4.2Joules of energy will rise the temperature of 1g of water by 1oC
In my experiment there was an temperature increase of 5.1oC. The total weight of the fluids used is 53 cm3. The density of water is 1g/cm3 so 53 cm3 of HCL and NaOH is equal to 53g
53g x 4.2j = 222.6 joules to raise the temperature of the solution by 1oC. The temperature of the solution was raised by 5.1oC so:
222.6 J x 5.1oC = 1135.26 J
This equation shows us that 1135.26 joules of energy were used in heating the water.
In standard conditions it takes 56 Joules to create 1 mol of water through neutralisation. This would mean that if all the acid and alkali were neutralised (chemically changed to water) then:
NaOH has a relative formula mass of 58.
H2O has a relative formula mass of 18
The relative formula mass of the products of the reaction is 76.
58+18=76
The % of water in this reactions products can be worked out like so:
- ÷ 76 x 100= 23.68% (2dp)
The water made by this reaction is equal to 23.68% of 53 cm3. This shows that 12.55mol of water made by this reaction.
53mol - 12.55mol= 40.45mol of NaCl in solution.
12.55mol x 56J =702.8 Joules used to create the water made by this reaction
This investigation shows that creating bonds in neutralisation gives out energy and is an exothermic reaction. This experiment was conducted with a high level of accuracy but the results show that some of the energy used (which was not detected) was wasted. If I were to repeat this experiment I would have used a more accurate thermometer which could have taken readings smaller than 0.1oC. This would have given more accurate results. The results recorded show a large waste of energy. This waste could be reduced by insulating the conical and therefore would show a better representation of what is going on in the reaction. I think when each extra 2cm3 of HCl was added to the solution it, gave the solution a small amount of time to cool. Another way to reduce this error would to use many different conicals with set amounts in, ie one flask for every reading. This would prove impractical in the lab as there are a limited supply. I think by conducting the experiment this way a higher amount of energy will have been shown to be released.
in water. H placed in water. A sodium Hydroxide molecule (NaOH) will release one Na+ and one (OH-) ion when placed in water. This means that one molecule of NaOH and one molecule of of HCl will neutralize each other and produce H20 and a neutral salt Sodium Chloride (NaCl). This means that that to neutralize HCl and NaOH at the same concentration, you need equal amounts of each. The reaction between acid and alkali is exothermic and produces heat. The heat is generated till the reaction takes place, which will occur till equal amounts of acid and alkali are mixed. On adding more acid than the equal amount of the alkali will have the effect of cooling the mixture. In this case the temperature will continue to rise for a longer period before the mixture starts to cooling.
On the other hand sulphuric acid molecule (H2SO4) will release two (H+) ions and one SO4- ions. This means that you need two molecules of NaOH to neutralise one molecule of H2SO4 and produce the neutral salt Sodium Sulphate Na2SO4. This meansthat need two molecules of NaOH to neutralize one molecule of H2SO4 and produce the neutral salt Na2SO4. This means that to neutralize H2SO4 and NaOH when they are at the same concentration, you need to use twice as much Sodium Hydroxide as Sulphuric. The neutralisation reaction between acid and alkali is exothermic and produces heat. The heat is generated till the reaction takes place. On adding more than half as much acid as the alkali will have the effect of cooling the mixture. In this case the temperature will continue to rise for a relatively shorter period before the mixture starts to cooling.
Sulphuric acid and Hydrochloric acid each release different amounts of Hydrogen ions. Acids which form one (H+) from each molecule are called Monoprotic acid and those which form two are called Diprotic acids.
We used two acids and two alkalis in order to answer the question whether the rule applies to all acids and alkalis and to just use one of each would not enable me to come to a general conclusion.
Prediction
Hydrochloric Acid
As hydrochloric acid is added to sodium hydroxide, the pH of the alkali will decrease as the acid is gradually added to it. If equal volumes of the same concentration of hydrochloric acid and sodium hydroxide are added to one other, the resulting solution is found to have a pH of 7 as the alkali has been neutralized and a neutral solution formed.
HCl + Na0H ---------- H20 + NaCl
The temperature would rise, until the point where there are equal amounts (H+) ions or (OH-) ions in the mixture. This is caused due to the fact that as the acid and alkali react, salt and water are produced giving out heat. The temperature of the mixture rises as the reaction continues. Equal amounts of acid and alkali are required to form a neutral solution as both hydrochloric acid and Sodium hydroxide have equal number of (H+) ions and (OH-) ions respectively. When the solution is neutral, the reaction will cease and from that point onwards the temperature will start to decrease. The drop in temperature is caused by adding a cool acid to the hot solution.
We predict the same results when hydrochloric acid is added to the control alkali, ammonium hydroxide. The equation for the reaction is as follows:
HCl + NH30H ---------- H20 + NH3Cl
Sulphuric Acid
As sulphuric acid is added to sodium hydroxide, the pH of the alkali will decrease as the acid is gradually added to it. You need two molecules of NaOH to neutralise one molecule of H2SO4 and produce the neutral salt Sodium Sulphate Na2SO4. If the same concentration of sulphuric acid is added to twice the volume of sodium hydroxide, the resulting solution is found to have a pH of 7 as the alkali has been neutralized and a neutral solution formed.
H2SO4 + 2Na0H &nbs ...
To calculate the energy released we have to use the formula: Energy change = Temperature change × total volume × specific heat capacity Concentration of Acid (M) Average Temperature Change (°C) ...
