1st Law of Thermodynamics

(Law of Conservation of Energy)

- Energy is Conserved

- Energy is not created or destroyed, it is transferred from one form to another

Work = Force x Distance

w = F x d

= -P x A x d

= -PΔV

Any energy that flows from the system to the surroundings has a _____ sign because the system has lost it

- E_final is smaller than _Einitial

Any energy that flows to the system from the surroundings has a _____ sign because system has gainted it

- E_final is larger than E_initial

A fuction or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.

- pressure, volume and temperature ARE state functions

- work and heat ≠ state functions

ΔV = Area x Distance

ΔV = A x d

Total kinetic energy (KE) of molecular motions

- heat = q

ΔE = q + w

= q - PΔV

Simpler to ignore work, so...

What would allow us to ignore work?

1. Work must = 0

2. ΔV = 0,

  - (a reaction might be carried out in a closed container with a constant volume)

This means no PV work is done and the energy chance is due ENTIRELY to heat transfer

The heat change that occurs @ constant pressure (q_p) is called enthalpy (H)

enthalpy = H

constant pressure = q_p

Enthalpy (H) of a system is the name given to the quantity E + PV, so...

q_p = ΔE + PΔV = ΔH

ΔE = ΔH

ΔH = q_p

q = q_h2o

q_h2o = (mass of water) x SpHt of H2O x (ΔT)

*Specific heat of H2O: 4.18J/g*c

*ΔT = T_final - T_initial

CH₄(g) + 2O₂(g) ---> CO₂(g) + 2H₂O(l)

Using folowing information, calculate ΔH°:

CH₄(g) + O₂ ---> CH₂O(g) + H₂O(g) ΔH° = -275.6 kJ

CH₂O(g) + O₂ ---> CO₂(g) + H₂O(g) ΔH° = -526.7 kJ

H₂O(l) ---> H₂O(g) ΔH° = 44.0 kJ

CH₄(g) + O₂ ---> CH₂O(g) + H₂O(g)       ΔH° = -275.6 kJ

CH₂O(g) + O₂ ---> CO₂(g) + H₂O(g)       ΔH° = -526.7 kJ

2[H₂O(g) ---> H₂O(l)]                2[ΔH° = -44.0 kJ] = -88.0 kJ

CH₄(g) + 2O₂(g) ----> CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

Calculate ΔH° in kilojoules for the synthesis of lime (CaO) from limestone (CaCO₃), an important step in manufacture of cement.

CaCO₃(s) ---> CaO(s) + CO₂(g)

ΔH°_f [CaCO₃ (s)] = -1207.6 kJ/mol

ΔH°_f [CaO (s)] = -634.9 kJ/mol

ΔH°_f [CO₂ (g)] = -393.5 kJ/mol

ΔH° = [H°_f (CaO) + ΔH°_f (CO₂)] - [ΔH°_f (CaCO₃)]

        = (1 mol)(-634.9 kJ/moll) + (1 mol)(-393.5 kJ/mol) - (1 mol)(-1207.6 kJ/mol)

        = 179.2 kJ

So, the reaction is endothermic by 179.2 kJ

Find approximate ΔH° in kilojoules for industrial synthesis of chloroform by reacion of methane with Cl₂

CH₄(g) + 3Cl₂(g) ---> CHCl₃(g) + 3HCl(g)

Bond dissociation energies for equation:

C - H D = 410 kJ/mol

C - Cl D = 330 kJ/mol

C - CL D= 243 kJ/mol

H - Cl D = 432 kJ/mol

ΔH° = [3 D_Cl-Cl + 4 D_C-H] - [D_C-H + 3 D_H-Cl + 3 D_C-Cl]

        = [ (3mol)(243 kJ/mol) + (4 mol)(410 kJ/mol)] - [(1 mol)(410 kJ/mol) + (3 mol)(432 kJ/mol) + (3 mol)(330 kJ/mol)]

        = - 327 kJ

So, the reaction is exothermic by approximately 330 kJ

Boyle's Law

(The relationship between Gas, Volume, and Pressure)

The volume of an ideal gas varies inversely with pressure (V↑ P↓)

*if pressure halved, volume doubles and vice versa

V = k(1/P) + 0 OR PV = k

Charles's Law

(The relationship between Gas, Volume and Temperature)

The volume of an ideal gas varies directly with absolute temperature

Ex: If gas tempt is doubled, volume is doubled, gas temp halved, volume halved

V/T = k at constant n and P