Chemistry - Specific Heat and Electron Configuration Notes



Chemistry - Specific Heat and Enthalpy Notes

I. Heat of Reaction

A. Energy is either lost or gained in every chemical reaction.

1. Reactions that absorb heat, need energy to occur, are called endothermic reactions.

2. Reactions that release heat are called exothermic reactions.

B. The change in heat that occurs in a reaction can be measured by carrying out the reaction in a calorimeter.

1. Heat is measured in joules in the SI System.

2. Specific heat is the amount of energy needed to raise one gram of a substance one degree Celsius. Every material has its own specific heat.

3. The change in heat is calculated by the following equation:

Change in H = (Mass in grams) (Change in temperature in Celsius) (Specific heat)

4. Example - How much heat is lost when a 25 g. piece of tin cools from 99.0 C to 19.0 C ? The specific heat of tin is .220 J / g. - C.

Change in H = (25 g.) ( 80.0 C) ( .220 J / g. - C.) = 440 J

*Read in Prentice Hall Chemistry pages 393 through 395.

C. The specific heat of a substance can be determined experimentally by using the heat change equation. In a closed system, the heat lost by one substance must equal the heat gained by another substance, or Change in H1 = Change in H2. Placing a hot brick in a bucket of cool water is an example. The brick loses heat that is then absorbed by the water.

D. Example - A 51 .0 g. piece of an unknown metal is dropped into a calorimeter containing 201 g. of water. The initial temperature of the metal was 121 C, initial temperature of the water was 21.0 C, and the final temperature of the system is 28.0 C.

The specific heat of water is always 4.18 J / g. - C.

(Mass1) (Change in T1) (Sp. Heat1) = (Mass 2) (Change in T2) ( Sp. Heat2)

(51.0 g.) ( 93.0 C) ( x) = ( 201 g.) (7.00 C) (4.18 J / g. - C)

4740 x = 5880

x = 1.24 J / g. - C

E. The same relationship can be used to determine the final temperature of a system.

Example - A 39.0 g. piece of iron ( sp. heat - .448 J / g. - C) at 81.0 C is dropped into a calorimeter containing 101 ml. of water at 21.0 C. What is the final temperature of the system?

(39.0 g.) ( 81.0 - x) (.448 J / g -C) = ( 101 g.) ( x – 21.0) ( 4.18 J / g - C)

17.5 ( 81.0 - x) = 422 ( x - 21)

1420 - 17.5 x = 422 x - 8860

10,300 = 440 x

23.4 C = x

*Read in Prentice Hall Chemistry pages 508 through 510.

II. Enthalpy

A. In every chemical reaction energy is either lost or gained. The energy contained in each substance (in its chemical bonds) is called enthalpy.

B. The enthalpy’s for each substance can be looked up in reference books. Only compounds have enthalpies, elements are not bonded as thus have no enthalpy.

C. Thermodynamics is the study of the flow of energy.

III. Calculation for Heat of Reaction.

A. Change in H = Enthalpy of Products - Enthalpy of Reactants

B. Steps for Enthalpy Problems.

1. Write a balanced equation for the reaction.

2. Look up the enthalpy’s of the products and reactants.

3. Multiply the enthalpy’s by the number of moles of the substance present.

4. Find the total enthalpy of the products and the total enthalpy of the reactants and subtract.

C. Example - What is the enthalpy for the reaction where zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas.

Zn + 2HCl ------------- ZnCl2 + H2

Zn = 0 ZnCl2 = - 487 kJ

HCl = -167 X 2 = -334kJ H2 = 0

H = - 487kJ - - 334 kJ = - 153 kJ

D. In many cases you are asked to find the enthalpy change when a specific amount of reactant is available. To solve this type of problem follow the normal steps for an enthalpy problem then multiply the answer by the number of moles of the reactant you are working with.

E. Example - What is the change in enthalpy when 8.0 grams of sulfur trioxide reacts with gasous water to form sulfuric acid.

SO3 + H2O --------- H2SO4

SO3 = -438 kJ H2SO4 = -908 kJ

H2O = -242 kJ

H = - 908 kJ - - 680 kJ = -228 kJ

8.0 g. / 80.1 g. = .10 moles X -228 kJ = -23 kJ

F. Reactions that have a negative enthalpy give off energy or are exothermic. Reactions that have a positive enthalpy need energy to take place and are called endothermic.

* Read in Prentice Hall Chemistry pages 511, 512, and 530 through 532.

IV. Entropy

A. As you have observed in your own room, unless energy is used all matter moves toward a state of disorder. Entropy is the quantitative measure of the disorder or randomness in the substances involved in a chemical reaction.

B. To calculate entropy, follow the same procedure as for enthalpy. Entropy values are found on the same table as those for enthalpy.

C. Example - What is the entropy for the reaction where mercury is dissolved in hydrochloric acid (aq) to form mercury II chloride and hydrogen gas.

Hg + 2HCl --------- HgCl2 + H2

Hg = 34 J HgCl2 = 85J

HCl = 56.5 J H2 = 131 J

S = 216 J - 147 = 69 J

V. Free Energy

A. A chemical reaction that proceeds on its own without any outside intervention is called a spontaneous reaction. Reactions that are spontaneous must give off energy (negative enthalpy) or the products must have more randomness (positive entropy).

B. Most reactions will have both positive entropy and negative enthalpy if they are spontaneous. Once in a while a reaction may have one measure indicate it is spontaneous and the other indicate it is not spontaneous. Is the reaction spontaneous or not.

C. Free energy is a concept that incorporates both entropy and enthalpy. It will indicate if a reaction is spontaneous even if enthalpy points one way and entropy the other way.

D. If the free energy for a reaction is negative the reaction is spontaneous, if it positive the reaction will not proceed on its own.

E. Free energy is calculated just like entropy and enthalpy using the values on the chart.

F. Example - What is the free energy for the decomposition of sulfurous acid into gaseous water and sulfur dioxide? Is the reaction spontaneous?

H2SO3 ------------- H2O + SO2

H2SO3 = -538 kJ H2O = -229 kJ

SO2 = -300 kJ

G = -529 kJ - - 538 kJ = 9.0 kJ No, G is positive.

* Read in Prentice Hall Chemistry pages 568 through 572.

VI. State of Matter Changes

A. The state of matter a substance is in depends on the separation of the atoms/molecules that compose the substance. If the atoms/molecules are tightly packed together, you have a solid. If the particles are slightly separated and can move over one another you have a liquid. When the particles are widely separated and do not interact you have a gas.

B. When a substance changes from a solid to a liquid energy is needed to move the particles apart. The energy to change a solid to a liquid is the heat of fusion. That amount of energy must be added to melt a substance or lost for a substance to melt (liquid to solid). For water the heat of fusion is 6.02 kJ/mole.

When a substance changes from a liquid to a gas energy is needed to move the particles apart. This energy is called the heat of vaporization. You must add the heat of vaporization boil or vaporize a substance and the heat of vaporization is lost when a substance condenses. The heat of vaporization of water is 40.7 kJ/mole.

The heat of vaporization or the heat of fusion do not affect the temperature of a substance. Temperature measures how fast the atoms/molecules move, not how far apart they are.

C. Example. How much energy is needed to change 125 ml. of water at 65.0 oC. to steam at 105 oC? Specific heat of steam is 1.70 J/g-oC.

100.0 oC – 65.0 oC = 35.0 oC

(125 g.)(35.0 oC)(4.18 J/g-oC) =18,300 J

125 g x 1 mole x 40700 J/mole =283,000 J

(125 g.)(5.00 oC.)(1.70 J/g-oC) = 1060 J

302,000 J Total

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