Download Sign of enthalpy changes Exothermic vs endothermic Acid

Sign of enthalpy changes
Reaction/transition thermochemistry
• In
I a reaction:
ti
Reactants (R)  Products (P)
• H heat of reaction
 H = H(products) – H(reactants)
 H < 0 means “” is exothermic (produces heat)
 H > 0 means ““”” iis endothermic
d th
i (absorbs
( b b h
heat)
t)
H
–
+
–TS
TS
heat is evolved
exothermic
heat is absorbed
endothermic
–
• Heat can be measured in a
((micro)calorimeter,
)
, and used to:
 Characterize the reaction
(Endothermic/exothermic? Molar H? Which forces
y involved?))
are likely
 Find the extent of reaction (how much product
formed?)
+
• ITC = isothermal titration calorimetry
Surroundings
warm up
•
•
•
•
•
Freezing
Condensation
Acid/base neutralization
Burning
Drug/protein binding (typically)
Surroundings
cool down
•
•
•
Melting
Evaporation
Crystal dissolution (typically)
http://microcal.com/technology/itc.asp
Exothermic vs endothermic
• Problem: Calorimetric data on
binding of several anti-HIV drugs to
their common target,
target HIV protease,
protease
at T=300K is presented in the table.
For how many drugs in the list, the
binding reaction is endothermic?
Drug
Nelfinavir
Indinavir
1
Saquinavir
2
Tipranavir
3
L i
Lopinavir
i
4
Atazanavir
5
Ritonavir
6
Amprenavir
• Answer: 3 drugs. Endothermic
means H > 0, which is the case for Darunavir
I di
Indinavir,
i S
Saquinavir,
i
i and
dN
Nelfinavir
lfi
i
H
binding
(kcal/mol)
3.1
18
1.8
1.2
-0.7
-3.8
38
-4.2
-4.3
-6.9
-12.7
Acid-base neutralization heat
•
•
•
Reaction
Equation
Hm(react) = molar H(react) = heat
produced/absorbed when…
Dissolution
Dsolid + H2O  Daq
1 mole of drug/substance
g
dissolves in
infinite amount of water
Neutralization
OH- + H+  H2O
1 mole of OH- interacts with 1 mole of H+
to form 1 mole of H2O
1:1 protein/ligand
binding
P + L  PL
1 mole of L binds to 1 mole of P to form
1 mole of PL complex
• H(react) = Hm(react)n
• n is the extent of reaction (# of moles):
Dissolution: n = amount of substance dissolved
Neutralization: n = amount of limiting reactant
Protein/ligand binding: n = amount of complex (not
necessarily equal to limiting reactant!)
• n = H/ Hm
Protein-ligand binding equilibrium
Problem: Molar enthalpy of neutralization reaction between HCl and
NaOH at 25C is -55.9 kJ/mol. When 25 mmol HCl was added to 10 ml
of NaOH solution in a calorimeter, the evolved heat was measured to be
1 kJ
kJ. What was the concentration of NaOH in the solution?





•
(Quantitative) H of reaction/transition
25 mM
25 M
0.55 M
1.79 M
17.9 mM
binding
Solution: If 25 mmol of HCl were completely neutralized by NaOH, H
would be equal to 0.025 mol x -55.9 kJ/mol ~ -1.4 kJ. However, we only
observed heat of 1 kJ.
 NaOH was a limiting reactant
 Only 1/55.9
1/55 9 = 0.0179
0 0179 mol of reaction product (water) was formed
 Therefore 17.9 mmol NaOH was present in the solution in the beginning
 This corresponds to NaOH concentration of 1.79 M.
Answer: 1.79M
1 79M
Note: Acid/base neutralization involves formation of H2O from OH- and H+;
covalent bonds are formed, exothermic.
dissociation
1 mole of protein
1 mole of ligand
unbound
1 mole of protein/ligand complex
bound
Enthalpy of drug binding - I
Protein-ligand binding equilibrium
•





