A P Chemistry Chapter 6 Thermochemistry 1 st

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A. P. Chemistry Chapter 6 Thermochemistry

A. P. Chemistry Chapter 6 Thermochemistry

1 st Law of Thermodynamics- total energy of the universe is constant (p. 244)

1 st Law of Thermodynamics- total energy of the universe is constant (p. 244) Definitions and State Functions • System- a system is the part of the universe that is being studied. In Chemistry, this is often just a particular reaction (p. 243) • Surroundings- everything else in the universe outside of the system (p. 243) • Open System- one where both energy and matter can be transferred to and from the surroundings • Closed system- one where only energy can be transferred to and from the surroundings

 • Isolated System- one where neither energy nor matter can be transferred to

• Isolated System- one where neither energy nor matter can be transferred to and from the surroundings • State Functions- those that depend only upon the initial and final state of a substance. Examples are /H, /S, and /G (p. 243) • Heat- the symbol used is q (p. 245) • Enthalpy- in thermochemistry, every substance is said to have a heat content or enthalpy (H). Most reactions involve an enthalpy change /H which measures the heat transferred (p. 248) (Under constant pressure, q = /H)

Problem: Calculate the work done on the system when 6. 0 L of a

Problem: Calculate the work done on the system when 6. 0 L of a gas is compressed to 1. 0 L by a constant external pressure of 2. 0 atm. (1. 0 x 103 J; obtaining a positive value for work means that work is done on the system by the surroundings. A positive work value means the system gains energy. ) Problem: A gas, initially at a pressure of 10. 0 atm and having a volume of 5. 0 L is allowed to expand at constant temperature against a constant external pressure of 4. 0 atm until the new volume is 12. 5 L. Calculate the work done by the gas on the surroundings. (w = -30 L-atm = -3. 0 x 103 J)

Problem: A gas is allowed to expand at constant temperature from a volume of

Problem: A gas is allowed to expand at constant temperature from a volume of 10. 0 L to 20. 0 L against an external pressure of 1. 0 atm. If the gas also absorbs 250 J of heat from the surroundings. What are the values of q, w, and /E? (q = 250 J w = -1. 0 x 103 J /E = q + w = -750 J)

Energy is measured in joules (J). Another unit of energy you may come across

Energy is measured in joules (J). Another unit of energy you may come across is the calorie (cal). 1. 000 calorie = 4. 184 J 1 kilocalorie = 1000 calories = 1 Cal 1 kilojoules = 1000 joules Example: Convert 1 joule to calories 1 J x 1 calorie/4. 184 J =

Specific Heat Capacity The specific heat capacity of a substance is defined as the

Specific Heat Capacity The specific heat capacity of a substance is defined as the amount of energy required to change the temperature of one gram of a substance by one degree Celsius. A useful relationship to use is Energy Change = (specific heat capacity)(mass)(temperature change) = mc/T (/T = Temp. final – temp. initial) The unit for specific heat capacity is usually J/go. C. The specific heat for water is designated as cp, and is 4. 184 J/go. C, or 1 cal/go. C. Generally, specific heat capacity has the symbol of c for other substances.

Example: The specific heat capacity for aluminum is 0. 900 J/g-o. C Calculate the

Example: The specific heat capacity for aluminum is 0. 900 J/g-o. C Calculate the energy needed to raise the temperature of 8. 50 x 102 gram block of aluminum from 22. 8 o. C to 94. 6 o. C. q = (8. 50 x 102 g)(0. 900 J/g o. C)(94. 6 o. C – 22. 8 o. C) = Example: Calculate the molar heat capacity of aluminum. 0. 900 J/g o. C x 27 g/ mol =

Enthalpy (p. 248) In thermochemistry, every substance is said to have a heat content

Enthalpy (p. 248) In thermochemistry, every substance is said to have a heat content or enthalpy. Enthalpy is given the symbol, H. Most reactions involve an enthalpy change, /H (delta H), where: /H = Hproducts –Hreactants /Horeaction = np/Hof products - np/Hofreactants (n = # of moles) (Data for /Hof is in Appendix 4, p. A 21 -A 24) Enthalpy is a molar dependent function. Example: problem #33. 283.

