Δ h every chemical reaction and change of physical state releases or absorbs heat
DESCRIPTION
Enthalpy. Δ H Every chemical reaction and change of physical state releases or absorbs heat. Goal – to determine whether heat is absorbed or released during a chemical reaction, therefore determine if exothermic or endothermic. Thermochemistry. - PowerPoint PPT PresentationTRANSCRIPT
ΔHEvery chemical reaction and change of physical
state releases or absorbs heat.Goal – to determine whether heat is absorbed or released during a chemical reaction, therefore
determine if exothermic or endothermic.
Thermochemistry
• Study of heat changes that accompany chemical reactions and phase changes
• Thermochemical equations are used to represent these reactions
• 4Fe(s) + 3 O2(g) → 2Fe2O3(s) + 1625kJ• 27kJ + NH4NO3(s) → NH4
+(aq) + NO3-(aq)
• First equation – exothermic- heat pack Second equation – endothermic – cold pack
Thermochemistry Terms
• System – specific part of the universe that contains reaction or process you want to study
• Surroundings – everything in universe other than system
• Universe = system + surroundings
Enthalpy
• Impossible to know the total heat content of a system because it depends upon many factors
• Chemists are interested in changes in energy during reactions
• For many reactions, energy lost or gained can be measured by calorimeter at constant pressure
• Enthalpy(H) – heat content of a system at constant pressure
• We measure change in H (heat content)
H0
H0
• Exothermic reaction a downhill change
• Endothermic reaction an uphill energy change
ΔH = H2 – H1
Products – reactants
Exothermic Reactions
reactants
productsEnt
halp
y (k
J)
• Is ΔH positive or negative?
• Why?
• Products - Reactants!
Endothermic ReactionsE
ntha
lpy
(kJ)
reactants
products
• ΔH is a positive #
Why is there a change in Enthalpy?
due to• energy required to break the
bonds in the reactants and • energy produced by forming the
chemical bonds in the products.
• A balanced equation can represent the energy absorbed or released
• Energy change for the reaction is called the • Enthalpy of Reaction
• and is represented by ΔHr
C(cr) + O2 → CO2 + energy (393.5kJ)∆Hr = -393.5kJ
Standard States
H = change in enthalpy• “ ° ” = Standard state enthalpy at
298.15K (25° C degrees) and 101.325 kilopascals (pressure)
• (No longer STP!)
• “f” = formation
ΔHºf
Elements in the Natural State
ELEMENTS ONLY
ΔHºf = 0
ALL OTHER COMPOUNDS
• Enthalpy of formation• found in standard tables• Table C-13 pg921
ΔHºf
Enthalpy of Formation
• Δ Hf represents the production of one mole from ...
• its free elements in their standard states. (units = kJ/mole)
Calculation of Enthalpy of Reaction
• ΔHr = ΣΔHºf (products) - ΣΔHºf
(reactants) • Σ (Sigma) is the symbol used to indicate
summation.• It means to ADD all the values of ΔH for all
the products and subtract ...• the sum of all the Δ H of the reactants
antsreactHproductsHH ffr tan
EXAMPLE 1
• Calculate the enthalpy change in the following chemical reaction
carbon monoxide + oxygen → carbon dioxide
• 1) Write a balanced equation
• 2CO (g) + O2 → 2CO2
2CO (g) + O2 → 2CO2
• 2) Calculate Δ Hf products
• 2 moles CO2 x (-393.5 kJ) = -787.0 kJ
• 3) Calculate Δ Hf reactants
• 2 moles CO + 1 mole O2 = 2(-110.5 kJ) + 1(0 kJ)
• =-221 kJ
•ΔHr =ΔHf (products) - ΔHf (reactants)
• Δ Hr = (-787.0 kJ) - (-221. kJ) = -566.0 kJ
• This means the 566.0 kJ are released in the reaction
tsreacproducts HHH tan tsreacproductsr HHH tan
Practice
• Compute ΔHºr for the following reactions:
2NO(g) + O2(g) → 2NO2(g)
-114.14kJ
__FeO(cr)+ O2(g) → __ Fe2O3(cr)
• -560.4kJ
4 2
• Solving for change in enthalpy we getΔHr = ΣΔHºf (products) - ΣΔHºf (reactants)
• Knowing the enthalpy of formation of each reactant and product, we can calculate the amount of energy produced or absorbed and predict whether a reaction will be exothermic or endothermic.
