Lab Report Nr 6 - lab 6 PDF

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lab 6...

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Thermochemistry: Heat of Neutralization and Hess’s Law Intro: When there is an energy change or most of the time described a change in hear content, energy is either released (exothermic) or absorbed (endothermic). Creation of new bonds in products side involves the release of energy, because the reactants also require consumption of energy in order for a chemical reaction to happen. The heat content of a system is the potential energy stored in chemical bonds, called the enthalpy. The change in enthalpy (∆H) happens when these chemicals react and there is a change in energy. Properties of the reactants and products play a role in the overall change in energy. The study of heat transferred during chemical reactions is called calorimetry. We use calorimeter to measure the changes in temperature. During this process the amount of heat released or absorbed is measure by changes in temperature.  A very important equation for calculating the heat associated with the temperature change is: q= msΔT q= heat, m= mass, s= specific heat (J g-1 oC-1), ΔT= Temperature change S is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. 

A very important equation is the relationship between enthalpy change and heat: ∆H = qp

∆H is the enthalpy change and q is the heat. Enthalpy is a property of a substance that can be applied to determine the heat absorbed or released in a chemical reaction. The “p” in this equation denotes that the reaction occurs at constant pressure. Many reactions are open to a constant atmospheric pressure. Hess’s Law: We cannot measure heat or enthalpy, so we have to calculate it in the example below. If a reaction is carried out in a series of steps, we can find ΔH as the overall process. Furthermore, adding the enthalpy change of each individual step. The example below explains the steps calculating enthalpy change to transform graphite into diamonds.

2 C (s,graphite) + 2O2 (g) → 2 CO2 (g)

ΔH1= -787 kJ

C (s,diamond) + O2 (g) → CO2 (g)

ΔH2= -396.5 kJ

C (s,graphite) → C (s, diamond)

ΔHoverall= __kJ

CAREFUL: We cannot add equations one and two to get the ΔHoverall of equation three because there would be 3 oxygen molecules in the reactants and 3 carbon dioxide molecules in the products and our overall desired reaction three does not contain any O2 and/or CO2 molecules. How to manipulate the chemical equations to get overall reaction of 3rd equation. 1st WAY. multiply/divide the chemical equation by a coefficient and/ 2nd WAY reverse a chemical equation When we use the first or second way, the corresponding ΔH must undergo the same thing. In this example, we can flip reaction.

C (s,graphite) + O2 (g) → CO2 (g).

ΔH1= -393.5 kJ

Divide by 2

CO2 (g) → O2 (g) + C (s,diamond)

ΔH2= 396.5 kJ

Flip the equation

C (s,graphite) → C (s, diamond)

ΔHoverall= 3 kJ

Add 1 and 2

For equation 2 we flipped the sign as well!!! For equation 1 we divided everything with 2, the ΔH1 as well. -393+396.5 = 3

Procedure: We will measure the enthalpy change that occurs in three separate exothermic acid/base reactions involving 1. a) NaOH and HCl 2. b) NaOH and CH3COOH 3. c) NH3 and HCl GOAL: We will calculate the respective enthalpy change for each reaction. Assuming the specific heat of each reactions is 4.18 J/g OC and the density of the solutions is 1.0 g/cm3. Materials Used: One thermometer Two styrofoam cups (calorimeter) One cup lid 50 mL graduated cylinder 250 mL beakers Stirring Rod Timer ----------------------------------------------------------1.0 M hydrochloric acid 1.0 M ammonia 1.0 M acetic acid 1.0 M sodium hydroxide Distilled water 

Write the objective of your experiment

We need to find the ΔH, an enthalpy change, once we add a base and acid together. Our goal is to get an Exothermic Reaction, a release of energy. 

