Research Question: How can the heat of reaction from two separate equations be combined using Hess's Law to determine the degree of reaction of a solution of sodium hydroxide with dilute hydrochloric acid? Say no to plagiarism. Get a tailor-made essay on “Why Violent Video Games Should Not Be Banned”? Get the original essay Introduction: Hess's law has been the basis of thermochemistry for the last century. It states that the total enthalpy change of a reaction is independent of whether the reaction takes place in one or more steps. Using Hess's law, the heats of reactions for individual steps can be combined algebraically in the same way that reactions can be combined to form the overall reaction, and therefore the overall heat of reaction. This law is extremely important in many chemical evaluations because it is often difficult, if not impossible, to determine the heat of reaction in a single step. The chemical reaction must be broken down into simpler reactions, which can be measured directly. The first and second reactions are dissociation reactions and the heat released is therefore called the heat of dissociation. The third reaction, which is the overall reaction, is a neutralization reaction between an acid and a base. The overall heat released is therefore called heat of neutralization. Reactions 1 and 2 will then be combined to form reaction 3, their heats of dissociation being manipulated in the same way in order to form the heat of neutralization of reaction 3. Hypothesis: Looking at the three equations considered, my prediction is that if The opposite of the heat released by reaction 1 is combined with the heat released by reaction 2, then the overall heat of neutralization (reaction 3) of NaOH and HCl would form. Variables: In this experiment, the variables with which I will test applying Hess's law will be solid sodium hydroxide and a dilute hydrochloric acid solution. These independent variables, when combined in the correct chemical and possibly algebraic form, should be able to combine to determine the overall enthalpy of a larger reaction. This overall enthalpy acts as a dependent variable, while the manipulated individual enthalpies act as independent variables. These variables will be tested in a polystyrene calorimeter to try to control the heat released by each of the reactions. Materials and equipment: 2 polystyrene calorimeters 100 ml graduated cylinder Solid sodium hydroxide (NaOH). 25 M hydrochloric acid solution (HCl) ThermometersDigital balanceGlass stirrer rod200 ml beakerProcedure:Reaction 1:Put 100 ml of room temperature distilled water into the beaker and place it inside the calorimeter. Record the water temperature when it has reached a constant temperature. Measure and record the mass of approximately 2.5 grams of solid sodium hydroxide and place it in the water of the calorimeter. Stir continuously and simultaneously monitor the temperature of the solution until the water temperature is reached. the solid is completely dissolved. Record the highest temperature reached during this process. Discard the solution safely and rinse the beaker with water. Repeat this process for three trials to obtain an average value for the enthalpy of the reaction. Reaction 2: Repeat steps 1 to 4, but instead of 100 ml of distilled water, use 100 ml of 2.5 M hydrochloric acid. Repeat step 6 for three tests to obtain an average value of the enthalpy of the reaction. For thetwo reactions, the same equation to find the heat released by the reaction will be used: , where m is equal to the mass of the substance (g), C is the specific heat of water (J/gK) and is the change temperature (K). The specific heat of water is assumed to be 4.184 J/gK. Reaction 1: Calculation of the average temperature change for reaction 1 using the three paths and their uncertainties: Test 1) Highest possible temperature change: 28.5 – 21.5 = 7 °C Lowest possible temperature change : 27.5 – 22.5 = 5oCTrial 2) Highest possible temperature change: 27.5 – 21.5 = 6 °CLowest possible temperature change: 26.5 – 22.5 = 4 oCTrial 3) Change highest possible temperature change: 27.5 – 21.5 = 6 oCLowest possible temperature change: 26.5 – 22.5 = 4 oCAverages) Highest temperature change: oCLowest temperature change: oCChange in average temperature: oCThe temperature change must be in Kelvin for the calculation of the heat released:K = oC + 273.15, therefore, the average temperature change in K is:278.45KCalculation of the heat released by reaction 1 using the three tests and their uncertainties: Test 1) Highest possible: J Lowest possible: J Test 2) Highest possible: J Lowest possible: J Test 3) Highest possible: J Lowest possible: J Averages) Highest: JLowest: JAverage: 2924.24 JFrom these calculations, the average heat gained by the The calorimeter (J) of reaction 1, the dissociation of the NaOH(s) in water, is 2924.24 J. Therefore, the average heat released by the system is 2924.24 J and = -2924.24 J. Reaction 2: Calculate the average temperature change for reaction 1 using the three paths and their uncertainties: Test 1) Highest possible temperature change: 31.5 – 21.5 = 10 oCLowest possible temperature change: 30.5 – 22.5 = 8 oCTest 2) Highest possible temperature change: 30, 5 – 21.5 = 9 oCLowest possible temperature change: 29.5 – 22.5 = 7 oCTrial 3) Highest possible temperature change: 30.5 – 21.5 = 9 oC Highest temperature change lowest possible: 29.5 – 22.5 = 7 oC Average) Highest temperature change: oC Lowest temperature change: oC Average temperature change: oC The temperature change must be in Kelvin for the calculation of the heat released : K = oC + 273.15, therefore the average temperature variation in K is: 281.45 KCalculation of the heat released by reaction 1 using the three tests and their uncertainties: Test 1) The highest possible: JThe most lowest possible: JTest 2) Highest possible: JLowest possible: JTest 3) Highest possible: JLowest possible: JAverages) Highest: JLowest: JAverage: 2963.59JFrom these calculations, the average heat gained by the calorimeter (J) of reaction 1, the dissociation of NaOH(s) in HCl(aq), is 2963.59J. Therefore, the average heat released by the system is 2963.59J and = -2963.59J. Hess's Law: Now that the reaction of the two constituents has been calculated, it must be manipulated using Hess's law to find the heat of neutralization of HCl(aq) and NaOH(aq).NaOH(s) Na+ (aq) + OH-(aq) = -2924.24 JNaOH(s) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = -2963.59JNa+(aq) + OH-(aq) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = ?If reaction number one is reversed into a reverse reaction, then the products and reactants of 1 and 2 will cancel algebraically to form the net ionic equation for reaction 3. Reaction 1 is then multiplied by -1 because it is the reciprocal reaction. Na+(aq) + OH-(aq) NaOH(s) = 2924.24 JNaOH(s) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = -2963.59JNa+ (aq) + OH-(aq) + H+(aq ) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = ?Now for reaction 3 you can -44,51..