C4-Thermochemistry

ө 4∆ f H ( NH3 , g) m ө + 5∆ f H (O 2 , g) m
ө ∆r Hm
=?
4NO(g ) + 6H 2O(g )
ө 4∆ f H ( NO, g ) m ө + 6∆ f H (H 2O, g ) m
2N 2 (g) + 5O 2 (g) + 6H 2 (g)
Thermochemistry
(see chapter 6, p192)
Thermochemisty – is the part of thermodynamics that deals with the relationship between chemical reactions and heats. (sometimes we are more interested in thermal
C(graphite) → C(diamond), ∆H = 1.895kJ / mol
θ
1 θ CO(g ) + O 2 (g ) → CO 2 (g) ∆H 1 2 θ C(graphite) + O 2 (g ) → CO 2 (g )∆H 2 C(diamond) + O 2 (g) → CO 2 (g) ∆H 3 2H 2 (g) + O 2 (g) → 2H 2O(g) 1 θ H 2 (g) + O 2 (g) → H 2O(g )∆H 5 2 ∆H 4
Only initial and final states matter. It doesn’t matter how you get there!
H 2O(l ) → H 2O( g ) ∆H vap (298 K ) = 44 kJ / mol How about ∆H vap (373K ) ?
-1
∆r H
θ
m Standard enthalpy change of the reaction
Thermochemical equations are always interpreted in terms of moles.
Standard state
Gas: ideal gas at a partial pressure of 105 Pa (1 bar). (previous standard pressure 1 atm) Solid and liquid: pure solid or pure liquid under standard pressure. Standard state does not specify the temperature. (Usually we use data measured at 25oC or 298K) Standard state is NOT STP!
Conclusion
aAபைடு நூலகம்+ bB → yY + zZ
∆ r H m (T ) = ∑ n p∆f H m (product) − Σn r ∆f Η m (reactant)
θ
θ
θ
1.Calculation of enthalpies of reaction by combination of the thermochemical equation
Example: Calculate the ∆Horxn (∆rHθ) for the reaction
C(s)+ 1 O2 (g) CO(g) → 2
energy)
Several terms System is the part of the universe in which a scientist is interested. (open system, closed system and isolated system) Surroundings: the rest of universe is referred to as surroundings. How to choose a system?
Initial State
∆ r H m,1
m
Final State
∆ r H m,2
Intermediate State ∆ r H = ∆ r H + ΔH m m,1 r m,2
or
∆r H = ∑ ∆r H
m
m, i
Application of Hess Law
Law of conservation of energy— the first law of thermodynamics
The energy can neither be created nor destroyed.
∆U = Q + W
Enthalpy： the thermal energy gained or lost when a change takes place under constant pressure is called the enthalpy change, △H.
State functions
Properties that depend only on the initial and final states of the system and not on how the system gets from one state to another. H,P,T,V are state functions (extensive and intensive properties?) Work and heat are not state functions. △H = Qp △H = Q ? Why do we use △H rather than Qp?
Q
Initial State
Q1
Final State
Q2
Intermediate State
Q = Q1 + Q2
1. Q means Qp most changes take place under the constant (atmospheric) pressure 2. Qv ≈ Qp
∆r H
ΔH = H final − H initial
Endothermic △H ＞ 0 Exothermic △H ＜ 0
Thermochemical equation
2H 2 (g ) + O 2 (g ) → 2H 2O(g )
ө ∆r H m
(298.15K ) = −483.64kJ ⋅ mol
θ θ
Standard enthalpy of formation
The standard enthalpy of formation of a substance,
∆H fθ (phase, T )
is the enthalpy change that results when one mole of the substance is formed from its elements with all substances in their standard states. The enthalpies of formation of the most stable elements in their standard states are zero.
ө ∆r Hm
=
ө ө 4∆ f H m ( NO, g ) + 6∆ f H m (H 2O, g ) ө ө − 4∆f H( NH3 , g) + 5∆f H(O2 , g) m m
[
]
= [4 × 90.25 + 6 × (− 241.82) − 4 × (− 46.11) − 0]kJ ⋅ mol−1 = −905.48kJ ⋅ mol−1
from the following information:
1.C(s) + O 2 (g) → CO 2 (g )
∆ r H m (1) = −393.51kJ ⋅ mol
∆ r H m (2) = −282.98kJ ⋅ mol
ө
θ
−1
2.CO(g) + 1 O 2 (g) → CO 2 (g) 2
θ
−1
So1ution 1
C(s) + O 2 (g )
ө ∆ r H m (1)
∆ r H m (3)
ө
1 2
O 2 (g ) + CO(g )
ө ∆ r H m ( 2)
Path 1
Path 2
θ
CO 2 (g )
θ
∆ r H m (1) = ∆ r H m ( 2) + ∆ r H m (3)
Energy is the ability to do work. Work is done when a force applied to some object moves the object. Kinetic energy is the energy of motion; potential energy is stored energy. (How to understand on the molecular level?) Heat is the thermal energy transferred that results from a difference in temperature. (transfer of kinetic energy)
θ θ
Hess Law
The thermal energy given off or absorbed in a given change is the same whether the change takes place in a single step or in several steps.