State the three acid/base exothermic chemical reactions in your experiment 1. Hydrochloric acid (acid) with Sodium Hydroxide (strong base)

2. Sodium Hydroxide (strong base) with Acetic Acid 3. Ammonia (weak base) with Hydrochloric acid 

State equation(s) you will use to find the change in enthalpy

1. NaOH + HCl 2. NaOH + CH3COOH 3. NH3 + HCl -----------------------------4. NH3 + CH3COOH

Data: Given: Concentration of Each Reactant (given): 1.0 M Specific Heat of Water for Each: 4.18 J/g oC Density of Water: 1.0 g/cm3 25 mL of each reactant

ΔH

ΔT

1.NaOH + HCl -> NaCl + H2O

-51.832

6.2 oC

2.NaOH + CH3COOH -> NaCH3COO+ H2O

-49.324

5.9 oC

3.NH3 + HCl -> NH4Cl

-53.504

6.4 oC

1.NaOH + HCl -> NaCl + H2O

We flip to cancel H2O in both sides = ΔH1 ,sign changes to pos. NaCl + H2O -> NaOH + HCl

2.NaOH + CH3COOH -> NaCH3COO+ H 2O

NaOH + CH3COOH -> NaCH3COO+ H2O

3.NH3 + HCl -> NH4Cl

NH3 + HCl -> NH4Cl

NH3 + CH3COOH -> NH4 + CH3COO

Na+Cl- + CH3COOH + NH3 -> Na+CH3COO- + NH+4ClCH3COOH + NH3 -> CH3COO- + NH+4

Calculations: To find the mass for each we have d=mass/volume => mass=volume * density= 1.0 * 25 mL= 20g= 0.025 kg for each To find ΔH of each equation we use the formula ΔH=q= msΔT After calculating for joules, we calculate to kJ, hence 1 J= 0.001 kJ

25mL= 0.025 L for each reactant Molarity= moles/volume => moles= molarity * volume= 0.1 M * 0.025L= 0.025 moles ΔH1= 50g *4.18 J/g oC *6.2 oC = 1295.8 J (joules) = 0.001 * 1295.8 = 1.2958 kJ / 0.025 moles = 51.832 kj/mol ΔH2= 50g * 4.18 J/g oC * 5.9 oC = 1233.1 J = 0.001 * 1233.1 = 1.2331 kJ / 0.025 moles = 49.324 kj/mol ΔH3= 50g * 4.18 J/g oC * 6.4 oC = 1337.6 J = 0.001 * 1337.6 = 1.3376 kJ / 0.025 moles = 53.504 kj/mol Since our 3 first chemical reactions are exothermic reactions, they will all be negative. Except for 1. which changed sign because we flipped the equation in order to cancel out. ΔHoverall = 51.832 + (-49.324) + (-53.504) = -50.996 kj/mol %error= We found the experimental value through calculations and we are given the theoretical value already which is -50.4. % error= E-T/T * 100= -50.99-(-50.4)/-50.4 * 100= 1.2%

Discussion and Conclusion: This experiment was successful and helped us meet our goal and objective set at the beginning of this experiment. Using calorimetry and Hess’s Law is a perfect combination to find overall heat of neutralization. We found the heat of neutralization for each equation using our information given, also using our formulas learnt in previous chapters. Was your objective met? Yes, my objective was met. How are calorimetry and Hess’s Law used to find the heat of neutralization of ammonia and acetic acid? The reaction between these two is ammonium acetate. And the reaction between an

acid and a base is called neutralization reaction. Also, important to mention delta H found is called heat of neutralization. Without using the formula of “q” we cannot use the Hess’s Law. Calorimetry helps us to find the heat and obtain the enthalpy information. If we have the information provided, we can conclude to Hess’s Law. Why are the heat of neutralizations values negative? The reaction between an acid and a base is always exothermic. The strongest the acid and the strongest the base, the more exothermic is the reaction going to be. Explain why it is it important to know if energy is being released or absorbed during a chemical reaction? That helps us when we put the negative or positive sign in order to proceed with Hess’s Law. If we do not have this information our calculations are not going to be accurate and may result incorrect. What are some sources of error in your set-up and how can they be reduced to achieve a greater accuracy? Errors can occur if we are not sure if energy is being released or absorbed. Other errors can possibly occur while preparing the mixtures. We have to be precise with the temperature we achieve at the beginning. That is why several tries are needed in order to obtain the exact temperature needed. We have to be patient with our timing during the performance of the experiment....