Thermodynamics

• Science of heat and energy flow in a chemical reaction

• Specific Heat (C)

• heat required to produce a given temperature change per gram of material

• J / (g * ºC) or J/ (g * ºK)

Enthalphy• ΔH = Hproducts -Hreactants

– heat transferred into a system at constant pressure

Phase Diagrams• q(plat) = m Δ H

• q(slant) = mCΔ t

• Heat of fusion - heat needed to melt

• Heat of Vaporization - heat needed to vaporize

Phase Diagram Example• Find the energy is required to turn 500g of ice at –50 deg C to steam at

200 deg C

• Csol = 2.1 Cliq = 4.2 Csteam = 2.0 (all in J/g*K)

• ΔHfusion = 333 J/g ΔHvaporization = 2256 J/g

• SOLUTION:

q(warm ice to 0 deg) = (500)(2.1)(50) = 5.3 x 10^4 J

q(melt ice) = (500)(333) = 1.67 x 10^5 J

q(warm water to 100 deg) = (500)(4.2)(100) = 2.1 x 10^5 J

q(vaporize water) = (500)(2256) = 1.13 x 10^6 J

+ q(warm steam to 200 deg) = (500)(2.0)(100) = 1.0 x 10^5 J

TOTAL ENERGY REQUIRED = 1.7x 10^6 J

Hess’s Law• If a reaction is the sum of two or more reactions, then the ΔH for

overall process is sum of ΔH’s for consituent reactions

• Example: Find ΔH for the reaction H20(l) --> H2(g) + O2(g) when given…

1) H20(l) --> H20(g) ΔH1 = 44 kJ

+ 2) H20(g) --> H2(g) + O2(g)

ΔH2 = 241.8 kJ

NET: H20(l) --> H2(g) + O2(g) ΔHnet= 285.8 kJ

Calorimetry• -q lost = q gained • qbomb = CbombΔt• Example: 1.00 g of Octane (C8H18) is burned in a calormeter which has

1.20kg water. The temperature of the bomb and water rises from 25 to 33.2 deg. C. If C(bomb)=837 J/K,calculate the heat transferred to 1 mol of octane.

• Solution:

q(water) + q(bomb) = -q(octane) ***watch out for - sign!!!!!

(1200)(4.18)(8.2) + 837(8.2) = -q(octane)

-48.1 kJ = q (octane) <----This is for 1g of octane

Do factor label to get energy per mole of octane-

Heat transferred /mol = (-48.1 kJ/mol)(114.2 g/mol) = -5.49 x 10^3 kj/mol

Bond Energy• ΔH (reaction) = sum of BE’s of REACTANTS - sum of BE’s of PRODUCTS

EnTrOpY• (S) - measure of randomness

• Progression of randomness:

– solid-->aqueus-->liquid-->gas (most random)

Gibbs Free Energy• ΔG = ΔH - TΔS NOTE:T in K, watch energy units (kJ or J)

• +ΔG = nonspontaneous , -ΔG = spontaneous– Spontaneous only means PRODUCT FAVORED

• Transition temperature of spontaneity:– T = ΔH / ΔS

Is the reaction spontaneous?

ΔS < 0 ΔS > 0

ΔH > 0 NEVER SPONTANOUS MAYBE (AT HIGH TEMPS)

ΔH < 0 MAYBE (AT LOW TEMPS)

ALWAYS SPONTANEOUS

• ΔG = -RTlnK ----relates w/equilibrium constant