Problem: The thermochemical equation for the combustion of propane is: C 3 H 8(g)

Problem: The thermochemical equation for the combustion of propane is: C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(l) /Ho rxn = -2220 k. J/mol How many k. J of heat are released when 0. 50 mol of propane reacts? How much heat is released when 88. 2 g of propane reacts? (q = -1. 1 x 103 J; q = -4440 k. J) Problem: Determining the Calorimeter Constant The combustion of benzoic acid is often used as a standard source of heat for calibrating combustion bomb calorimeters. The heat of combustion of benzoic acid has been accurately determined to be 26. 42 k. J/g. When 0. 8000 g of benzoic acid was burned in a calorimeter containing 950 g of water, a temperature rise of 4. 08 o. C was observed. What is the heat capacity of the bomb calorimeter (the calorimeter constant)? (Ccal = 1. 21 x 103 J/o. C)

Enthalpy Level Diagrams Endothermic- /H is positive, enthalpy of products > enthalpy of reactants.

Enthalpy Level Diagrams Endothermic- /H is positive, enthalpy of products > enthalpy of reactants. Energy must be put into the reaction for it to occur. (beaker gets cold)(p. 243) Exothermic- /H is negative, enthalpy of products < enthalpy of reactants. Energy is released from the reaction as it occurs. (beaker gets hot)(p. 243)

When discussing the enthalpy of a substance, it is necessary to state the conditions

When discussing the enthalpy of a substance, it is necessary to state the conditions under which the enthalpy is measured. Usually enthalpy changes are stated under standard conditions (p. 260, table), i. e. , • Gases at 1 atm pressure • All solutions at unit concentration (1 M) • All substances in their normal physical state at standard temperature and pressure • Standard temperature (273 K) A reaction which is spontaneous will tend toward conditions of lower enthalpy, more negative values of /H.

Standard Enthalpy of Formation (/Hfo)(p. 260)- defined as the enthalpy change when one mole

Standard Enthalpy of Formation (/Hfo)(p. 260)- defined as the enthalpy change when one mole of a substance is formed from its elements, in their standard form, under standard conditions. 2 C(graphite) + 3 H 2(g) + ½ O 2(g) C 2 H 5 OH(l) /Hfo = -279 k. J/mol Practice: CH 3 Br(l) CH 3 COC 2 H 5(l) Na. NO 3(s) /Hfo may be negative or positive. If the enthalpy change is negative, then the energy is released and the reaction is exothermic. If it is positive, then energy is taken in and it is endothermic.

Example: Problem #61, 285 Example: problem #67, p. 286 Problem: Using /Hof values in

Example: Problem #61, 285 Example: problem #67, p. 286 Problem: Using /Hof values in the back of your textbook, calculate the standard enthalpy change for the incomplete combustion of ethane: C 2 H 6(g) + 5/2 O 2(g) 2 CO(g) + 3 H 2 O(l) (/Hrxn = -993. 7 k. J/mol)(note: this problem and the values used to solve it were obtained from another book; the values may vary slightly. )

Standard Enthalpy of Combustion(/Hco)- defined as the enthalpy change when one mole of a

Standard Enthalpy of Combustion(/Hco)- defined as the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. Since energy is released in such a reaction, /Hco will usually be negative. C 2 H 6(g) + 3 ½ O 2(g) 2 CO 2(g) + 3 H 2 O(l) /Hco = -1565 k. J/mol Remember that the combustion of hydrocarbons or organic molecules in air (O 2) will produce carbon dioxide and water. This becomes more complicated for other compounds.

Hess’s Law(p. 256)- the enthalpy change during a reaction depends only on the nature

Hess’s Law(p. 256)- the enthalpy change during a reaction depends only on the nature of the reactants and products and is independent of the route taken. In algebraically manipulating the equations for Hess’s Law, 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 in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, then the value of /H is multiplied by the same integer.

Example: problem #53, p. 285

Example: problem #53, p. 285