• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Validating Hess's law

Extracts from this document...

Introduction

Research Question: Is the enthalpy change for a given chemical change the same whether the reaction takes place in a single stage or via several stages, provided the initial and final conditions are the same. Introduction: Hess's Law (1840) states that for a given chemical change the enthalpy change is the same whether the reaction takes place in a single stage or via several stages, provided the initial and final conditions are the same. We will test the validity of this law using the reaction between sodium hydroxide and hydrochloric acid. The reaction between solid sodium hydroxide and dilute hydrochloric acid can be carried out in two ways. Method 1 NaOH(s) + HCl(aq) NaCl(aq) + H2O(l) ?H?1 Method 2 NaOH(s) NaOH(aq) ?H?2 then NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ?H?3 According to Hess's Law ?H?1 = ?H?2 + ?H?3 In calculating the enthalpy of reaction values in each of the above and following cases, it is assumed that: a. The density of the solutions is 1 gcm-3 b. ...read more.

Middle

+ HCl(aq) NaCl(aq) + H2O(l) Initial Temperature = 26.5�C ?0.1�C Maximum Temperature = 48.1�C ?0.1�C ?T = Maximum Temperature - Initial Temperature = 48.1�C - 26.5�C = 21.6�C ?0.2�C % Error in ?T = The enthalpy change ?H?1 is given by multiplying the mass of HCl (m) by its specific heat capacity (Cp) and the change in temperature (?T). Since this reaction takes place in solution, which we assume to be mostly water. Hence the mass of the solution will be 50g and the specific heat capacity is given as 4.2J/g/�C. The calculation for the mass of the solution is as follows: Mass = Density � Volume (Density given as 1g/cm3) Mass = 1g/cm3 � 50cm3 Mass = 50g Hence, ?H?1= m � Cp � ?T ?H?1 = 50g � 4.2J/g/�C � 21.6�C ?H?1 = 4536J ?0.926% ?H?1 = 4.536kJ ?0.042kJ Mass of NaOH Pellets: 2.20g ? 0.01g % Error in mass = 2.20g of NaOH produces 4.536kJ ?0.042kJ 1 mole (40g) will produce: ? ...read more.

Conclusion

Mass = 1g/cm3 � 50cm3 Mass = 50g Hence, ?H?3 = m � Cp � ?T ?H?3 = 50g � 4.2J/g/�C � 9.9�C ?H?3 = 2079J ?2.02% ?H?3 = 2.079kJ ?0.042kJ Volume of NaOH: 0.05dm3 ? 0.0001dm3 % Error in volume = 0.05dm3 of NaOH produces 2.079kJ ?0.042kJ 1 mole (1dm3) will produce: ? 2.22% 1 mole (40g) will produce 41.6kJmol-1 ? 0.924kJmol-1 Conclusion and Evaluation: According to Hess's Law ?H?1 = ?H?2 + ?H?3 So using the results found above we see that: ?H?1 = 82.5kJmol-1 ? 1.14kJmol-1 ?H?2 + ?H?3 = 38.7kJmol-1 + 41.6kJmol-1 = 80.3kJmol-1 ? 1.944kJmol-1 It can be seen that ?H?1 is almost equal to ?H?2 + ?H?3 but due to experimental errors such as heat being lost to the surroundings and not having exact readings they are not equal. Hess's law has therefore been validated but the results would have been much more accurate if there was an insulating capsule around the polystyrene cup so as to prevent any heat from being lost to the environment and by having more accurate apparatus in terms of measuring cylinders and thermometers. ?? ?? ?? ?? ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate 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

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Free essay

    Vitamin C in Fruit Juices

    and reconstituted FCOJ almost always have higher levels of vitamin C and is above the 100% US RDA values. This is most likely due to blending of early-season fruit with late season fruit. Canned single strength orange juice will have lower vitamin C levels due to heating during the canning process.

  2. Hesss Law Lab, use Hesss law to find the enthalpy change of combustion of ...

    The percentage error for the three trials is as under Percentage error for the first trial: = (table value - experimental value) = 602- 561 = 40 = Percentage error for the second trial: = ( table value - experimental value)

  1. Enthalpy Change Design Lab (6/6)How does changing the initial temperature (19C, 25C, 35C, and ...

    One, it would be tedious and unfeasible to wait for the temperature of both solution of 1.00 mol dm-3 KOH(aq) and HCl(aq) to drop to lower degrees, such as 10�C. Also, if a lower temperature is selected, then it complicates to process of finding the maximum temperature measured during the reaction of the solution.

  2. Enthalpy and Hess law

    Then using a measuring cylinder 50 ml of 2M of HCl was measured and poured into a can. A thermometer was used to record the temperature of the hydrochloric acid in the can. Afterwards NaOH was added into the can.

  1. experiment Hess Law

    mole = = 0.85 % Total = 9.25% ?Hrxn = -240.25 kJmol-1 � 9.25 % Part B:- Mass of magnesium oxide = 0.4985 � 0.0001g Volume of 0.5 M of HCl = 50.0 � 0.5 cm3 Reactant Initial Temperature, � 0.5�C Highest Temperature, � 0.5�C MgO + HCl 28.0

  2. To determine the standard enthalpy of formation of Magnesium Oxide using Hess Law.

    To carry out the arithmetic calculations required, values of the reaction enthalpy of the all the three reactions were needed but since the value for ?HH2O had already been provided to us by the lab in-charge, all that was needed to be determined were the values for ?HX and ?HY.

  1. Chemistry Internal Assessment Hesss Law

    This experiment could be improved, if the experiment was carried out in a more insulated container, which would allow less heat to be gained from the surroundings. Uneven stirring The frequency of stirring between each trial was altered as different people stirred the solution, this would have caused the rate of reaction to change between trials.

  2. Research question: how to convert NaOH to NaCl by two different routes , and ...

    50.00 cm3 of HCl is added to the solution. 3. Temperature of the solution is collected each 30 seconds . 4. The end product will be 100.0 cm3 of NaCl with the concentration of 1 mol dm-3 Route (B): 1.

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