# Finding the enthalpy of decomposition of Sodium Hydrogencarbonate

September 25, 2017 September 1st, 2019 Free Essays Online for College Students

I decided to add 2.76g of sodium carbonate to the hydrochloric acid solution at the fourth minute, consequently there is no result for this time in my result’s table that follows-

(The start temperature of the acid was 20.5ï¿½C)

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TIME

(MINUTES)

TEMPERATURE

(ï¿½C)

0

20.5

3

20.4

5

26.6

6

26.6

7

26.6

8

26.4

9

26.2

10

26.0

PART TWO: THE REACTION OF SODIUM HYDROCARBONATE WITH HYDROCHLORIC ACID

I decided to add 3.65g of sodium hydrogencarbonate to the hydrochloric acid solution at the fourth minute, consequently there is no result for this time in my result’s table that follows-

(The start temperature of the acid was 20.5ï¿½C)

TIME

(MINUTES)

TEMPERATURE

(ï¿½C)

0

20.5

3

20.4

5

11.8

6

11.8

7

12.0

8

12.1

9

12.3

10

12.4

In order to find the exact change in temperature for each part of the experiment I will have to draw a graph and extrapolate back from my line of best fit for each set of results.

ANALYSIS

If I assume that the relative formula mass (Mr) of the acid is 50, I can calculate the molarity of the acid and hence the enthalpy change per mole of acid. I can then use this value to deduce what acid took part in this reaction.

In order to find the molarity I first have to work out the number of moles of acid present in the reaction. To do this I will use the formula:

Number of Moles of acid = Mass of acid (g)

Mr

Putting my values in I get:

Number of Moles of acid = 12.33

50

= 0.2466 moles of acid

Now I can calculate the molarity using the formula:

Molarity of acid = moles of acid x 1000

Volume

Putting my values in I get:

Molarity of acid = 0.2466 x 1000

250

= 0.9864 mol dm

From this information I can now start to calculate to enthalpy change per mole of acid using the equation:

q = mc?t

Where q = the heat energy change (joules)

m = the mass of liquid present

c = the specific heat capacity (4.18 joules per gram per degree)

?t = the rise in temperature extrapolated from my graph

Putting my values in I get:

q = ?

m = 100cm (50cm + 50cm )

c = 4.18

?t = 4.6ï¿½C

Which, when arranged in the formula gives me:

q = 100 x 4.18 x 4.6

so q = 1922.8 J

As this reaction used 50cm of acid, the number of moles present in the solution is given by the equation :

Moles of acid present in solution = volume x molarity

1000

Which gives me:

Moles of acid present in solution = 50 x 0.9864

1000

= 0.04932 moles

Hence the enthalpy change per mole of acid is:

1922.8 J

0.04932 moles

= 38986 J

= 38.99 kJ

The reaction is exothermic and is carried out at constant pressure, so the enthalpy change per mole of acid is -38.99 kJ mol

However this value is only reliable to 3 significant figures, this makes it -39.0 kJ mol

Throughout the experiment I tried to keep my results as accurate and reliable as possible by doing things such as washing the thermometer after taking the temperature, and stirring the mixture for the same amount of time before taking each reading. I also recorded the temperature of the acid and alkali before starting the experiment (these values are shown in my results table) this made sure that they where the same temperature and therefore made

my experiment more reliable. However there where errors that I could not do anything about. These are percentage errors and every piece of equipment has them. In order to see how accurate my experiment was I am going to calculate the main percentage errors. These are for the measuring cylinder, the thermometer, the graduated flask, the balance and then the overall percentage error.

For the measuring cylinder it is: 0.5 x 100

50

= 1.0 %

for the thermometer: 0.1 x 100

4.6

= 2.17 %

250

= 0.05 %

For the balance: 0.01 x 100

12.33

= 0.08 %

My overall percentage error is: 1.0 + 2.17 + 0.05 + 0.08

= 3.30 %

As you can see from my calculations my experiment was not very accurate in a number of places. My greatest sources of error were the heat loss from the apparatus, the heat lost to the surroundings, and measuring cylinder that I used to measure out my solutions.

EVALUATION

Looking at my results you can see that the general trend, after the acid was added, is that the temperature decreases by around 0.1ï¿½C every minute. However if you look at the result taken on the 7th minute on graph A it doesn’t fit onto the line of best fit, hence it is an

anomalous result. This result could of occurred because the thermometer only read to the nearest 0.1ï¿½C so it was hard to say what the exact temperature was as ideally the experiment should have been done with a thermometer that measured to the nearest 0.05ï¿½C at least. Another reason could be that a draught went past during that minute which affected my results, my experiment was susceptible to this as it was not insulated and I could not do anything to stop this happening. To lower this error I could use a stronger acid and alkali in the experiment as this would make the temperature change bigger and therefore lower the error of the thermometer.

My greatest source of error was the amount of heat lost from the apparatus during the experiment, this was due to the fact that the equipment had no kind of insulation to stop heat loss. If I was to conduct the experiment again, to reduce this error, I would insulate the equipment with cotton wool as this helps to keep in the heat. I would also place a lid on the top of the polystyrene cup to reduce the heat lost out of the top of the cup. I could even decide to conduct the experiment in a thermos flask, as this would insulate the solution better than the polystyrene cup and reduce the heat losses from every direction. My other source of error was the measuring cylinder that I used to measure out 50cm of acid and alkali. To reduce this error if I did the experiment again I would use a burette or a graduated pipette, as these pieces of equipment are more accurate and therefore the percentage error would be reduced making my experiment more accurate and reliable. To make my experiment even more reliable I could do repeat experiments and then take an average value for my enthalpy change from these results. This would also indicate to me how accurate I had been while conducting the experiments as close repeat results would indicate that I had been accurate and reliable.

My experimental result is also lower than the actual enthalpy change as my apparatus lost vast amounts of heat as they had no insulation of any kind. This made my overall enthalpy change -38.99kJ mol which would mean an acid similar to HCN was used in the experiment when actually I think that the enthalpy change should have been nearer -57 kJ mol and the acid used was probably more like citric acid.

Finally to improve the results, if I conducted the experiment again, I would use data logging to gather my results so that they would be more reliable as a computer would be reading the temperature not a person so no human errors would occur in gathering the results.

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