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by: Nick Manning

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# Notes on Thermochemistry and the beginning of Quantum Mechanics CHEM - 10060 - 001

Marketplace > Kent State University > Chemistry > CHEM - 10060 - 001 > Notes on Thermochemistry and the beginning of Quantum Mechanics
Nick Manning
KSU
GPA 4.0

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Week of 3/7 to 3/1; Dr. Leslie reviewed what we went over already, added a lot of new stuff on thermochemistry and how to do different problems, and started a whole new chapter on Quantum Mechanics...
COURSE
GENERAL CHEMISTRY I
PROF.
TBA
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Quantum Mechanics, thermochemistry, endothermic, exothermic, atomic structure
KARMA
25 ?

## Popular in Chemistry

This 9 page Class Notes was uploaded by Nick Manning on Friday March 11, 2016. The Class Notes belongs to CHEM - 10060 - 001 at Kent State University taught by TBA in Fall 2015. Since its upload, it has received 23 views. For similar materials see GENERAL CHEMISTRY I in Chemistry at Kent State University.

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Date Created: 03/11/16
Thermochemistry Review **quiz Monday 3/14** System (atoms bonded)  new bonds Thermochemistry measures a change of energy in the system, impossible to see in the system, yet we can see it manifest as work or heat (measured as ⧍T) qp= ⧍H = Enthalpy; heat at constant pressure exothermic- surroundings heat up ⧍T surroundings endothermic- surroundings give heat to system (cool down) ⧍T surroundings specific heat capacity- amount of heat to bring substance up by 1 degree E = Ek+ E p Heat energy = (mass of solution)(specific heat capacity)(temperature final – temp initial) q= mc⧍T ⧍H (Enthalpy Change) ⧍H = q p The reaction is the system while the H2O or other solvents is the suroundings EXAMPLE 9.55 g of NaOH is dissolved in 100.0 g of water. The temperature started at 23.6 degrees and rose to 47.4 degrees. The specific heat capacity is 3.94 J/g*K. Calculate q of the reaction and the change in H of the reaction in kJ/mol of NaOH. The mass of the solution is 100.0+9.55 and with sig figs that equals 109.6. qsoln = m * c * change in T qsoln = 109.6 * 3.94 * (47.7 – 23.6) qsoln = 109.6 * 3.94 * 24.1 qsoln = 10277 J or 10.3 kJ q rxn = -10.3 kJ So, to do the change in H, we already have kJ, we need the mols. 9.55 g NaOH * 1 mol NaOH/40.0 g NaOH = .239 mol NaOH -10.3 kJ / .239 mol = -43.1 kJ/mol STANDARD ENTHALPIES AND ⧍H F Since Change in H is a state function, only the final and initial matter. ⧍H values are measured in standard state conditions. Standard state is the most stable form at:  gases at 1 atm  solutions at 1 M concentration T usually is 25ᵒ C = 298 K  ⧍Hᵒ indicates ⧍H under standard conditions. Technically, we can never measure absolute H, only the change of H (⧍H), which tells us relative H for products and reactants. ⧍Hᵒ feans the standard heat of formation; the change in H with the formation of 1 mol of a pure substance from its elements in most stable forms. ⧍Hᵒ f 0 for pure element in standard states - It also measures stability of an element, 0 = stable but the farther away a number is from 0 means its that much more unstable. ⧍Hᵒ flso depends on the physical state of the substance (solid, liquid, or gas). ⧍Hᵒrxn= ∑m⧍Hᵒ (produfts)- ∑ n⧍Hᵒ (reactantsf Standard heats of formation Coefficients from balanced equation This looks like a really intimidating equation, but it’s pretty easy. It just means that the change in enthalpy of the reaction (⧍Hᵒrxn) equals the sum of all the numbers you get when you multiply the standard heat formation of the products of the reaction by their coefficients, minus the sum of the standard heat formation of the reactants of the reaction by their coefficients. Okay, maybe that still sounds a little confusing… I guess an example would be the easiest way to show this. EXAMPLE Calculate the total ⧍H of the below reaction by using the given ⧍Hᵒfvalues. NH 3g) = -45.9 kJ NO (g) = +90.3 kJ H 2 (g) = -241.8 kJ H O (l) = -285.8 kJ 2 4NH (g) + 5O (g)  4NO (g) + 6H O (g) 3 2 2 -45.9 0 +90.3 -241.8 4(-45.9) 4(90.3) 6(-241.8) -184 361 -1451 ⧍Hᵒrxn = (361-1451) – (-184) ⧍Hᵒrxn = -90.6 kJ Note that the O2counts as 0 because it is a stable element. And you also have to make sure that you use the correct state of matter for each element, Dr. Leslie said she would put two in her choices but you just have to pick the right state (if it asks for gas, use the gas, not that hard). QUANTUMTHEORY AND ATOMICSTRUCTURE Rutherford’s model -could not explain stability of atom (why wouldn’t the atom just collapse on itself if e- and protons attract?) Relied on classical physics, particles and waves as being different Waves of Light -light is a form of electromagnetic radiation -has electrical and magnetic components that move 90 degrees to each other through space Waves Wavelength- distance of one full wave (2 peaks or 2 troughs) Frequency- # of wavelengths that go by in one second - Inversely proportional to wavelength - Unit: Hz (hertz), cycles per second Amplitude- distance from middle point; depends on intensity. Can be viewed as height of trough or -higher amplitude = higher intensity (sound, brightness, etc.) Speed = frequency x wavelength For all EM radiation in a vacuum, c = 3.00 x 10^8 (speed of light) White light disperses into individual colors when passed through a prism, because each color has specific wavelength and frequency Waves undergo refraction when hitting new medium Waves exhibit diffraction, particles do not - Can result in interference patterns - Constructive interference - Destructive interference Classical Physics: Light is a wave Waves do not behave like particles because particle don’t Undergo refraction, show diffraction, or dispersed into wavelength and frequencies. - Energy (E) is related to intensity, intensity to amplitude Shifting viewpoints Color, temperature, and therefore E are related - Smoldering coal, heated metal, fires, etc. E is related to wavelength and frequency E=hv v=frequency h= Planck’s constant Dual nature of light: - Light exhibits typical wave behavior but also exhibits particle behavior. Visible light cannot pass through solid objects. Light travel in a vacuum, but sound does not, Matter is quantized- is composed of discrete, individual particles. Elements are composed of individual atoms; and compounds can only be built off of full atoms, not fractions of atoms. Atoms are composed of smaller particles; some are fundamental, meaning they can’t be broken down any more. EM Radiation is also quantized; particles called photons, which are fundamental (can’t be broken down more) QUANTUM THEORY AND ATOMIC STRUCTURE (CONT) EM radiation is quantized, made of photons, with a “rest mass” of zero, meaning they only have mass when they are moving Higher amplitude = higher intensity = more photons (not higher E) Photoelectric Effect -probed atomic structure by the interaction of light with metals - Light Hit metal in a vacuum with certain frequency and proved electrons in light can knock e- off of metal. There’s a threshold for these electrons, if the light doesn’t have a certain frequency (v) it cannot break the threshold. If v is way more than the threshold amount, then the velocity of the e- bouncing off increases.

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