5.111 Lecture Summary #17 Friday, October 17, 2014
Reading for today: Sections 8.8, 8.12, 8.13, 8.15, and 8.16 (same sections but in Chapter
7 in 4th ed): Entropy and Gibbs Free Energy, and Free-Energy Changes in Biology.
Reading for Lecture #18: Sections 10.1-10.9 (same sections but in Chapter 9 in 4th ed): Chemical
Equilibrium
Topics: Thermodynamics
I. Effect of temperature on spontaneity
II. Thermodynamics in biological systems
A. Hydrogen-bonding
B. ATP-coupled reactions
I. EFFECT OF TEMPERATURE ON SPONTANEITY
Consider the decomposition of sodium bicarbonate at 298 versus 450. K.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
∆Hº = 135.6 kJ/mol ∆Sº = kJ/ (K •mol)
∆Gr° = ∆Hr° – T(∆Sr°)
At T = 298K ∆Gº = 135.6 kJ/m ol -298 K( kJ/ (K mol)) = kJ/mol
The reaction is at room temperature.
But at baking temperatures of 350ºF or 450.K
∆Gº = 135.6 – (450.)(0.334) = ________ kJ/mol
The reaction is at baking temperature.
When ∆Hº and ∆ Sº have the sign, it is possible to control spontaneity with T.
Assuming that ∆Hº and ∆Sº are independent of T, a reasonable first-order assumption,
then ∆ Gº is a function of T.
1
, ∆Gº = ∆Hº – T∆Sº or ∆Gº = – ∆Sº(T) + ∆Hº
Calculate T* (at which ∆Gº = 0 ) for the decomposition of sodium bicarbonate.
∆Gº = ∆Hº – T∆Sº
0 = ∆Hº – T*∆Sº T* =
T* = kJ/mol = K
kJ/(Kmol)
Consider the plot of temperature dependence when both ∆Hº and ∆Sº are negative
∆Gº = ∆Hº – T∆Sº
2
Reading for today: Sections 8.8, 8.12, 8.13, 8.15, and 8.16 (same sections but in Chapter
7 in 4th ed): Entropy and Gibbs Free Energy, and Free-Energy Changes in Biology.
Reading for Lecture #18: Sections 10.1-10.9 (same sections but in Chapter 9 in 4th ed): Chemical
Equilibrium
Topics: Thermodynamics
I. Effect of temperature on spontaneity
II. Thermodynamics in biological systems
A. Hydrogen-bonding
B. ATP-coupled reactions
I. EFFECT OF TEMPERATURE ON SPONTANEITY
Consider the decomposition of sodium bicarbonate at 298 versus 450. K.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
∆Hº = 135.6 kJ/mol ∆Sº = kJ/ (K •mol)
∆Gr° = ∆Hr° – T(∆Sr°)
At T = 298K ∆Gº = 135.6 kJ/m ol -298 K( kJ/ (K mol)) = kJ/mol
The reaction is at room temperature.
But at baking temperatures of 350ºF or 450.K
∆Gº = 135.6 – (450.)(0.334) = ________ kJ/mol
The reaction is at baking temperature.
When ∆Hº and ∆ Sº have the sign, it is possible to control spontaneity with T.
Assuming that ∆Hº and ∆Sº are independent of T, a reasonable first-order assumption,
then ∆ Gº is a function of T.
1
, ∆Gº = ∆Hº – T∆Sº or ∆Gº = – ∆Sº(T) + ∆Hº
Calculate T* (at which ∆Gº = 0 ) for the decomposition of sodium bicarbonate.
∆Gº = ∆Hº – T∆Sº
0 = ∆Hº – T*∆Sº T* =
T* = kJ/mol = K
kJ/(Kmol)
Consider the plot of temperature dependence when both ∆Hº and ∆Sº are negative
∆Gº = ∆Hº – T∆Sº
2
