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Principles of Chemical Science_Thermodynamics Now What Happens When You Heat It Up - Lec17
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Lecture 17: Thermodynamics: Now What Happens When You Heat It Up? 1. Thermodynamics 1. Effect of Temperature on Spontaneity 2. Thermodynamics in Biological Systems 1. Hydrogen-bonding 2. ATP-coupled Reactions

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  • April 13, 2023
  • 5
  • 2014/2015
  • Class notes
  • •&tab;prof. catherine drennan
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  • chemical

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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/(Kmol)

Consider the plot of temperature dependence when both ∆Hº and ∆Sº are negative

∆Gº = ∆Hº – T∆Sº




2

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