chapter 6 Flashcards

Question

sign conventions for q (heat)

Answer 1

+ system gains thermal energy | - system loses thermal energy

Answer 2

+ work done on a system | - work done by a system

Answer 3

+ energy flows into the system | - energy flows out the system

Answer 4

• If we only focus on P-V work, the work done by a chemical reaction can be calculated when gases are involved w = -ΔnRT w = -PΔV & PV = nRT

Answer 5

During a chemical reaction the internal energy (ΔU) is a | measure of all of the energy, heat and work

Answer 6

If we can calculate the change in temperature and volume that occurs then we can calculate the ΔU q = C x ΔT w = -PΔV = 0 • If the reaction takes place at constant volume, then all of the energy change is associated with a reaction is evolved as heat

Answer 7

• Bomb calorimetry - known mass of a substance is burned in a container with constant volume qcal = -qr

Answer 8

When a reaction occurs open to the atmosphere, energy can evolve as heat and work • For many reactions we are more concerned with the heat given off and not the work • We define enthalpy as the sum of the internal energy and the product of pressure and volume ΔH = ΔU + PΔV • Internal energy, pressure, and volume are all state functions, so therefore, enthalpy is also a state function

Answer 9

``` Most reactions involve very little P-V work ΔH = ΔU + PΔV ΔH = ΔU + ΔPV ΔH = ΔU + Δ(nRT) ΔH = ΔU + ΔngRT ```

Answer 10

For liquid and solid phase reactions volume changes only | a little ΔH ≃ ΔU

Answer 11

When the moles of gas are the same for a gas-phase reaction ΔH = ΔU • When the moles of a gas phase reaction change, PΔV ≠ 0, but q is usually much bigger than PΔV so ΔH ≃ ΔU

Answer 12

``` ΔH = ΔU + PΔV Remember: ΔU = q + w and PΔV = -w Therefore: ΔH = ΔU + PΔV ΔH = (qp + w) + PΔV = qp + w - w ΔH = qp The sign on ΔH tells you the flow of energy either to or from the system ```

Answer 13

If a reaction gives off heat as thermal energy does it come from the thermal energy of the reactants? No üThermal energy comes from the kinetic energy of the system üExothermic reactions get warmer, endothermic reactions get colder

Answer 14

üIn exothermic reactions if the heat released to the surroundings came from thermal energy of the reactants the system would get colder! üExcess thermal energy is generated from the potential energy of the system - energy mostly from the electrostatic forces between protons and electrons that compose atoms and molecules

Answer 15

Can represent enthalpy of reaction as ΔrH • It is an extensive property, and therefore depends on the amounts of reactants that react • For non-stoichiometric amounts you need to adjust the heat released/absorbed accordingly

Answer 16

• Coffee cup-calorimetry is open to the atmosphere • A known mass of solid (system) is added to a known mass of liquid (surroundings) • Or two volumes of solutions are mixed together - where the dissolved species reacting is the system and the aqueous water is the surroundings • Any change in temperature of the surroundings is measured and we can relate the heat change of solution to the heat change of the solid/reactants qsoln = msoln x Cs,soln x ΔT = -qrxn

Answer 17

1. When a chemical reaction is multiplied by a factor, ΔrH is multiplied by the same factor: 3. If a chemical reaction can be expressed as a sum of series of steps, then ΔrH for the overall reaction is the sums of the steps A + B ➞ C + D ΔrH 2A + 2B ➞ 2C + 2D 2ΔrH 2. When a chemical reaction is reversed, the sign of ΔrH is reversed: A + B ➞ C + D ΔrH C + D ➞ A + B -ΔrH

Answer 18

Hess’s Law states that if a chemical equation can be expressed as the sum of a series of steps, then ∆rH for the overall equation is the sum of the ∆rH for each step

Answer 19

Thermodynamic variables, such as ∆H, vary somewhat with conditions • Therefore, a set of specific conditions called standard states were established ΔHº - º sign indicates values were measured with chemicals in standard states

Answer 20

- For a gas, the standard state is 1 bar and ideal behavior - For a substance in aqueous solution, the standard state is 1M concentration - For a pure substance (element or compound), the standard state is the most stable form of the substance at 1 bar and the temperature of interest (usually 25°C)

Answer 21

• Just as there are ΔrH for reactions, there is also an enthalpy associated with the formation of individual compounds, ΔfH • Standard enthalpies of formation are derived from formation equations ü1 mole of a substance is formed from its elements üFractional coefficients are often used with reactants to obtain 1 mole of products C(s) + 2H2(g) ➞ CH4(g) ΔHfº = -74.9 kJ Na(s) + 1/2Cl2(g) → NaCl(s) 2C(s) + 3H2 + 1/2O2 → C2H5OH(l) ∆!! ° = -411.1 kJ ∆!! ° = -277.6 kJ

Answer 22

For an element in its standard state, =0 üThe standard state for molecular elements (e.g: Cl2, H2, O2) is the molecular form, not for single atoms • Most compounds have a negative ∆!

Answer 23

``` • We can use ∆H° f values to determine the ∆H°rxn for any reaction • Imagine the reaction occurring in two steps: 1.reactants decompose into their elements 2.Each product forms from its elements Decomposition of reactants to elements is the reverse of the formation reaction, so the standard enthalpy change is -∆H ``` hess' law

Answer 24

``` ΔH° reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants) Determining ∆H° rxn from ∆H° f Σ is the symbol for sum n is the number of moles of component (from balanced chemical reaction ```

Answer 25

• Bond energy is the amount of energy it takes to break 1 mol of bonds in the gas phase • Bond energies are all positive, indicating you must put energy into a bond to break it • Larger bond energies reflect stronger chemical bonds N-N triple bond 946 kJ/mol O-O double bond 498 kJ/mol C-H single bond 414 kJ/mol

Answer 26

Not all bonds are created equally • The bond energy of a C-H bond will vary depending on what else is bonded to the C üCH4 438 kJ/mol üCHF3 446 kJ/mol üCHCl3 401 kJ/mol üCHBr3 402 kJ/mol • Reported bond energies are averages of all of the specific bond energies for a large number of compounds

Answer 27

For simple reactions we can estimate enthalpies using the bond energies taking into account the number and type of bonds broken and the number and type of bonds formed ∆ rH = Σ (ΔH bonds broken) − Σ(ΔH bonds formed)