
Results
Initial Temperature
Temperature after Reaction
Change in Temperature
22.8
9.1
3.7
22.4
7.5
4.9
22.5
8.2
4.3
22.4
8.6
3.8
22.9
7.7
5.2
Average change in temperature = 4.38 °C
Calculations to find DH
DH = mcDT
DH = 100*4.18*4.38
DH = 1831 J
DH = 1.831 kJ
In order to find the number of Joules per mole given off, I must use the number of moles of solid that I used in my experiment to multiply the value produced by substituting values into the above equation.
I used 1/18 of a mole of solid in my experiment; I must therefore multiply by 18 in order to find the number of Joules per mole.
DH = 1.831*18
DH = 32.96 kJ/mol
Evaluation
The value that my teacher has given me for DH is 16.4 kJ/mol. Unfortunately, my calculated value is quite considerably larger than it should be; in fact it is more than double the predicted value. I must now try to explain why my value is so far out and offer some solutions in order to improve the experiment if I was to do it again. I must first explain that, even with my attempts to eliminate heat loss, I expected that my value would be lower than the predicted value as some heat would still be lost to the surrounding air when the reaction takes place. However, as I have found that instead of my prediction being correct and my value being below the accurate value as expected it is higher. I think this could be due to several factors. As ammonium chloride is extremely soluble, if I put slightly too much solid in through slight miscalculation or poor weighing technique, or even inaccurate reading of the thermometer e.g. not giving the thermometer enough time to adjust to the temperature of the water. This could explain the greater temperature rise. However, I feel that an error this big can't simply be explained by human error. As I assumed that the cup was clean I neglected to rinse it out before the experiment, I also only rinsed the cup after each set of results was taken. I believe that the mistakes may have been due to a combination of errors and that there may have been something in the cup before I used it that affected the reaction, or else more probably, some solid was left adhering to the side of the cup after each experiment. This would result in more ammonium chloride reacting with the water than I had actually put in resulting in a greater drop in temperature as more solid reacts, and also explaining the fluctuation in my results, as different amounts of solid would be left behind after rinsing every time. I would also suggest that perhaps the distilled water that I used was not 100% pure and perhaps this could have had an effect on the reaction, the cup was also washed after each experiment with tap water that contains impurities and therefore could also affect my results by a small margin. I believe that a combination of these errors led to the large error in my value for DH.
If I was to repeat this experiment I would be more careful. I would make sure that each container used was cleaned by distilled water as opposed to tap water before use and that I used a fresh container for each new set of results. I would ensure that my measurements were as accurate as I could make them. I think I should also take slightly more ammonium chloride and water to reduce any affect that a slight error in weighing of the substance made.
In conclusion, my experiment has shown me simply that the reaction between ammonium chloride and water is endothermic and a reasonable drop in temperature takes place when the ammonium is dissolved. It has also taught me that I need to be more careful in my experimental technique if I wish to obtain good results in my experiments.
