Causes of Change

Ch.11

(11-1) Governing Principles

• Nature favors rxns that proceed toward lower E & greater disorder

• Heat: total KE of particles– Quantity of thermal E– Joules

• Temp.: avg. KE of particles– Intensity of thermal E– Kelvin

Calorimeter

• Measures heat absorbed or released in a rxn

• Exo = E released

• Endo = E absorbed

Calorimeters

Specific Heat Capacity

• (Cp): amt of heat needed to raise the T of 1 g of a substance by 1ºC– Units = J/g· ºC– Water = 4.18 J/g· ºC

• q = CpmΔT– m = mass (g)– ΔT = Tfinal - Tinitial = temp. change (ºC)– q = heat (J)

Specific Heat Practice

How many joules are needed to raise the T of 300 g of Al from 20 °C to 70°C if the Cp of Al is 0.902 J/g• °C ?

1. List the eq.q = CpmΔT

2. Substitute & solveq = (0.902 J/g• °C)(300 g)(70°C - 20°C) = 13,530 J

Law of Heat Exchange

• Heat flows from hot to cold

• The law:– Heat lost = heat gained– Heat lost by a metal will be gained by the

surrounding H2O (measured w/ calorimeter)

Heat Exchange Practice

Find the Cp of 100 g of an unknown metal when it’s removed from H2O at 100°C & placed into 200 g of H2O at 20°C. The final T of the mixture is 23.5°C.

1. Write eq.Heat lost (by metal) = heat gained (by water)

q(metal) = q(water)

CpmΔT = CpmΔT

Heat Exchange Practice

2. Substitute & solveCp(100 g)(100°C – 23.5°C) =

(4.18 J/g• °C)(200 g)(23.5°C - 20°C)

Cp = 0.382 J/g• °C

Note: this side must remain +, Therefore ΔT = Ti - Tf

(only in these types of problems)

Molar Heat Capacity

• (C): heat required to inc. the T of 1 mol of a substance by 1 K– Units = J/K·mol– Table 11-1, p.389

• q = nCΔT– n = moles

Molar Heat Capacity Practice

If C of H2O is 76 J/K·mol, calculate the amt of heat E needed to raise the T of 90.0 g of H2O from 35°C to 45°C.

1. List eq.

q = nCΔT

Molar Heat Capacity Practice

2. Substitute (make sure to convert g to mol) & solve

q = (90 g x 1 mol ) )(76 J/K·mol)(45°C - 35°C)

18.02 g

= 3,792 J

(11-2) Thermodynamics

• Study of E flow

• Thermo = “heat”

• Dynamics = “motion”

Entropy

• Total disorder in a substance or system

• Molar entropy (S): quantity of entropy in 1 mol of a subst.– Units = J/K·mol

• ΔS > O (+), disorder inc.

• ΔS < O (-), disorder dec.

Enthalpy

• E “inside” an atom or molecule• Molar enthalpy (H): total E content of a

system– Units = kJ/mol or J/mol

• ΔH > O (+), endo• ΔH < O (-), exo

• ΔH = q = nCΔT = CΔT n n

Enthalpy Practice

How much does the molar enthalpy change when a 92.3 g block of ice is cooled from –0.2°C to –5.4°C?

1. List eq. ΔH = q = CΔT

nCan’t usew/out q

Enthalpy Practice

2. Find C for ice in Table 11-137.4 J/K·mol

3. Convert °C to K-0.2°C + 273 = 272.8 K-5.4°C + 273 = 267.6 K

4. Subst. & solve ΔH = (37.4 J/K·mol)(267.6 K – 272.8 K) = -194 J/mol Most ΔH are

in kJ/mol

Properties of Matter

• Extensive property: depends on amt. of subst. – S, H, m, V, C

• Intensive property: does not depend on amt. of material – D, P, T

(11-3) Change of State

• S & H change dramatically during a state change

• Heat of fusion (ΔHfus): heat absorbed when 1 mol of a subst. melts – Molar enthalpy of fusion

• Heat of vaporization (ΔHvap): heat absorbed when 1 mol of a liquid vaporizes– Molar enthalpy of vaporization

TE

MP

ER

AT

UR

E (C

)T

EM

PE

RA

TU

RE

(C

)

HEAT ADDED OVER TIME HEAT ADDED OVER TIME

SOLIDSOLID

LIQUIDLIQUID

GASGAS

meltingmelting

freezingfreezing

condensationcondensation

vaporizationvaporization

Heating CurveHeating Curve

0°

100°

ΔHfus

ΔHvap

(s)

(l)

(g)

Gibbs Energy

• Molar Gibbs E (G): “free E”; determines spontaneity of a rxn– Units = kJ

• Spontaneous rxns occur w/out outside assistance– ΔG < O (-), spont. rxn– ΔG > O (+), nonspont. rxn

• ΔG = ΔH - TΔS, (T in K)

Gibb’s Practice

If ΔH° is 41.2 kJ/mol & ΔS° is 0.0418 kJ/K is the following rxn spontaneous at 25°C?

H2 + CO2 H2O + CO

1. List the eq.ΔG = ΔH – TΔS

2. Subst. & solveΔG = 41.2 kJ/mol – (298 K)(0.0418 kJ/K)

= 28.7 kJ, nonspontaneous

(11-4) Hess’s Law

• Overall enthalpy change in a rxn is = to the sum of the individual steps

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

• Actually occurs in 2 steps:– CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

– 2H2O(g) 2H2O(l) H = -88 kJ

• ΔH = -802 kJ + -88 kJ = -890 kJ

Heat of Rxn

• E absorbed or released during a chemical rxn

• Std. heat of formation (ΔHºf): change in enthalpy when 1 mol of a cmpd is produced from free elements– Table A-13, p.802 (check state of matter)

Standard Conditions

• (°) are generally:– Temp: 298 K or 25°C

– Pressure: 1 atm or 760 mmHg

Equations

• ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants)

• ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants)

• ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)

Enthalpy Practice

Calculate ΔHº for the following rxn. Is the rxn exo. or endothermic?

H2(g) + CO2(g) H2O(g) + CO(g)

1. List the eq.

ΔHº = ∑ΔHºf(products) - ∑ΔHºf(reactants)

Enthalpy Practice

2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)

ΔHº = [(1 mol)(-241.8 kJ/mol) + (1 mol)(-110.5 kJ/mol)]

- [ (1 mol)(0 kJ/mol) + (1 mol)(-393.5 kJ/mol)]

= 41.2 kJ, endothermic

Entropy Practice

Calculate ΔSº for the following rxn. Does the rxn proceed toward a more ordered or disordered state?

H2(g) + CO2(g) H2O(g) + CO(g)

1. List the eq.

ΔSº = ∑ΔSºf(products) - ∑ΔSºf(reactants)

Entropy Practice

2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)

ΔSº = [(1 mol)(188.7 J/K•mol) + (1 mol)(197.6 J/K•mol)]

- [ (1 mol)(130.7 J/K•mol) + (1 mol)(213.8 J/K•mol)]

= 41.8 J/K, disorder

Gibb’s Practice

Calculate ΔGº for the following rxn. Is the rxn spontaneous at 25°C?

H2(g) + CO2(g) H2O(g) + CO(g)

1. List the eq.

ΔGº = ∑ΔGºf(products) - ∑ΔGºf(reactants)

Gibb’s Practice

2. Using Table A-13, subst. & solve (account for # of mols (coef.) of each cmpd)

ΔGº = [(1 mol)(-228.6 kJ/mol) + (1 mol)(-137.2 kJ/mol)]

- [ (1 mol)(0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]

= 28.6 kJ, nonspontaneous