Download Ch 6 Thermochemistry

Ch 6 Thermochemistry
Heat and Energy
- Thermochemistry is the study of quantities of heat (q)
absorbed or evolved (given off) by a chemical reaction.
- Heat is energy that flows into or out of a system due to a difference
between the system’s T and the surrounding’s T.
- Energy is the capacity to move matter.
- Total Energy is the sum of kinetic, potential, and internal energies.
Etotal = Ek + Ep + U
- Kinetic Energy (Ek) is due to motion. It is a function of mass (m) and velocity (v).
Ek = (1/2)mv2
- Potential Energy (Ep) is due to position in a field of force.
The gravitational constant (g) is 9.81 m/s2, and the height (h) is in meters.
Ep = mgh
- Internal Energy (U) is the combined kinetic and potential energies
of particles (molecules) within a system.
Ex 6.1 Ek = (1/2)mv2 = (1/2)(0.143 kg)(33.5 m/s)2 = 80.2 kg·m2/s2 = 80.2 Joules
Law of Conservation of Energy
- Energy may be converted between forms, but total quantity of energy is constant.
- Energy is never created or destroyed, it only changes forms.
Heat of Reaction
- Heat of Reaction is the heat required to keep T of a reacting system constant.
- An Exothermic reaction evolves (releases) heat, so that T increases.
q < 0 and system gives away heat to the surroundings (T > 0)
- An Endothermic reaction absorbs heat, so that T decreases.
q > 0 and system draws heat from the surroundings (T < 0)
Heat of Reaction at Constant Pressure
- Enthalpy (H) is a measure of a system’s heat energy. H = U + PV
- Enthalpy is a state function.
That is, it depends only on present state (T and P), not on history.
- For a reaction at constant pressure, the change in enthalpy is equal to the heat of reaction.
qp = H
- Since
H = U + PV
we also have
U = qp − PV
Thermodynamic Equation for a Reaction
- Uses phase subscripts, has molar basis, and H is written
after the reaction for the molar amounts as written.
- Rules for writing the equation:
1. If reaction qty’s are multiplied by any factor, then H is as well.
2. Reverse reaction has opposite sign for H (that is, reaction is multiplied by −1)
Ex 6.2 Writing a Thermodynamic Equation
- Sodium hydrogen carbonate, NaHCO3(aq) reacts with hydrochloric acid, HCl(aq).
- The reaction releases carbon dioxide, CO2(g) and absorbs 12.7 kJ per mole of NaHCO3.
NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2O(L) + CO2(g) H = + 12.7 kJ
Ex 6.3 Multiplying a Reaction by Factors
-
1
(2) * [ 2H2(g) + O2(g)  2H2O(L)
H = − 572 kJ
]=
1
H2(g) + (2)O2(g)  H2O(L) H = − 286 kJ
-
Multiply by -1 to get the reverse reaction:
1
H2O(L)  H2(g) + (2)O2(g) H = + 286 kJ
Stoichiometry and Thermochemistry
- The quantity of heat is proportional to the quantities of reactant and products.
- To determine the amount of heat, first convert mass (g) to moles by dividing by molar mass.
Then, total q = (moles)(
𝐇
𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭
)
Ex 6.4 Find heat released when 9.05 × 105 g of ammonia is produced.
- N2(g) + 3H2(g)  2NH3(g) H = − 91.8 kJ
-
1 mol NH3
q = (9.05 × 105 g) (
17.0 g
− 91.8 kJ
) (2 mol NH ) = − 2.45 × 106 kJ
3
Heat Capacity
- Heat Capacity (C) is the quantity of heat needed to raise T of an entire system by 1 oC.
Q = C(T)
The units for C are J/oC.
- Molar Heat Capacity (Cmolar) is the quantity of heat needed to raise T of one mole by 1 oC.
Q = (Cmolar)(n)(T)
The units for (Cmolar) are J/(mol oC).
- Specific Heat is the quantity of heat needed to raise T of one gram by 1 oC.
Q = (s)(m)(T)
-
Units for s are usually (g
J
o C)
, but can be
calories
(g o C)
For H2O, s = 1.00 calories/(g oC) = 4.18 J/(g oC), where 1 cal = 4.18 J
This means 1 cal (or 4.18 J) will raise T of 1 g of H2O by 1 oC.
See Table 6.1 in the text for specific heats and heat capacities of other substances.
.