Exothermic Reactions
reactants
products
Ent
halp
y (k
J)
115 kJ
75 kJ
• Δ Hr= 75 kJ -115 kJ
• = - 40 kJ
• Negative Δ Hr for EXOTHERMIC
• Δ H = 247 kJ - 122 kJ• = 125 kJ
• Positive Δ Hr for ENDOTHERMIC
Endothermic ReactionsE
ntha
lpy
(kJ)
reactants
products 247 kJ
122 kJ
Spontaneous or Nonspontaneous That is the Question ?
• Suppose Δ H is negative Δ Hf products < Δ Hf reactants
• It’s EXOTHERMIC
• When Δ Hr is negative reaction tends to be
• SPONTANEOUS • Spontaneous means that it will occur without any
outside influence
• How about Δ H positive?
Hess’s Law
• the enthalpy change for a reaction is the sum of the enthalpy changes for a series of reactions that add up to the overall reaction.
Hess’s Law
N2 + 2O2 ↔ 2NO2 41kJ
N2O4 ↔ 2NO2 35kJ
N2 + 2O2 ↔ N2O4
Reverse the second reaction to get:
Hess’s Law
N2 + 2O2 ↔ 2NO2 ΔH= 41kJ
2NO2 ↔ N2O4 ΔH= 35kJ
N2 + 2O2 ↔ N2O4
Hess’s Law
N2 + 2O2 ↔ 2NO2 ΔH= 41kJ
2NO2 ↔ N2O4 ΔH= 35kJ
N2 + 2O2 ↔ N2O4 Reversing the second reaction reverses the sign of the enthalpy of the reaction
Hess’s Law
N2 + 2O2 ↔ 2NO2 ΔH= 41kJ
2NO2 ↔ N2O4 ΔH=-35kJ
N2 + 2O2 ↔ N2O4 Now add the two
Hess’s Law
N2 + 2O2 ↔ 2NO2 ΔH= 41kJ
2NO2 ↔ N2O4 ΔH=-35kJ
N2 + 2O2 ↔ N2O4 6kJ
End
Entropy
• (S) is the measure of the degree of disorder in a system.
• A spontaneous process is one that occurs in a system left to itself. No external action is needed to make it happen.
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• Increase in disorder;
• Decrease in disorder;
∆S > 0
∆S < 0
GIBBS FREE ENERGY
• Gibbs free energy indicates whether a reaction will occur or not.
G H TS
G H T S
T temp K
( .( ))
• Exergonic reactions (spontaneous)
• Endergonic reactions (nonspontaneous)
G0
G0
• Endothermic reactions occur spontaneously when T Δ S is large.
• The thermodynamic definition of a system in equilibrium is; when Δ H and Δ S have the same sign, and there is some temperature at which Δ H and T Δ S are numerically equal.
• A spontaneous reaction proceeds toward equilibrium.
• Chemical potential energy, G, is least at equilibrium.
• Enthalpy, entropy, and free energy depend on temperature. ( We will only work at 298.15 K and 100.00 kPa-standard states)
GIBBS FREE ENERGY CALCULATIONS
• Δ H values are relative; free elements are considered to have
• change in enthalpy is found by;
H f 0
H H Hr f products f reac ts ( ) ( tan )
• entropy changes and Gibbs free energy is computes as follows;
S H H
G G G
r f products f reac ts
r f products f reac ts
( ) ( tan )
( ) ( tan )
• 1. Organize the data you will use from the appropriate table.
• 2. Multiply each Δ Gf0 by the number of
moles from the balanced equation. (Always make sure the chemical equation is balanced). Substitute these values into the equation used to determine Δ Gf
0 .
• The Gibbs free energy decreases in a spontaneous reaction because the system is changing to a more stable state.
• To find
1. Multiply each by the number of moles from the balanced equation.
2. Substitute these values into the equation.
S
S