Thermochemistry Chapter 9 Lecture 3 Flashcards
Thermodynamic state of a system
Specifying values of a set of measurable properties sufficient to determine all other properties.
e.g. n, P, and T determine states of an ideal gas. V=nRT/P d=m/v = nM/v = PM/RT
State function
unique value for a specified state of a system.
State function: Extensive examples
Volume, Mass, Internal Energy, Enthalpy
State function Intensive
Density, Pressure, Temperature
Path dependent Extensive
Heat, Work
Path dependent Intensive
Heat, Capacity
First Law of Thermodynamics
(Uf-Ui) = q + w = q -P (Vf-Vi)
Change in the internal energy of the closed system (U) is equal to the amount of work + heat supplied to the system.
Enthalpy H
Extensive state function that is the sum of the internal energy and pressure-volume product: H = U + PV or (Hf-Hi) = (Uf-Ui) + P(Vf-Vi)
What happens when there is constant pressure for Enthalpy?
Only heat can be gained or loss by function H=qp
Standard Enthalpy of reaction
(Hrxn) Change that occurs when all the reactants and products are in standard states.
Examples:
Gas: pure gas at 1 atm
liquid: pure liquid at 1 atm
Solid: pure solid at 1 atm
Solute: At concentration of 1M (mole/liter)
Hess’s law
Total enthalpy change = reaction is the sum of all changes
Standard Enthalpy of formation
(Hf) is enthalpy change in the formation of one mole of a substance in the standard state from the reference forms of its elements in standard states.
Ex: (Hf) of H2O(l) = -285.8 Kj/ mol
Allotropes
Different structured form of elements within the same physical state:
Ex: O2 & O3
what does it mean for (Hf-Hi)<0
They are stable because the enthalpy of a compound that is lower than the constitute elements and its decomposition need energy input as heat.
Usually exothermic rxns
what does it mean for (Hf-Hi)>0
They are less stable because when they are greater than constitute elements, they release heat upon decomposition.
How can you calculate the change in Enthalpy?
Subtract the Enthalpies formation that are in the product from the enthalpy’s formation in reactant side
Product- Reactant = (Hf-Hi)
Exothermic reactions in relation to Q
They are overall having a negative enthalpy reaction and q is negative because it exits the system.
why is change in G >0 equal product in favored?
If G < 0 they are stable enough for reactants to turn into products the inverse happens if G <0
what are the rules of S universe in relation to 0?
Suniv > 0 = product is favored
Suniv < 0 = reactant is favored
Cause less entropy overall means the reactant was probably better in having more randomness.
If the surrounding is doing work on the system in chemical reaction
The moles of gas decrease
What happens when chemical energy increases?
Chemical bonds break down.
Finding Enthalpy vaporization
Hvap = (-R ln ( P2/P1) )/ 1/T2 - 1/T1
Finding Pressure
P2 = P1 e^ Hvap/R (1/T2 - 1/T1)
Constant of mole —> L
1mole =22.4L
Finding Temp change in Hvap
T2= [1/T1 - Rln(P2/P1)/Hvap)^-1