binding
dissociation
Problem: 0.3 mL of 1mM solution of a drug is added to a tube with a
protein solution. Some of the drug reacts with the protein to form 1:1
protein-drug complex. The reaction produces a heat of –2.4 x 10–3 cal.
Estimate the molar enthalpy of binding
binding.
•
–8 kcal/mol
–12 kcal/mol
24 kcal/mol
3.6 x 10–3 cal/mol
impossible to tell because the extent of reaction is not known
Solution: The total molar amount of drug is 0.3 mol – but how much of it
reacted? And how much product (complex) was formed, in moles?
 If n = 0.3 mol, Hm = H/n = –2.4 x 10–3 cal / (0.3 x 10–6 mol) = –8 kcal/mol
 If n = 0.2 mol, Hm = H/n = –2.4 x 10–3 cal / (0.2 x 10–6 mol) = –12 kcal/mol
 If n = 0.1 mol, Hm = H/n = –2.4 x 10–3 cal / (0.1 x 10–6 mol) = –24 kcal/mol
•
1 mole of protein
1 mole of ligand
unbound
mole of protein
protein/ligand complex
0.4
0 4 mole of 1unbound
bound
0.4 mole of unbound ligand
0.6 mole of bound complex
EQUILIBRIUM
Enthalpy of drug binding - II
•
Problem: 0.3 mL of 1mM solution of a drug is added to a tube with a
protein solution, and the drug reacts entirely with the protein to form a
1:1 protein-drug complex (no unbound drug is left). The reaction
produces a heat of –2
2.4
4 x 10–3 cal.
cal Estimate the molar enthalpy of
binding.





•
–8 kcal/mol
–12 kcal/mol
12 kcal/mol
3.6 x 10–3 cal/mol
impossible to tell because the extent of reaction is not known
Solution:
 The total molar amount of drug is 0.3 mol
 All of it reacted, so 0.3 mol of complex was formed (n = 0.3 mol)
 Hm = H/
H/n = –2.4
2 4 x 10–33 call / (0.3
(0 3 x 10–66 mol)
l) = –8
8 kkcal/mol
l/ l
•
•
Answer: –8 kcal/mol
Note: It is important that all 0.3 mol of drug was bound
Dissolution heat and T changes
•
Problem: After dissolving some amount of Lidocaine HCl in 1 ml pure
water at room temperature, the ampule cooled down by 1.04 K. How
much drug was dissolved? Molar enthalpy of Lidocaine HCl dissolution
is 43
43.5
5 kJ/mol
kJ/mol. Consider the density and the heat capacity of the solution
approximately equal to that of pure water.





•
1 mole
100 mmoles
10 mmoles
1 mmole
100 moles
Solution: Two processes: (1) dissolution of Lidocaine and (2) heating
water
 Specific heat capacity of water is 4.184 J/(g K)
 To decrease temperature of 1 ml of water by 1
1.04K,
04K heat of 4
4.184
184 J/(g K) x
1.04K x 1g ~ 4.35 J must be absorbed
 This corresponds to dissolution of 4.35 J / 43.5 kJ/mol ~ 0.1 mmole of the
drug
•
Answer: 100 moles
Answer: impossible to tell because the extent of reaction is not known
 However, we know that the reaction is exothermic
 And that Hm  –8 kcal/mol
H, Cp, and heating matter
• Cp is sample heat capacity at constant P
J/K or cal/K
• Cp,m is molar heat capacity
J/(K mol) or cal/(K mol)
Cp = nCp,m
p m where n is the number of moles
• Molar Hm = Cp,m T
• Total H = Cp T = nCp,m
p m T
• Specific heat capacity (Cp per unit mass) may be
useful. WHAT IS IT FOR H2O?
• Calculation assumes that Cp is not changing within
the temp. range of the experiment.
• T = H/(nC
H/(n Cp,m)
• Cp,m= H/(nT)
Heat capacity of gases
• Cp is (molar/specific/sample) heat capacity at
constant P
• PV = nRT:
isochoric/isovolumetric heating: V = const, P
increases (e.g.
(e g heating in a closed container)
isobaric heating: P = const, V increases
• Useful to have Cv (molar/specific/sample) at
constant volume
• For liquid and solid samples, Cp ~ Cv
• For gaseous samples, Cp ~ Cv + nR
Molar: Cp,m ~ Cv,m + R
DF vs Cv,m
Experimentally measured Cv,m at 298K
Gas
CV,m,
J/(mol·K)
CV,m/R # DF
He
●
12.5
1.5
3
Ne
●
12.5
1.5
3
Ar
●
12.5
1.5
3
Kr
●
12.5
1.5
3
Xe
●
12.5
1.5
3
H2
●–●
20.18
2.43
5
CO
●–●
20.2
2.43
5
N2
●–●
20.8
2.50
5
O2
●–●
21.03
2.53
5
Cl2
●–●
24.1
3.06
~6
~7
Br2 (v)
●–●
28.2
3.39
H2O (v)*
●●●
28.49
3.43
~7
CO2
●-●-●
28.5
3.43
~7
CH4
●●●
●
27.1
3.26 ~ 6-7
DF and heat capacities
• # of DF determines heat
capacity of a substance at
constant volume:
– Each rotational/translational
DF contribute R/2 to Cv,m
– Each (excited) vibrational
mode contributes up to R to
Cv,m
• Monoatomic gases have 3
DF: Cv,m = 3/2R
• Diatomic gases have 5 DF
b l
below
Tvib
• P
Problem:
bl
E ti t the
Estimate
th specific
ifi heat
h t capacity
it att constant
t t
pressure and room temperature of nitrogen gas, N2, in J/(g
K). Assume that vibrational modes of N2 molecules are not
excited
it d att room ttemperature.
t
MW(N2) = 28 g/mol.
/ l





