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Thermochemistry : The Study of heat change in chemical reactions

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Topic 6: Part I

Thermochemistry

Thermochemistry: The Study of heat change in chemical reactions

- Energy – the capacity to do work or produce heat
- Types of Energy:
- Radiant – solar energy, comes from the sun
- Thermal – energy associated with the random motion of atoms and molecules
- Chemical – energy stored within the structural units of chemical substances
- Potential – energy available by virtue of an object’s position

- Every chemical reaction obeys two fundamental Laws: (1) The Law of Conservation of Mass, and (2) The Law of Conservation of Energy

- Energy is the capacity to do work or produce heat.
- Law of Conservation of Energy:
energy can be converted from one form to another, but not created or destroyed.

- Potential – energy of position or composition
- Kinetic – energy of motion

- Temperature reflects the avg KE of the molecules.
- Heat involves the transfer of energy between two objects.

- Change in energy (∆E) is calculated using heat (q) and work (w).
- It is a state function – meaning it is independent of the pathway, or how you get from point A to B.
- Work is a force acting over a distance.
- Heat is energy transferred between objects because of temperature difference.

- The Universe is divided into two parts:
- the system (sometimes a reaction)
- and the surroundings.

- The system is the part we are concerned with.
- The surroundings are the rest.
- Exothermic reactions release energy to the surroundings.
- Endothermic reactions absorb energy from the surroundings.

Heat

Potential energy

Heat

Potential energy

- Every energy measurement has three parts.
- A unit ( Joule, J= kg∙m2 / s2 ).
- A number indicating how many
- A sign to tell direction.
- negative - exothermic
- positive - endothermic

Surroundings

System

Energy

DE <0

Surroundings

System

Energy

DE >0

- Heat given off is negative.
- Heat absorbed is positive.
- Work done by system on surroundings is negative.
- Work done on system by surroundings is positive.
- Thermodynamics- The study of energy and the changes it undergoes.

- The energy of the universe is constant.
- Law of conservation of energy.
- q = heat
- w = work
- DE = q + w
- Take the system’s point of view to decide signs.

- abbreviated H
- H = E + PV (that’s the definition)
- E = internal energy of a system
- P = pressure of the system
- V = volume of the system

- at constant pressure.
- DH = DE + PDV

- the heat, q, at constant pressure can be calculated from
- DE = q + w (w = - PDV)
- Substitute for work (w) and solve for heat (q)

- q = DE + P DV = DH
- Therefore q = DH at constant pressure
- We use DH as a measure of the change in enthalpy or heat in joules of a system.

- Heat is a measure of change in enthalpy of a system at constant P.
- Heat of a reaction = change in enthalpy
- For a chemical reaction:
- ∆H = Hproducts – Hreactants

- + ∆H = heat absorbed by system (endothermic)
- - ∆H = heat lost by system (exothermic)

- 6.4 When one mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ∆H for a process in which a 5.8g sample of methane is burned at constant pressure.

- Measuring heat.
- Use a calorimeter.
- Two kinds
- Constant pressure calorimeter (called a coffee cup calorimeter)
- heat capacity for a material, C, is the measure of the energy needed to raise the temp. of an object 1oC.
- C= heat absorbed/ DT = DH/ DT

- specific heat capacity, s, is the heat capacity given per gram of substance (C/g).
- molar heat capacity = C/moles
- the heat of reaction, q = s x m x DT
- S = specific heat capacity
- m = mass of substance
- DT = change in temperature

- A coffee cup calorimeter measures DH.
- An insulated cup, full of water.
- The specific heat of water is 4.18J/ºC∙g
- Heat of reaction= DH = s x m x DT

- If 50ml of 1.0M HCl at 25oC is reacted with 50ml of 1.0M NaOH at 25oC. The final temperatrue of this exothermic reaction is 31.9oC.
- Calculate the change in heat
- Calculate ∆H in kJ/mol

- When 1.00 L of 1.00 M Ba(NO3)2 solution at 25oC is mixed with 1.00L of 1.00M Na2SO4 solution at 25oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC.
- Assuming the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/oC ∙ g and that the density of the final solution is 1.0g/mL, calculate the enthalpy change per mole of BaSO4 formed.

- Constant volume calorimeter is called a bomb calorimeter.
- Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
- The heat capacity of the calorimeter is known and tested.
- Since DV = 0, PDV = 0, DE = q

- thermometer
- stirrer
- full of water
- ignition wire
- Steel bomb
- sample

- intensive properties not related to the amount of substance.
- density, specific heat, temperature.
- Extensive property - does depend on the amount of stuff.
- Heat capacity, mass, heat from a reaction.

- Enthalpy is a state function meaning it is independent of the path.
- We can add equations to come up with the desired final reaction and add the associated DH values to get the final ∆H.
- Two rules using Hess’s Law:
- If the reaction is reversed the sign of DH is changed
- If the reaction is multiplied by a constant, so is DH for that reaction.

Overall Reaction:

N2 + 2O2 → 2NO2∆H1 = 68kJ

Reaction steps added:

N2 + O2 → 2NO ∆H2 = 180kJ

2NO + O2 → 2NO2 ∆H3 = -112kJ

____________________________________________________________________________________________________________________________________________________

N2 + 2O2 → 2NO2∆H2 + ∆H3 = 68kJ

O2

NO2

-112 kJ

H (kJ)

180 kJ

NO2

68 kJ

N2

2O2

- If a reaction is reversed the sign of ∆H is also reversed.
- The magnitude of ∆H is directly proportional to the quantities of reactants and products.
- If coefficients are multiplied by an integer, then ∆H is multiplied by the same integer
- Examples: 6.7 and 6.8

- The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions)
- Symbol DHº
- When using Hess’s Law, work by adding the equations up to make it look like the answer.
- The other parts will cancel out.

- Hess’s Law is much more useful if you know lots of reactions.
- Made a table of standard heats of formation. The amount of heat needed to form 1 mole of a compound from its elements in their standard states.
- Standard states are 1 atm, 1M and 25ºC
- For an element it is 0
- There is a table in Appendix 4 (pg A21)

- Need to be able to write the equations.
- What is the equation for the formation of NO2 ?
- ½N2 (g) + O2 (g) ® NO2 (g)
- Have to make one mole to meet the definition.
- Write the equation for the formation of methanol CH3OH.

- We can use heats of formation to figure out the heat of reaction.

- Using the standard enthalpies of formation listed in the appendix, calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water.
- 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O (l)
- Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction:
- 2Al (s) + Fe2O3(s) → Al2O3(s) + 2Fe (s)