Question

In this experiment, you will determine and compare the quantity of heat energy released in three exothermic chemical reactions through application of Hess’s law.

Reaction 1: NaOH(s) → Na⁺(aq) + OH^-(aq) + x₁ kJ

Reaction 2: NaOH(s) + HCl(aq) → H₂O(l) + Na⁺(aq) + OH^-(aq) + x₂ kJ

Reaction 3: NaOH(aq) + HCl(aq) → H₂O(l) + Na⁺(aq) + OH^-(aq) + x₃ kJ

In order to accurately measure the heat released in each reaction, we will be using a calorimeter. As discussed in the textbook, a calorimeter is a device used to measure the amount of heat involved in a chemical or physical process; the calorimeter used in this experiment will be a Styrofoam cup.

In the reactions above, the variables x₁,x₂, and x₃ are the heats that are evolved during the reactions. The change in temperature that occurs for each reaction will be used to calculate the energy released in kJ/mol of NaOH used. We will assume for our calculations that any heat transferred to the Styrofoam cup and surrounding air will be negligible; we will also assume that the specific heat of water is 4.184 J/g∙°C. Thus, the applicable equation is simply

qreaction=-m∙Cwater∙∆T

There are several objectives of this lab:

1. Explore the technique of calorimetry, and
2. Calculate and interpret heat and related properties using typical calorimetry data.

for reaction 1: 50.0 mL of H2O was transferred to a foam cup. Waters initial temp was 25 deg, then 1.00 g of solid pellets NaOH was added weighing at 1.0000g. Highest temperature was 30.3 deg.

for reaction 2: 50 mL 0f the 0.50 M HCI solution was added to foam cup. Initial temp was 25 deg, then 1.00 g of solid pellet NaOH was added weighing 0.5496 g. Highest temp was 37.0 deg

for reaction 3: 25.0 mL of the 1.0 M HCI solution was added into the foam cup. initial temp was 25 deg. 25.0 mL of the 1.0 M NaOH solution was added to the foam cup, initial temp was 25 deg. after mixing the two solutions the highest temp was 31.7 degrees.

1. The energy, x₁, in Reaction 1 represents the energy of solution for one mole of solid NaOH. Look at the net ionic equations for Reactions 2 and 3, and make a similar statement concerning the significance of x₂ and x₃ and their connection to Hess’s law. Record your statements in the ‘Notes’ section below for review later.
   
2. Find the difference between the value of x₂ and the sum of x₁ plus x₃.; let x₄ be equivalent to the sum of x₁ and x₃ and let x₅ be the difference between x₂ and x₅. This calculation scheme is shown below. Make notes about any similarities or differences between the values in your notes.
   
   x₄ = x₁ + x₃
   x₅ = x₂ – x₄
   
3. Calculate the percent difference between x₂ and x₄ according to the equation below. (Assume x₂ to be the accepted value.) Record the calculation in your notes.
   
   % difference=experimental value-accepted valueaccepted value×100%

% difference=x4-x2x2×100%

Thermochemical Data
	Tinitial (°C)	Tfinal (°C)	ΔT (°C)	moles NaOH	qreaction (kJ)	ΔHrxn
Reaction 1
Reaction 2
Reaction 3

*Note: You can copy and paste your data table into your report later.

Insert the net ionic equations here:

Reaction 1:

Reaction 2:

Reaction 3:

Answer 1

reaction 1: 50.0 mL of H₂O . density of water = 1g/ml m (water ) = 50 g

; 1.00 g of solid pellets NaOH

Water  initial temp was 25 deg ; Highest temperature was 30.3 deg.

mass = 50 g

Q (reaction ) =  -m* Cwater *∆T = - 50g *4.184 J/g∙°C * 5.5 C = -1150.6 J = -1.1506 kJ

ΔH_rxn = -1.1506 kJ

net ionic equations here: NaOH (aq) -----> Na⁺ (aq) + OH^-(aq)

reaction 2: 50.0 mL 0.50 M HCI solution  .

; 1.00 g of solid pellets NaOH

mass = 50 g

Water  initial temp was 25 deg ; Highest temperature was 37.0 deg.

mass = 50 g

Q (reaction ) =  -m* Cwater *∆T = - 50g *4.184 J/g∙°C * 12 C = -2510.4 J = -2.5104  kJ

ΔH_rxn = -2.5104  kJ

net ionic equations : H⁺ (aq) + OH^-(aq) -------> H₂O (l)

reaction 3: 25.0 mL of the 1.0 M HCI solution. density of solution  

then 1.0 M NaOH solution was added to it.

total volume = 50 ml , so mass = 50 g

Water  initial temp was 25 deg ; Highest temperature was 31.7 deg.

Q (reaction -3 ) =  -m* Cwater *∆T = - 50 g *4.184 J/g∙°C * 6.7 = -1401.64 J = = -1.4016 kJ

   ΔH_rxn = -1.4016 kJ

net ionic equations: H⁺ (aq) + OH^-(aq) -------> H₂O (l)