• Solution:
 N2 is a diatomic gas, so each molecule has 5 degrees of freedom (bond
vibrations are not excited).
 Molar heat capacity at constant volume is Cv,m
v m = 5/2R ~ 20.8 J /(mol K).
 Molar heat capacity at constant pressure is
Cp,m = Cv,m + R = 20.8 J /(mol K) + 8.314 J /(mol K) = 29.1 J /(mol K)
 Specific heat capacity at constant pressure is
Cp,m / MW = 29.1 / 28 = 1.04 J/(g K).
– for H2, CO, N2, Tvib >> 298K
• When T  Tvib, vibrational
DF’s appear
* Value for H2O at 373K
– for Cl2, Br2, I2, Tvib  298K
0.44 J /(g K)
~1 J /(g K)
20 8 J /(g K)
20.8
29.1 J /(g K)
impossible to tell
• Answer: ~1 J/(g K)
Thermodynamic cycle
Thermodynamic cycle & Hess law
• The enthalpy change for a
reaction carried out in a series
of steps is equal to the sum of
the enthalpy changes for the
individual steps.
• In a reaction:
Reactants (R)  Products (P)
• Enthalpy
py change
g for the
reaction can be found from
enthalpies of formation of R
and P:
Hf(R) + Hr = Hf(P)
Products
• Problem: Standard enthalpy of formation of ammonia at
25C is -80.8 kJ/mol for its aqueous form and -45.9 kJ/mol
for its gaseous form. What is the enthalpy of dissolution of 2
moles of gaseous ammonia in water at 25C?
 +34.9 kJ
 –34.9 kJ
 +69.8 kJ
 –69.8 kJ
 +126.7 kJ
 –126.7 kJ
Hr
Reactants
Hf
(P)
Hf
(R)
• Solution: Molar enthalpy of dissolution is the difference
between standard enthalpies of formation:
 -80.8-(-45.9) = -34.9 kJ/mol.
 The enthalpy of dissolution of 2 moles is 2 x -34.9 = -69.8 kJ.
 Ammonium dissolution is exothermic!
Elements
• Answer: -69.8 kJ.
Kirchhoff’s Law
Kirchhoff’s Law
• Relates enthalpies at different temperatures
• HT2 – HT1 = Cp(T2 –T1),
•
assuming that Cp is approx. constant within T





• HT2 is HT1 plus energy needed to heat the sample.
• Find enthalpy at a different temp:
HT2 = HT1 + Cp(T2 –T1)
•
• Find heat capacity within the given temp. range:
1.11 kJ
~1 J
37 J
-10 kJ
1179 kJ
Solution 1: fHT1, fHT2, Kirchhoff's Law  molar hear capacity of CO2
at constant pressure and 20
20-50C:
50C:
 Cp,m = (-391.89 kJ/mol+393 kJ/mol)/30K ~ 37 J/(mol K)
 Heating a 3 mole sample by 10C will require:
T x n x Cp,m = 10 K x 3 mol x 37 J/(mol K) ~ 1.11
1 11 kJ of heat
Cp = (HT2 – HT1) / (T2 –T1) = H / T
• Find a new temp given enthalpy:
Problem: Molar enthalpies of formation of gaseous CO2 are -393
393 kJ/mol
at 293K and -391.89 kJ/mol at 323K. Approximately how much heat is
needed to raise the temperature of a 3 mole sample of CO2 from 30C to
p
40C at constant pressure?
•
Solution 2: Heating 3 moles by 10C is ~ the same as heating 1 mole
by 30
•
Answer: 1.11 kJ
T2 = T1 + (HT2 – HT1) / Cp
 simply
p y take the difference fHT2 - fHT1 = 1.11 kJ