Research Question: How can the heat of reaction of two separate equations be combined using Hess's law to determine the heat of reaction of sodium solution sodium hydroxide with diluted hydrochloric acid? no to plagiarism. Get a tailor-made essay on "Why Violent Video Games Shouldn't Be Banned"? Get an original essay Introduction: Hess' law has been the basis of thermochemistry for the past century. It states that the total enthalpy change of a reaction is independent of whether the reaction occurs 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 thus the overall heat of reaction. This law is extremely important in many chemical evaluations, because it is often difficult or 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 therefore the heat released is called the heat of dissociation. The third reaction, which is the overall reaction, is a neutralization reaction between an acid and a base, and therefore the overall heat released is called the neutralization heat. Reactions 1 and 2 will then be combined to form reaction 3, their heats of dissociation will be similarly manipulated to form the heat of neutralization of reaction 3. Assumptions: Looking at the three equations in consideration, my prediction is that if the opposite of the heat released by reaction 1 is combined with the heat released by reaction 2, so the overall heat of neutralization (reaction 3) of NaOH and HCl would be formed.Variables: In this experiment, the variables I will be testing with apply the law of Hess will be solid sodium hydroxide and 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 in an attempt to control the heat released by each of the reactions. Materials and equipment: 2 x 100 mL polystyrene graduated cylinder calorimeters Solid sodium hydroxide (NaOH). 25 M hydrochloric acid solution (HCl) Thermometers Digital scale Glass stirring wand 200 ml beaker Procedure: Reaction 1: Place 100 ml of distilled water at room temperature 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 in the calorimeter. Mix continuously and simultaneously observe the temperature of the solution until a constant temperature is reached. the solid is completely dissolved. Record the highest temperature reached during this process. Discard the solution safely and rinse the glass with water. Repeat this process for three trials to obtain an average value for the enthalpy of the reaction. Reaction 2: Repeat steps 1-4, but instead of 100 ml of distilled water, use 100 ml of 2.5 M hydrochloric acid. Repeat step 6 for three trials to obtain an average value for the enthalpy of the reaction. For both reactions, the same equation will be used to find the heat released by thereaction: , where m is equal to the mass of the substance (g), C is the specific heat of water (J/gK) and is the temperature change (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 all three routes and their uncertainties: Test 1) Maximum possible temperature change: 28, 5 – 21.5 = 7 oCLowest possible temperature variation: 27.5 – 22.5 = 5oCTrial 2) Maximum possible temperature variation: 27.5 – 21.5 = 6 oCMinimum possible temperature variation: 26.5 – 22, 5 = 4 oCtrial 3) maximum possible temperature variation: 27.5 – 21.5 = 6 oCminimum possible temperature variation: 26.5 – 22.5 = 4 oCaverages) maximum temperature variation: oClowest temperature temperature variation: oCA average temperature variation: oCThe temperature variation must be expressed in Kelvin for the calculation of the heat released:K = oC + 273.15, therefore the average temperature variation in K is:278.45KCalculation of the heat released by reaction 1 using all three tests and their uncertainties:Test 1) Highest possible: JLowest possible: JTest 2) Highest possible: JLowest possible: JTest 3) Highest possible: JLowest possible: JMean ) Highest: JLowest: JAverage: 2924.24 JFrom these calculations, the average heat gained by the calorimeter (J) of reaction 1, the dissociation of NaOH(s) in water, is 2924.24 J. Therefore, the average heat released by the system is 2924.24 J e = -2924.24 J. Reaction 2: Calculation of the average temperature change for reaction 1 using all three paths and their uncertainties:Test 1) Maximum possible temperature change: 31.5 – 21.5 = 10 oCLowest possible temperature variation: 30.5 – 22.5 = 8 oCTrial 2) Maximum possible temperature variation: 30.5 – 21.5 = 9 oCLowest possible temperature variation: 29.5 – 22.5 = 7 oCTrial 3) Maximum possible temperature variation: 30.5 – 21.5 = 9 oCMinimum possible temperature variation: 29.5 – 22.5 = 7 oCaverage variations) Maximum temperature variation: oCLowest temperature change: oCAverage temperature change: oCThe temperature change will need to be in Kelvin to calculate the heat released:K = oC + 273.15, therefore, the average temperature change in K is:281.45KCalculation of heat released by reaction 1 using all three tests and their uncertainties:Test 1) Highest possible: JLowest possible: JTest 2) Highest possible: JLowest possible: JTest 3) Highest possible: JLa lowest possible: JAverage) Highest: JLowest: JAverage: 2963.59JFrom these calculations, the average heat gained by the calorimeter (J) from reaction 1, the dissociation of NaOH(s) into HCl(aq), is 2963 ,59J. Therefore, the average heat released by the system is 2963.59J and = -2963.59J. Hess's Law: Now that you have calculated the value of both constituent reactions, you need to manipulate it 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 in the reverse reaction, then the products and reactants of 1 and 2 will cancel out a each other algebraically to form the net ionic equation of 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 it can be determined by adding the values ​​of reactions 1 together and 2. -35.39 JThe accepted value for the heat of neutralization between acids and bases is -44.51 J, so it can be used to -44,51.