Enthalpy of Neutralisation of Water Temperature Probe

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The Enthalpy of Neutralization of Water

Experimental ProblemDetermine the products formed by the decomposition reaction of hydrogen peroxide.

Educational Purpose

To introduce stoichiometric calculations and chemical reactions.

Background

Thermodynamics, or the relation between heat and energy is an extremely old science, dating

back to the nineteenth century when in 1824, Carnot published his Reflections on the Motive

Power of Fire, which detailed his research and presented a well-reasoned theoretical treatment

for the perfect (but unattainable) heat engine, now known as the Carnot cycle

This was a few years before Rudolf Clausius who also stated the second law in a paper

published in 1854 as,” Heat can never pass from a colder to a warmer body without some other

change, connected therewith, occurring at the same time”

The fundamental thermodynamic variables that are measured in the laboratory are the familiar

variables of pressure, temperature and volume and from these are calculated through various

mathematical manipulations, the thermodynamic functions, energy, E; enthalpy, H; Gibb’s free

energy, G; and Helmholtz free energy, A.

The First Law of Thermodynamics is the law of conservation of energy applied to macroscopicsystems and introduces the concepts of work and heat and system and surroundings. Since

energy is a state function (i.e. the value of the property is dependent on the initial and final states

only), while both work and heat are path functions (i.e. the value of the property is dependent on

the path taken to go from the initial to the final state), the function E can be mathematically

defined by

dU = q + w

Experimental Problem

Determine the enthalpy of neutralization of water.

Educational Purpose

To introduce thermodynamic calculations and chemical reactions.To introduce Vernier Logger Pro as a laboratory data collection method

To study regression analysis

What To Turn In

1. The prelab and postlab

http://library.thinkquest.org/C006011/english/sites/thermo3.php3?v=2






where dU is the change in energy of the system when the system undergoes a heat transfer of q

units to the system and w units of work are done on the system.

Similarly, enthalpy, H can be defined by,

dH = dU + PdV

where P and V are the usual laboratory variables of pressure and temperature and U has already

been defined as the energy.

The other thermodynamic variables, the Helmholtz free energy, A, and the Gibb’s free energy,

G, can be likewise defined in terms of U and H and the laboratory variables of temperature,

pressure and volume through the relationships,

dA = dU – TdS

for isothermal processes, where S is the entropy of the system and

dG = dH – TdS

once again restricted to isothermal processes.

In the laboratory, temperature changes in chemical reactions are easily measured and have been

used extensively to elucidate various thermochemical values. This has been achieved through theuse of Hess’ Law which states that the heat evolved or absorbed in a chemical process is the

same whether the process takes place in one or in several steps. This is also known as the law of

constant heat summation. To illustrate Hess's law, This example problem demonstrates strategies

and how to use Hess's Law to find the enthalpy change of a reaction using enthapy data from similar

reactions. Diagrammatically one can show a set of reactions as below,

0 kJ ------------ C(graphite) + O2

| |

-110 kJ | |

V |

CO + 0.5 O2 ----- |

| | -393 kJ| |

-283 kJ | |

| |

V V

------------ CO2

For reactions taking place in solution and under laboratory conditions, (Constant P, varying T

and V), the basic definitions ΔH involving an energy term dU is not suitable and must be altered

so that fundamental properties can be measured. In the chemical environment where one deals

with mixing reagents it is volume that is altered and temperature that changes as a consequence

of reaction. The variation of ΔH used in this case depends on knowing the heat capacity

(Cp)(energy required to raise 1mol of reagent by 10K) of all reagents (reactants and products)

involved in the reaction and the equation used is

ΔHrxn = ΔH prod - ΔHreact

Where ΔH of each individual component is calculated by ΔH = mCp ΔT



Problem:

To evaluate the enthalpy of formation of CO

C + O2 -> CO2, dH ° = -393 kJ/mol

CO + 1/2 O2 -> CO2, dH ° = -283 kJ/mol.

Subtracting the second equation from the first gives

C + 1/2 O2 -> CO, dH ° = -393 -(-283) = -110 kJ/mol

Problem:

What is the value for ΔH for the following reaction?

CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)

Given:

C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol

S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol

C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol

Solution

Hess's law says the total enthalpy change does not rely on the path taken from beginning to end.

Enthalpy can be calculated in one grand step or multiple smaller steps.

To solve this type of problem, we need to organize the given chemical reactions where the total

effect yields the reaction needed. There are a few rules that must be followed when manipulating

a reaction.

1. The reaction can be reversed. This will change the sign of ΔHf .2. The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the

same constant.

3. Any combination of the first two rules may be used.

Finding a correct path is different for each Hess's law problem and may require some trial and

error. A good place to start is to find one of the reactants or products where there is only one

mole in the reaction.

We need one CO2 and the first reaction has one CO2 on the product side.



C(s) + O2(g) → CO2(g), ΔHf = -393.5 kJ/mol

This gives us the CO2 we need on the product side and one of the O2 moles we need on the

reactant side.

To get two more O2 moles, use the second equation and multiply it by two. Remember to

multiply the ΔHf by two as well.

2 S(s) + 2 O2(g) → 2 SO2(g), ΔHf = 2(-326.8 kJ/mol)

Now we have two extra S and one extra C molecule on the reactant side we don't need. The third

reaction also has two S and one C on the reactant side. Reverse this reaction to bring the

molecules to the product side. Remember to change the sign on ΔHf .

