• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5

# The aim of this experiment is to find the enthalpy change for the decomposition of sodium hydrogen carbonate.

Extracts from this document...

Introduction

Assessed Practical: Planning Introduction: The aim of this experiment is to find the enthalpy change for the decomposition of sodium hydrogen carbonate. 2NaHCO = Na2CO3 + CO2 + H2O Using the enthalpy change of the following reactions. Sodium Hydrogen Carbonate: NaHCO3 + HCl = NaCl + CO2 + H2O Sodium carbonate: Na2CO3 + 2HCl = 2NaCl + CO2 + H2O Apparatus Sodium hydrogen carbonate Sodium carbonate Polystyrene Cup x 2 Measuring cylinder 50cm� x 2 Weighing scale Weighing boats Thermometer degrees Spatula HCl acid 2M Prediction Background Information Hess' Law states that the enthalpy change for a reaction is the same whether the reaction occurs directly or in steps. This is a direct consequence of the fact that enthalpy, is a state function. One of the applications of Hess' Law is to determine the enthalpy change for a reaction by combining other reactions to get the desired reaction, then combining the enthalpy changes for the reactions to get delta H for the reaction under consideration. An exothermic enthalpy change is always given a negative value, as the energy is lost from the system to the surroundings. An endothermic enthalpy change is always given a positive value as the energy is gained by the system from the surroundings. ...read more.

Middle

The Mr is 84. I will now use the following formula to find the mass of NaHCO3. Mass = Moles x Mr 0.1 X 84 Mass = 8.4g The mss of Sodium hydrogen carbonate I need is 8.4g. I will now find the amount of Sodium Carbonate. In this reaction I am using twice as much HCl acid so I will need to divide it by two top ensure I have the same no of moles in each reaction 100 x 2 = 0.2 moles 1000 0.2/ 2 = 0.1 moles. The Mr of Na2CO3 is 106. I will need to divide this by 2, because the stoichoimetry is 1:2. So if I have divided HCl acid by 2 I will also need to divide the mass of sodium carbonate by 2. 106 x 0.1 = 10.6 10.6 = 5.3g 2 The mass of Sodium carbonate I need is 5.3g. Method. Measure out approx 8.4g of sodium hydrogen carbonate into a weighing boat. Then fill the 50cm� measuring cylinder with HCl acid to the mark. Then pour the HCl acid into the polystyrene cup. Measure the temperature for a couple of minutes so you will get a steady temperature and then pour in some of the sodium hydrogen carbonate, a couple of seconds later add the rest of the sodium hydrogen carbonate. ...read more.

Conclusion

I could have also repeated the experiment twice so I could see that if the experiment was right, by comparing the end points. Results NaHCO3 Initial temperature 21 Final Temperature 11 change in temperature -10 Mass of HCl -50g Specific heat capacity- 4.2Jg�� K�� ?T- 10?C ?H= 4.2g��?C�� x 50g x 10?C = 2100J We then divide this by the no of moles used = 2100 = 21000J 0.1 21000 = 21KJ mol �� 1000 ?H= - 21KJ mol �� Na2CO3 Initial temperature 21 Final Temperature 28 Change in temperature 7 Mass of HCl used - 50g Specific heat capacity - 4.2g��?C�� ?T- 7?C ?H= 4.2g��?C�� x 50g x 7?C = 1470J We then divide by the no of moles used 1470 = 29400 0.05 29400 = 29.4KJ mol�� 1000 ?H= - 29.4 KJ mol�� It is a minus as it is a endothermic reaction. 2NaHCO3 = Na2CO3 + CO2 + H2O In order to get the enthalpy change for the above equation I would need to multiply the sodium hydrogen carbonate reaction by 2 and multiply the sodium carbonate reaction by -1. and add the two numbers together. Sodium hydrogen carbonate ?H= 21 x 2 = 42KJmol�� Sodium carbonate ?H= - 29.4 x -1 = 29.4KJ mol�� 42 + 29.4 = 71.4 KJ mol�� The enthalpy change for the decomposition of sodium hydrogen carbonate is 71.4 KJ mol��. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Aqueous Chemistry essays

1. ## How much Iron (II) in 100 grams of Spinach Oleracea?

present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20. 0.000302916 mol dm-3 X 20 = 0.00605832 mol dm-3 Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II)

2. ## An Experiment to Determine the Enthalpy Change for the Decomposition of Calcium Carbonate.

There are 2 moles of HCl (hydrochloric acid) in 1000cm� so there are 0.1moles of HCl (hydrochloric acid) in 50.0cm�. So I will use 50.0cm� of HCl (hydrochloric acid). So in this experiment I would want to use 5.0 grams of CaCO3 (calcium carbonate) and 50.0cm� of HCl (hydrochloric acid)

1. ## Indirect determination of enthalpy change of decomposition of sodium hydrogen carbonate by thermochemical measurement ...

Thermometer (graduated in 0.5?C divisions) ? Accurate weighing scales (correct to three decimal places) ? Weighing bottle (plastic, with lid) ? 25 cm3 burette ? Lab clamp ? Graph paper ? Dry sodium hydrogencarbonate (3.0g) ? Dry anhydrous sodium carbonate (2.0g)

2. ## Application of Hess's Law

NaHCO3 = (1672+2090+2194.5)/3 = 1985.5 Joules For 1 mole of NaHCO3 = 1985.5 Joules x 10 = 19,855 Joules Calculations for ?H2: Mass of Na2CO3 used = 6.32 grams, 6.38 grams and 6.33 grams Volume of HCl (2M) used = 50cm3 For ?H2 the temperature changes in order were 0.5?C,

1. ## The aim of this experiment is to answer the following question: What is the ...

= 0.01moles Now the difference between 0.01 and 0.0048 is equal to is 0.0052. Therefore the moles of ester and water formed is 0.0052moles. Now I can use these values to work out the equilibrium constant. As the equilibrium constant is found from concentrations and concentration is equal to moles/volume

2. ## Determining the enthalpy change for different chemical reactions.

DATA COLLECTION 2HCl (aq) + Na2CO3 (s)� 2NaCl (aq) + CO2 (g) + H2O (l) Mass of tube + sodium carbonate 28.17 g Mass of empty test tube 25.37 g Mass of sodium carbonate used (m) 2.80 g Temperature of acid initially 21.8 oC Temperature of solution after mixing 22.0 oC Temperature change during reaction (?T)

1. ## Copper carbonate decomposition

'At room temperature, 25?C and atmospheric pressure at 1 atmosphere 1 mole of ant gas will occupy a volume of 24dm3.' 1. This is useful because we have to use this to work out how much gas should be evolved from a specific measurement.

2. ## Investigation of the carbonate - bicarbonate system

consumed during the phenolphthalein indicator method. Methyl orange indicator method measures the buffering capacity of the neutralized solution. The bicarbonate (HCO3-) ion initially present with those produced during the half reaction is completely neutralized. Methyl orange produces a colour change at pH range 3.5 - 4.5.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to