Ex 6.5 Raise T of 15 g H2O from 20.0 oC to 50.0 oC
- T = (50.0 oC) – (20.0 oC) = 30.0 oC
J
)(15 g)(30.0 oC) = 1.88 × 103 J
-
Q = (s)(m)(T) = (4.18 g
-
Q = (1.88 × 103 J)(1 kJ / 1000 J) = 1.88 kJ
oC
Calorimeter
- A calorimeter determines H (in kJ/mol) that is absorbed (positive) or released (negative)
by measuring of surroundings (which is often water).
- See Figs 6.12 and 6.13.
Ex 6.6 Combustion of 0.562 g of graphite raises calorimeter T from 25.00 to 25.89 oC
- Combustion releases heat, so H of reaction is negative.
- C = 20.7 kJ/ oC and q = − C(T) = – (20.7 kJ/ oC)(25.89 – 25.00 oC) = – 18.4 kJ
- (0.562 g)(1 mol / 12.01 g) = 0.0468 mol of carbon
- H = (– 18.4 kJ) / (0.0468 mol) = – 390 kJ per mole of C
- C(graphite) + O2(g)  CO2(g) + 390 kJ/mol (heat is released)
Hess’ Law of Heat Summation
- If a reaction can be written as the sum of other reactions as its components,
then the overall H is the sum of the H’s for individual component reactions.
-
2 [ C(graphite) + O2(g)  CO2(g) H = − 393.5 kJ ]
−1 [ 2 CO(g) + O2(g)  2 CO2(g) H = − 566.0 kJ ]
─────────────────────────────────
Sum:
2 C(graphite) + (2−1) O2(g) – 2 CO(g)  (2 −2) CO2 (g)
Becomes: 2C(graphite) + O2(g)  2CO(g)
H =
2(− 393.5 kJ) − (− 566.0 kJ) = − 221.0 kJ overall
Ex 6.7 Manipulate Reactions (with Factors) to Get the Desired Sum
-
1
Divide equation #1 by 2. Then, add in #2 without a factor. Finally, multiply #3 by (− 2).
3
1
H = (2)(−1658.8 kJ) = − 842.9 kJ
W(s) + (2)O2(g)  WO3(s)
H =
C(graphite) + O2(g)  CO2(g)
5
− 393.5 kJ
1
WO3(s) + CO2(g)  WC(s) + (2)O2(g)H = (− 2)(− 2391.8 kJ) = 1195.9 kJ
───────────────────────────────────────────────────────
5
5
Sum: W(s) + C(graphite) + (2)O2(g) + WO3(s) + CO2(g)  WC(s) + (2)O2(g) ) + WO3(s) + CO2(g)
Simplify: W(s) + C(graphite)  WC(s)
H = − 842.9 − 393.5 + 1195.9 = − 40.5 kJ
Standard Enthalpy of Formation (Hof)
- Determined from reactions which form substances from elements at standard conditions.
- Degree sign means reaction is at standard state (298 K and 1 atm).
- Each element is in its reference form, which is the most stable form at standard state.
- Different forms of an element are called allotropes. Graphite and diamond are both
allotropes of carbon. Graphite, however, is the reference form because it is more common.
Table 6.2 and Appendix C have Hof’s in kJ/mol
- For an element in its reference form, Hof is zero.
- Note phase subscripts.
Different phases have different Hof’s because phase changes require heat.
- Values for ions in table are relative to their ionic reactions with H+1(aq).
- To find Ho for a chemical reaction, use Hof of reactants and products.
Horeaction = nHof (products) – mHof (reactants)
Multiply the Hof values by stoichiometric coefficients (n and m) in the balanced reaction.
Ex 6.8 Finding Enthalpy of Vaporization (physical change)
CS2(L)  CS2(g)
(n = m = 1)
o
o
o
- H vap’n = H f (CS2(g)) – H f (CS2(L)) = (116.9 – 89.7) kJ/mol = + 27.2 kJ/mol
- This means 27.2 kJ are required to vaporize one mole of liquid CS2.
Ex 6.9 Finding Enthalpy of a Reaction (chemical change)
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) (with Pt catalyst)
o
- H = nHof (products) – mHof (reactants)
- Ho = 4Hof (NO(g)) + 6Hof (H2O(g)) – 4Hof (NH3(g)) – 5Hof (O2(g))
- Ho = 4(90.3) + 6(– 241.8) – 4(– 45.9) – 5(0) = – 906 kJ (per 4 mol NH3)