CS2(l) → C(s) + 2 S(s), ΔHf = -87.9 kJ/mol

When all three reactions are added, the extra two sulfur and one extra carbon atoms are cancelled

out, leaving the target reaction. All that remains is adding up the values of ΔH f .

ΔH = -393.5 kJ/mol + 2(-296.8 kJ/mol) + (-87.9 kJ/mol)

ΔH = -393.5 kJ/mol - 593.6 kJ/mol - 87.9 kJ/mol

ΔH = -1075.0 kJ/mol

Answer:

The change in enthalpy for the reaction is -1075.0 kJ/mol.

Problem

How much energy must be absorbed by 20.0 g of water to increase its temperature from 283.0 °C

to 303.0 °C? (Cp of H2O = 4.184 J/g °C)

Solution

ΔH = mCp ΔT = 20.0 x 4.184 x (303.0 – 283.0) = 1673.6 J

Problem

When 10mL of 1.00M AgNO3 solution is added to 10mL of 1.00M NaCl solution at 250C in a

calorimeter, a white precipitate of AgCl forms and the temperature of the aqueous mixture rises

to 32.60C. Assuming that the specific heat of the aqueous mixture is 4.18J/g/

0C, that the density



of the mixture is 1.00g/mL and that the calorimeter absorbs a negligible amount of heat,

calculate H (in kJ) for the reaction.

Solution

Because the temperature rises in the reaction, heat must be liberated and H must be negative.

The amount of heat evolved during the reaction is equal to the amount of heat absorbed by the

mixture:

Heat evolved = (mass) x (specific heat) x (Temp change)

Mass = 20mL x (1.00g/mL) = 20g

Specific Heat = 4. 18J/g/0C

Temp Change = 32.60C -25.0

0C = 7.6

0C

Heat Evolved = 20(g) x 4. 18J/g/0C x 7.6

0C = 6.4 x 10

2J

According to the balanced equation, the number of moles of AgCl produced equals the numberof moles of Ag

+ reacted:

Moles of Ag+ = (10mL)(1.00mol/1000mL) = 1.00x10

-2 mol Ag

+

Moles AgCl = 1. 00x10-2

mol AgCl

Heat evolved per mole of AgCl = 6.4 x 102J/ 1. 00x10

-2 mol AgCl

= 64 kJ/mol AgCl

Therefore ΔH = -64kJ

Problem

From the equations below derive the enthalpy for the reaction

dH °f

/(kJ/mol)

(1) H2(g) + 0.5 O2(g) -> H2O(l) -285.8

(2) C(graphite) + O2(g) -> CO2(g) -293.5(3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) -890.4

1. Calculate the enthalpy of the reaction

CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g)

Use the data from the following reactions

C + 2H2 -> CH4



CH4 + 2O2 -> CO2 + 2H2O dH o = -890 kJ/mol

H2O(l) -> H2O(g) dH o = 44 kJ/mol at 298 K

Objective

In our experiment we shall use the dH °f of the various components present in the reaction

mixture in order to compute the heat of neutralization of water.

1. Measure the change in temperature of the reaction between NaOH and HCl

2. Calculate the enthalpy of neutralization of water

3. Compare the calculated value with the accepted value

4. We shall use dH °f(NaOH) = -426.7kJ/mol, dH °f(HCl) = -92.3kJ/mol,

dH °fNaCl) = -411.0kJ/mol

Procedure: Work in pairs

1. Connect a Temperature probe to Channel 1 of the Vernier computer interface

2. Start the Logger Pro program on your computer.

3. Place a stir plate on the ring stand and plug into the stir function at a low speed.

4. Measure out 50 mL of 0.2M NaOH into the insulated beaker and place on stir plate.

5. Place a clean, dry stir bar into the beaker and make sure that the solution does not

splash

6. Suspend the temperature probe from the ring stand, lower into the NaOH solution.

CAUTION: do not let the probe touch the stir bar

7. Rinse and then fill a buret with 0.2M HCl and position above the beaker.

Conduct the experiment.

a. Your monitor should show temperature on the y axis and volume on the x-

axis.

b. Setup data collection parameters: Goto ‘Experiment’ Data Collection

Mode ‘Time based’ Duration 350sec Sampling rate 5 seconds per

sample. Done

c. Click *Collect* to begin the data collection and to obtain the initial

temperature of NaOH.d. When the initial temperature is stable, (15 sec, 4 or 5 readings at the same

temperature) start adding HCl

e. The temperature of the solution will start to rise.

f. Add HCl solution 2mL at a time until 50mL have been added.

g. After the run is over analyze the data with ‘Analyze’

i)Find the initial temp of the NaOH solution using the R= function



ii) Take the sloping down portion of the curve and interpolate to the y-axis

to find the best ‘final temp.’

iii) Use these and information given to you to find H for the

neutralization of H2O.

h. Submit it with your postlab. Make sure that all information is clearly marked,

all axes are labeled, the graph is titled and the difference in temperature

shown.

Postlab

1. What are Ti and Tf for your reaction solutions and what are the R 2 values for each?

2. For each run, what is the calculated value of Hneutralization(H2O)?

3. Take the average of the values you have obtained - what is the % error in your value?

4. Calculate H0

for the fermentation of glucose to make ethyl alcohol, the reaction that

occurs during the production of alcoholic beverages:

C6H12O6(s) 2C5H5OH(l) + 2CO2(g)Using the heats of formation information: H

0f (C6H12O6) = -1260 kJ,

H0

f (C2H5OH) = -277.7kJ , H0

f (CO2) = -393.5 kJ

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Enthalpy of Neutralisation of Water Temperature Probe