Chem Chapter 5 Notes
Chem Chapter 5 Notes Chem 111
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This 5 page Class Notes was uploaded by Bailey Wilhoit on Thursday September 22, 2016. The Class Notes belongs to Chem 111 at University of South Carolina - Columbia taught by Stefik in Fall 2016. Since its upload, it has received 9 views. For similar materials see Chemsitry in Chemistry at University of South Carolina - Columbia.
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Date Created: 09/22/16
Chemistry Lecture and Textbook Notes Chapter 5 5.1 Vocabulary ● Thermochemistry: study of the relationship between heat and chemical reactions ● Kinetic energy: energy possessed by matter because it is in motion ○ Thermal energy: energy in the form of random motion of particles in any sample of matter ○ Heat: energy that cause change in the thermal energy of a sample ● Potential energy: energy derived from position or condition of matter ○ Chemical energy: form of potential energy derived from forces that hold the atoms together in compounds ● System: sample of matter on which we focus; generally the atoms involved in a chemical reaction ● Surroundings: all other matter in the universe ● Work: application of force across some distance; expressed in joules Important to Know: ● Thermochemistry studies the relationships and exchange of energy between the system and the surroundings ● As temperature rises, thermal energy increases ● An example of potential energy is a skateboarder on top of a ramp. The potential he has to go down the ramp, coupled with gravity, is potential energy. ● Law of Conservation of Energy ○ The energy in the universe is constant during both chemical and physical reactions ○ Total energy does not change, but the forms of energy may transform 5.2 Vocabulary ● Enthalpy: measure of energy in a system ○ Change in enthalpy (????H) is equal to the heat absorbed or given off by the system ● Exothermic reaction: when the system transfers heat to surroundings (loses heat) ● Endothermic reaction: absorbs heat from reaction (gains heat) Important to Know: ● A thermochemical equation is an equation where ????H is given Ex: N2g) + 2 (g) → 2NO (g); ????H= +181.8 kj ● When the delta symbol precedes another symbol is means “final value initial value” ????H= H(final) H(initial) ● Enthalpy changes assume that the coefficients are moles of the substances ● Calculate enthalpy change when 5.00 mol N (2 reacts with O2g) to make NO. ○ N2(g) + 2 (g) → 2NO(g) ○ ????H= +181.8 kJ 5.3 Vocabulary ● Calorimetry: experimental measurement of heat released or absorbed by a chemical reaction ● Calorimeter: device used to measure heat ● Specific heat: heat needed to increase the temperature of 1 gram of matter by 1 K. Has units of J/g x K (or ℃) Important to Know ● The relationship between heat (q), mass (m), and change in temperature (????T) is q=mC ????T s ○ Where q= mass (grams) of sample, C = secific heat of sample, and ????T= T(final) T(initial) ● Example: ○ How much heat is must be added to 120 g of aluminum (C = 0s00 J/gx℃) from table in notes) to raise the temperature from 23.0 ℃ to 34.0 ℃? ○ ● The law of conservation of energy requires that the sum of the heat (q) of the surroundings and the heat (q) of the system is 0. ○ So, qsurr ????H = 0. ● Example: a 50.0 g sample of an acid solution is added to a 50.0 g base solution. The temperature of the liquid increases from 18.20 ℃ to 21.30 ℃. Calculate q for the neutralization reaction, assuming the specific heat of the reaction is the same as water (4.184 J/g℃) ○ 5.4 Vocabulary ● State Function: property of the system that is fixed by the present conditions and is independent of the system’s history ● Path functions: property that depends on how the particular change took place Important to know: ● Enthalpy is a state function ● Energy level diagrams can represent enthalpy change for a reaction ○ H2g) + 1/2O 2) → H O(2 ????H= 285.8 kJ ○ ● Hess’s Law ○ States that when two or more thermochemical reactions are added, the enthalpy change of the resulting equation is the sum of those for the added equations ○ Example: Calculate ????H for 3C H 2(g2→2 H (l) 6 6 ○ Given: ■ 2C H (g) + 5O (g) → 4CO (g) + 2H O (l) ????H= 1692 kJ 2 2 2 2 2 ■ 2 C66g) + 15O 2) → 12CO (g) 26H O (l)2H= 6339 kJ ■ In the first equation we see that acetylene (C H ) 2 2 the correct side of the final equation (the reactant side). However the coefficients are not the same. To get from 2 acetylene to 3, we multiply the entire first equation by 1.5, resulting in: ● 3C22g) + 15/2O 2) → 6CO (g)2 3H O (l2 ■ In the second equation we notice that Benzene (C H ) is n6 6 the right side of the equation. We need it on the product side, and we need it reduced. So we mulitiply by ½. The negative will reverse the equation so benzene is on the correct side, and the ½ will give us the correct coefficient. The new equation is: ● 6CO (g) + 3H O(l) → C H (l) + 15/2O (g) 2 2 6 6 2 ■ Notice how the oxygens, carbon dioxides, and waters all cancelled each other out to leave acetylene on the reactant side and benzene on the product side. ■ Now, we multiply the respective ????H by their multiplying factors. So the first equation will be: ● 1692 x 1.5= 2538 kJ ■ And the second equation would be: ● 6339 x ½= 3169 kJ ■ Now all we do is add the two enthalpy products together ● 2538 + 3169 = 631 kJ 5.5 Vocabulary ● Standard state: the standard state of a substance at a specified temperature is the pure form at 1 atm pressure ● Standard enthalpy of formation: enthalpy change when one mole of a substance in its standard state is formed from the most stable form of its constituent elements in their standard states Important to know: ● Symbol for standard enthalpy is ????H ° f ● Standard enthalpy change of the reaction is found using: ○ ????H rxn= ∑n ????H °froducts] ∑m ????H °[reafants] ■ Where n and m are the coefficients of the equation ○ Example: ■ 6CO (2 + 6H O2) → C H 612)6 6O (g) 2 ■ We are given: ● CO 2 393 ● H O = 286 2 ● C6126= 1268 ■ We multiply these values by the corresponding coefficients ● CO 2 393 x 6= 2358 ● H2 = 286 x 6= 1716 ● C6126= 1268 x 1= 1268 ● Note: if the substance is on the reactant side, you will multiply by a negative number and if it is on the product side you will multiply by a positive number ■ Now we just add these products together ● 1268 + 1716 + 2358= 2806 kJ